Dr. Maher Atteya ______________________________________________ Chapter 10: Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals Molecular geometry: - Molecular geometry is the 3-D arrangement of atoms in a molecule. - Molecular geometry affects the chemical and physical properties of the molecule (i.e. density, boiling points, melting points, etc.) - Molecular geometry is the geometry on the central atom related to terminal atoms. - The model which handles the molecular geometry is called ―VSEPR‖ model where V stands for Valence S stands for Shell E stands for Electron P stands for Pair R stands for Repulsion - Valence shell electrons are electrons found in the outermost shell. (# of valence electrons = # of the group in the periodic table) - Valence electrons are the only electrons involved in the chemical reactions. VSEPR: - Valence electron pairs repel each other. - In a polyatomic molecule, the repulsion of electron pairs can occur between two or more bonds of the central atom and the surrounding terminal atoms. - The repulsion between electrons in different bonding pairs causes them to remain as far apart as possible. - The geometry that the molecule finally assumes leads to minimize the repulsion (the molecule assumes more stability). - The source of the repulsion between electron pairs is electrostatic repulsion force. VSEPR – Rules - Double bond, single bond and triple bond are treated the same (as one unit). - Formal charges are not shown. VSEPR can be used to any resonance structures. - In determining VSEPR model, follow the steps below:

a. Draw Lewis structure of the compound. a. Count # of bonding pairs (central atom / terminal atoms). a. Count # of lone pairs (nonbonding); (around the central atoms ONLY) a. Look at the tables 10.1 and 10.2 (pages 369 and 375, respectively) in the textbook and figure out the electron bonding pair geometry and the molecular geometry of the whole molecule. a. Note that in cases where there are no lone pairs (nonbonding pairs), the electron pairs geometry will equal the whole molecular geometry of the molecule. a. VSEPR model is used ONLY for covalent bonds but not for ionic bond. VSPER Types One can distinguish between two types: a. VSEPR model with no lone pair on the central atom. Examples are given in table 10.1 (page 369). Example: CO2 Lewis Structure: # e valence electrons = C + 2(0) = 4 + 2(6)=16 ::O = C = O:: # of bonding electron pairs = 2 (remember, single, double around the central atoms and triple bonds are treated equally and is considered as one unit each)) # of lone pair electrons = 0 VSEPR-designation = AB2E0 Where: A = central atom B = terminal atoms = 2 # of terminal atoms, in this case (two) E = # lone pair electrons (here E = 0) Because there are no lone pair electrons, the geometry of the electron pairs equals the geometry of the compound. Comparing table 10-1: CO2 is linear With angle OCO of 180 degrees Geometry of the electron bonding pairs is linear Geometry of the molecule (molecular geometry) is linear also b. VSEPR model with lone pair on the central atom: Follow the above mentioned rules and consider the lone pairs as a separate unit (a unit by itself) Example:

With help of VSEPR model, determine the molecular geometry of CIF3 First draw a Lewis structure of CIF3 # of Valence electrons = 7 + 3(7) = Cl + 3(F) = 28 e- ::F: | :Cl: __ :F:: | ::F: Lewis structure reveals a central atom with the following: - # of bonding pairs = 3 (3 single bonds) - # of non bonding pairs (lone pairs) = 2 - total # of electron pairs = 3+2 = 5 - VSEPR designation = AB3E2 where E2 = two electron pairs. Now, comparing table 10.2 (page 375) Geometry of electron pair is trigonal bipyramidal Geometry of the molecule (molecular geometry) is T-shaped Important Note: In a molecule in which the central atom has one or more lone pairs the following relationship is held: lone pair vs. lone pair repulsion > lone pair vs. bonding pair repulsion > bonding pair vs. bonding pair repulsion Why is this relationship true? Because lone pair occupies a larger space and is not restricted or attracted by the nuclei, which leads to more repulsion among themselves. Extra Examples: 1. Determine the geometry of H20 using the VSEPR model. Solution: Draw Lewis Structure: # of valence electrons = 2(H) + 1(O)= 2(1)+6 = 8e- H—:O:—H # of bonding pairs = 2 (2 single bonds) (around central atom)

# of lone pairs = 2 (around central atom) VSEPR designation = AB2E2 Comparing table 10.2 page 375, The geometry of the electron pairs = tetrahedron The molecular geometry = bent 2. Determine the geometry of NO3- using VSEPR model. Solution: Draw Lewis Structure: # of valence electrons = N + 3(O)+ 1(negative charge = 5 + 3(6) + 1 = 24e- # of bonding pairs around central atom = 3 (2 single bond + 1 double bond) # of lone pairs around central atom = 0 VSEPR designation = AB3 Comparing table 10.1 page 369 The geometry of the electron pairs = trigonal planar The molecular geometry = trigonal planar (no lone pairs are present around central atom) Dipole Moments: Dipole moment is the quantitative measure of the polarity of a bond. More electronegative element tends to pull the electron density towards it. H ------> F The arrow shows the shift of electron density towards fluorine (more electronegative) which leads to charge separation. Hydrogen will have a partial positive charge (positive end)and fluorine will have a partial negative charge (negative end). Partial charge confirmed by electrical field (turning it on/off) Generally speaking: Compounds that possess dipole moments are said to be polar and compounds that do not possess dipole moments are said to be non polar. Examples: Diatomic molecules: CO, NO, HCl, HF, etc…. all have dipole moments and hence are polar. Exceptions for the general rule: 1) A Diatomic molecule with the same element has no dipole moment and is

referred to as a nonpolar compound. Examples include: F2, H2, O2, Cl-2, Br2, etc. 2) In molecules with more than two atoms (three atoms and more), the polarity of bonds and the molecular geometry determine whether or not the molecule has a dipole moment or not. Example 1: CO2 ::O = C = O:: (linear, according to the VSEPR model) The CO2 compound does have two electronegative atoms (oxygen) and is expected to have two dipole moments. However, these two dipole moments are identical (same magnitude) and are in opposite directions so that they cancel themselves (CO2 is symmetric). Hence, the compound is not polar. Therefore, the geometry of the compound is very important to determine the polarity. Generally speaking, linear compounds with the same electronegative atom are not polar. Example 2: Compare SO2 with CO2 polarities! While CO2 is linear nonpolar, SO2 is not linear (2 bonding pairs + 1 lone pair) :O: = .S. = :O: According to table 10.2, SO2 is bent and hence is polar. Note: Oxygen is more electronegative than sulfur, therefore the lone pair electrons are more electronegative than oxygen. The dipole moment is small in magnitude because oxygen atoms are working in the opposite direction of the lone pair. Example 3: Compare the polarities of NH3 versus NF3 .. | N__ H |\H H

(dipole moment = 1.46 Debye) .. | N __ F::: |\F::: F::: (dipole moment = 0.24 Debye) NH3 is more polar than NF3, Why? In the case of NH3: Nitrogen is more electronegative than hydrogen and the lone pair electrons are more electronegative than nitrogen. All dipole moments are working in the same direction (helping each other and not against each other). Therefore, the resultant dipole moment is higher (dipole moment = 1.46 D). In the case of NF3: Nitrogen is less electronegative than fluorine and the lone pair electrons are more electronegative than nitrogen atoms. The dipole moments are working against each other; hence the resultant dipole moment will decrease (dipole moment = 0.24 D). Example 4: Compare the polarities of the following compounds. Cl Cl | \ C = C | / H H versus: H Cl | \ C = C | / Cl H Determining the dipole moments and their directions (noting that chlorine is more electronegative, followed by carbon and at last by hydrogen) Cl Cl | \ C = C | / H H

All dipole moments are All dipole moments are working together. (Dipole moment = 1.89 D). This compound is polar. All dipole moments are All dipole moments are working against each other H Cl | \ C = C | / Cl H ----------------------------------------------------- Valence Bond Theory: A theory that explains the existence of the single, double, and triple bonds. On the other hand, VSEPR model does not distinguish between single, double, and triple bonds. It treats all bonds equally, which reflects the major disadvantage of this model. - Valence bond theory is based on the hybridization of atomic orbitals. - Hybridization: the mixing of atomic orbitals in an atom (generally the central atom) to form a set of hybrid orbitals responsible for forming single, double, & triple bonds. - Hybridization is made of a combination of s, p, and d orbitals. - The only exception for this is H2 (H - H) In H2 there is an overlap of the 1s orbital from each hydrogen atom to form the single bond. - Hybridization occurs with the outermost orbitals where the valence electrons can be found. There are five different types of hybridization: sp, sp2, sp3, sp3d, and sp3d2 1) sp3 hybridization Example: CH4 H: 1s1 orbital C: 2s2 2p2 orbitals Carbon is the central atom. sp3 hybridization is around the central atom. Mechanism of hybridization: [Consider only the central atom which is the carbon atom in this case]:

(ground state): 2s2 2p2 -------> (excitation): 2s1 2p3 (excited state) -------> (hybridization) --------> sp3 Hybrid orbital sp3 is made of 1s orbital mixed with 2px1 2py1 2pz1 - sp3 hybrid orbitals will make 4 single bonds (1s + 3p = 4 single bonds) - sp3 hybrid orbitals geometry is tetrahedral with an angle of 109.5 degrees - CH4 has a carbon atom with sp3 hybrid orbital and 4 hydrogen atoms with 1s orbitals - Therefore, the single bond is made of sp3 hybrid orbitals overlapping with each 1s orbital of the hydrogen atoms. - The single bond is called sigma and is made of 1s (from hydrogen)and sp3 overlapping (at the central atom). H | C - H |\H H Each single (sigma) bond between the hydrogen and the carbon is 1s—sp3. There are 4 sigma single bonds. 2) sp2 hybridization Example: H2C=CH2 H: 1s1 orbital C: 2s2 2p2 orbitals Each carbon atom is considered as the central atom. Mechanism of Hybridization: {Consider only the central atoms which are the two carbon atom in this case]: (ground state): 2s2 2p2 -------> (excitation): 2s1 2p3 (excited state) -------> (hybridization) --------> sp2 + 1p Hybrid orbital sp2 is made of 1s orbital mixed with 2px1 2py1. Only 2pz1 is left unhyhridized. 2pz1 makes the double "pi" bond. The five single bonds are made of the following: - 4 single bonds made between 1s of hydrogen and sp2 of carbon atoms [4

sigma single bonds: (1s – sp2)] - 1 sigma single bond between the two sp2 - sp2 carbon atoms. - 1 double bond called ―pi‖ bond is made of the nonhybrid p orbitals of the two carbon atoms The geometry of sp2 hybrid orbital of each central atom (carbon atom) is planar with the angle of 120 degrees 3)sp – hybridization Example: H |_ C=CH H: 1s1 orbital C: 2s2 2p2 orbitals Each carbon atom is considered as the central atom. Mechanism of Hybridization: {Consider only the central atoms which are the two carbon atom in this case]: (ground state): 2s2 2p2 -------> (excitation): 2s1 2p3 (excited state) -------> (hybridization) --------> sp2 + 2p Hybrid orbital sp is made of 1s orbital mixed with 2px1. Only 2px1 and 2pz1 are left unhyhridized. 2px1 and 2py1 make the two double "pi" bonds. The five single bonds are made of the following: - 2 single bonds made between 1s of hydrogen and sp of carbon atoms - 1 sigma single bond between the two sp - sp carbon atoms. - 2 double bonds called ―pi‖ bonds which are made of the nonhybrid p orbital of the two carbon atoms The geometry of sp hybrid orbital of each central atom (carbon atom) is linear with the angle of 180 degrees. ------------------------------------------------------- Hybridization of bonding/nonbonding orbitals:

==================================== Procedure: [Note that this procedure determines the overall geometry around the central atom and not the whole molecule] 1. Draw Lewis structure of the molecule 2. Predict the overall arrangement of the electrons pairs (both bonding and lone pair) 3. Treat all lone pairs, single double and triple bonds the same [i.e. one unit each] 4. Use the following examples to determine the geometry around the central atom 5. To determine the geometry of the whole molecule or compound use VSEPR method. Remember that Valence Bond Theory does not give the overall geometry of whole structure of the molecule or the compound. Examples: BeCl2 Sum of bonding/nonbonding: 2 # of Hybrid orbitals: 2 Type of Hybrid. on Central atom: sp angle: 180 degrees Geometry of Central atom: Linear Examples: BF3 Sum of bonding/nonbonding: 3 # of Hybrid orbitals: 3 Type of Hybrid. on Central atom: sp2 angle: 120 degrees Geometry of Central atom: Planar Examples: CH4 Sum of bonding/nonbonding: 4 # of Hybrid orbitals: 4 Type of Hybrid. on Central atom: sp3 angle: 109.5 degrees Geometry of Central atom: Tetrahedral Examples: PCl5 Sum of bonding/nonbonding: 5 # of Hybrid orbitals: 5 Type of Hybrid. on Central atom: sp3d

angle: 90 degrees/120 degrees Geometry of Central atom: Trigonal bipyramidal Examples: SiF5 Sum of bonding/nonbonding: 6 # of Hybrid orbitals: 6 Type of Hybrid. on Central atom: sp3d2 angle: 90 degrees/90 degrees Geometry of Central atom: Octahedral Example: Determine the hybrid orbital on the central atom and its geometry and determine the geometry of the whole compound: H | C=O:: | H a. The hybrid orbital on the central atom (carbon atom) The carbon atom is the central atom does not have any lone pairs. The sum of the bonding pair sets (or units) are 3. Therefore three orbitals are needed and hence the hybrid orbital is sp2 (compare the above examples). b. The geometry around the central atom is planar because the hybrid orbital is sp2. c. The geometry of the compound as whole is determined by VSEPR as follows: - number of the bonding electron pairs = 3 - number of the non bonding electron pairs = 0 - VSEPR expression is = AB3 (No lone pairs) Comparing table 10.1 the overall geometry (the molecular geometry) is planar as well. Please NOTE the following: 1. This is NOT always true if the central atom has lone pairs VSEPR ONLY will determine the outcome of the geometry of the whole compound or molecule. 2. Hydrogen has no hybrid orbital at all. It shares its orbital by making an overlapping orbital with the central atom. 3. If one decides to determine the hybrid orbital on the terminal oxygen, then one has to count all electron pairs bonding or non bonding and compared them with the number of electrons. Oxygen has 1 bonding pair (the double bond) and 2 lone pairs. The sum of

the electron pairs (bonding/non bonding) is 3. Therefore 3 orbitals are needed. Hence the hybrid orbital on oxygen is sp2 as well. ------------------------------------------------------- Selected problems from the textbook: Chapter 10 Please work on these problems. If you have any question, contact your instructor! 10.7;10.9;10.12;10.19;10.20;10.21;10.36;10.37;10.38;10.40;10.41;10.67;10.78;10.87 ------------------------------------------------------------------ Selected problems and answers: Chapter 10 1. Using VSEPR model, predict the geometry of: a. SiBr4 Answer: Lewis structure: Br::: | Si-Br::: |\Br::: Br::: Silicon is the central atom has 4 bonding electron pairs but no lone pairs. Comparing this with the result of table 10.1, the geometry on the central atom is equal the geometry of the whole molecule which is Tetrahedral. b. NO3^-1 (nitrate ion) Answer: Lewis structure: O::: | N = O:: | O::: 2 single bonds + 1 double bond (3 sets). [No lone pairs are around the central atom]. The VSEPR shorthand expression is AB3 The geometry of the whole molecule is Trigonal planar.

c. CS2 Answer: Lewis structure: ::S = C = S:: 2 double bonds. [No lone pairs are around the central atom]. The VSEPR shorthand expression is AB2 The geometry of the whole molecule is Linear. d. C2H4 Answer: Lewis structure: H2 - C = C - H2 1 double bond and 2 single bonds. [No lone pairs are around the central atom]. The VSEPR shorthand expression is AB3 The geometry of the whole molecule is Trigonal planar. Note that there are two central atoms (2 carbon atoms)and they are equivalent. The geometry on one of them will determine the overall geometry of the molecule. ------------------------------------------------------- 2. Using the VB Theory, predict the hybridization state on the central atom underlined. Predict the geometry of the molecule. Predict all types of bonds involved. a. C2H4 (carbon is underlined as central atoms) Answer: Lewis structure: H2 - C = C - H2 1 double bond and 2 single bonds. [No lone pairs are around the central atom]. Each atom will make three bonds (2 single bonds + 1 double bond). Therefore each carbon atom will make three orbitals. The hybrid orbital is sp2 (3 orbitals). In order to predict the geometry of the molecule, one should use VSEPR model. Each carbon atom will have 3 sets around it (2 single bonds + 1 double bond). The VSEPR shorthand expression is AB3 The geometry of the whole molecule is Trigonal planar. Note that there are two central atoms (2 carbon atoms)and they are

equivalent. The geometry on one of them will determine the overall geometry of the molecule. The bonding between carbon - carbon atoms is as follows: - single (sigma) bond between carbon - carbon atoms: This single bond is formed by sp2 (carbon) - sp2 (carbon). - double (pi) bond between carbon - carbon atoms: This double bond is formed by p(z)(carbon) - p(z) (carbon). Note that the p(z) orbital is not a hybrid. It is unused p orbital of the carbon atoms. The bonding between carbon - hydrogen atoms is as follows: - single (sigma) bond between carbon - hydrogen atoms: This single bond is formed by sp2 (carbon) - s (hydrogen). There are single (sigma) bonds sp2 (carbon) - s (hydrogen). Note that the overall geometry is determined based on the VSEPR model: each central carbon atom has three sets (3 bonding): 2 hydrogen - carbon and 1 carbon - carbon. (treat single, double and triple bonds as the same). 3 sets (no lone pair) ----> geometry is trigonal planar. ------------------------------------------------------- b. BeF2 (beryllium is underlined as a central atom) Lewis structure is as follows: :::F - Be - F::: There are two single bonds of beryllium connecting to fluorine atoms. Therefore, there are 2 sets (2 orbitals) and hence the overall geometry is linear. There are no lone pairs. The two single (sigma) bonds are made of the following: sp (beryllium) - sp3 (fluorine) [Note that beryllium has two bonds and hence needs two orbitals (sp)] [Note that each fluorine atom has 4 sets (3 sets of lone - unpaired - pairs of electrons and 1 set of single bond connecting to the fluorine atom (sp3)] Again the geometry is determined by the central atom ONLY! 2 sets, no lone pairs ------> linear ------------------------------------------------------- c. SiH4 ( Si is underlined as a central atom) The Lewis structure is as follows: H | Si - H |\ H H Silicon (Si) forms 4 single (sigma) bonds, no lone pair. Therefore, the

geometry is tetrahedral. The four single bonds are made of the following: sp3 (Si) - s (hydrogen) [Note that the hybrid orbital on the Silicon central atom is sp3 (4 orbitals)] ------------------------------------------------------- 3. Using VB Theory, predict the hybridization state on the central atoms and terminal atoms. Predict the geometry of the molecule! Predict all types of bonds involved! Cl:::: | C = O:: | Cl::: Carbon atom is the central atom forming three bonds: 2 single bonds and 1 double bond. The overall geometry of the molecule is trigonal planar (3 sets, no lone pair) The hybrid orbital of carbon central atom is sp2 (3 sets means 3 orbitals) The bonding of carbon central atom is as follows: 2 single (sigma) bonds: sp2 (carbon) - sp3 (chlorine) [Note that the hybrid orbital on chlorine atoms is determined by the number of the sets formed. Chlorine has 3 sets of lone pair and 1 set single bond connected to carbon atom. Therefore, the 4 sets of chlorine are represented as sp3 hybrid orbital)] 1 double (pi) bond between carbon and oxygen. This double bond is made of the following: sp2 (carbon) - sp2 (oxygen) p(z)(carbon) - p(z)(oxygen) [Note that the hybrid orbital on oxygen atom is determined by the number of the sets formed. Oxygen has 2 sets of lone pair and 1 set double bond connected to carbon atom. Therefore, the 3 sets of oxygen are represented as sp2 hybrid orbital)] [The p(z) is unused p orbital found in carbon and oxygen]. 4. Using VB Theory, predict the hybridization state on the central atom and terminal atoms. Predict the geometry of the molecule! Predict all types of bonds involved! :N _= C - S:::- Carbon atom is the central atom. Carbon is connected to nitrogen by a triple bond and is connected to sulfur by a single bond.

Therefore there are 2 sets (no lone pair) and hence the overall geometry of the whole molecule is linear. [Note that the triple bond and the single bond are treated equally; each is worth 1 set only]. The hybrid orbital on carbon atom is sp (linear). The single bond between carbon and sulfur atom is as follows: sp (carbon) - sp3 (sulfur) [Note that the hybrid orbital on sulfur is sp3 because sulfur atom has 4 sets: 3 lone pairs + 1 single (sigma) bond]. triple bond between carbon and nitrogen atoms is made of the following: sp (carbon) - sp (nitrogen) p(y)(carbon) - p(y)(nitrogen) p(z)(carbon) - p(z)(nitrogen) Note that both p(y) and p(z) of carbon and nitrogen atoms are unused p orbitals and not hybrid orbitals]. ------------------------------------------------------- 5. Predict whether the following compounds are polar (having dipole moment) or not! Compare the strength of their dipole moments! :O: || C |\ H H Dipole moments' direction: From Lewis structure indicates that carbon is the central atom. The dipole moment direction will be from the hydrogen atoms to the carbon atoms (inward). These two dipoles are equal each other in strength but oppose to each other. They cancel themselves out. The other dipole is working is the carbon - oxygen. The direction of this dipole is towards the oxygen atom. Hence the compound is strong polar. F::: | C - H |\ H H The two dipoles formed by carbon - hydrogen are inwards and working against each other and hence are canceled. The third hydrogen makes a dipole with carbon atom at the same direction of the dipole formed by the carbon - fluorine atoms. Since fluorine atom is very strong electronegative atom one therefore expect that this compound is very strong polar.

N: ||| C | H The two dipoles are working with each other: direction from hydrogen to carbon and from carbon to nitrogen atom. This compound is polar but its polarity is moderate because nitrogen is less electronegative compared with oxygen and fluorine atoms. N: ||| C | F::: The two dipoles: carbon to fluorine and carbon to nitrogen are working each other and hence are canceling each other. Because of the electronegativity difference between fluorine atom and nitrogen atom, this compound is expected to be very week polar to none polar. _______________________________________________ Chapter 11: Properties of Liquids: Intermolecular forces between liquids. Intermolecular forces: - are attractive forces between molecules (usually two or more) - are responsible for the bulk properties of matter (solids and liquids) such as boiling point, freezing points, melting point, density etc. Intramolecular forces: - are forces within the molecule itself. - are responsible for holding the atoms together - are responsible for stabilizing individual molecules - chemical bonding such as single (sigma), double (pi) and triple bonds discussed in relation to the valence bond theory and VSEPR model in chapter 10 are example of the intramolecular forces holding the molecule together. Note that intermolecular forces are much weaker than intramolecular forces. Why is that? Answer: It takes more energy to break a chemical bond, but it takes very less energy to evaporate a liquid to a gas.

Note also that melting points of substances increase with increasing strength of the intermolecular forces. Types of intermolecular froces: There are six types: 1. Dipole - Dipole forces: are attractive forces between polar molecules due to electrostatic forces. The greater these electrostatic forces, the stronger the dipole - dipole attractive forces. Example: HF <---> NH3 2. Ion - Dipole forces: Anions or cations are attracted to a dipole forces of a polar compound. Example: NO3 - (anion) <------> H2O (dipole) K+ (cation) <---------> NH3 (dipole) 3. Dipole - Induced dipole: 4. Ion - Induced dipole: Induced dipole is a temporary dipole (instantaneous dipole) due to the closeness of a none polar atom to a dipole or ion, which leads to instant (temporary) arrangement of the atoms producing an induced dipole. Example: H2O <------> Fe (dipole - induced dipole) NO3- <---------> Ni (ion - induced dipole) 5. Dispersion forces: - Generally dispersion forces are formed due to instant dipoles in non polar compounds (temporary rearrangement of the non polar atoms because of its close position to a polar compounds or ions). - However, dispersion forces can be found in polar compounds as well. Hence the dispersion forces are found in polar and none polar compounds. - Dispersion forces increase with increasing polarizability. - Dispersion forces increase with increasing number of electrons of the atoms of the compound and with increasing molar mass of the compound. - As a result, melting points increases as number of electrons increases in non polar compounds (large molar mass). Example: Melting points / dispersion forces increase as follows in the following compounds: CH4 < CCl4 < CBr4 < CI4 With CI4 with highest melting point and highest dispersion forces (Iodine atom is the largest compared with other atoms). Example: Compare the melting points of the two compounds: CH3F and CCl4

CH3F is polar (can have dipole - dipole, dipole - induced dipole and dipole - ion dipole attractive forces). CCl4 is non polar (can have dispersion forces only!) But the CCl4 has higher melting point than CH3F because CCl4 has higher molar mass than CH3F. Example: What type of intermolecular forces exist between the following pairs. Include all forces. 1. H2S and HBr: dispersion, dipole - dipole 2. Cl2 and CBr4: dispersion only 3. I2 and NO3-: dispersion, ion - induced dipole 4. NH3 and C6H6: dispersion, dipole - induced dipole 6. Hydrogen bonding: It is a dipole - dipole attraction between the hydrogen atom in a polar bond (bond that has N,O,F as in N-H, O-H, F-H) with an electronegative atom (such as N,O,F) of another polar compound. Example: H - F ..... H - F H - O ..... H - NH2 H / - Hydrogen bonding is responsible for higher boiling points of smaller molar mass compounds of HF, H2O, NH3. - This is due to the interaction of lone pair electrons of the electronegative atom and hydrogen nucleus. -------------------------- Phase Change: Solid -------> liquid (melting) Liquid ------> solid (freezing) Liquid ------> vapor (boiling) Vapor -------> liquid ( condensation) Solid -------> vapor (sublimation) without undergoing liquid phase. Vapor -------> solid (deposition)without undergoing liquid phase. Solid ----> liquid (melting): In this process, one speaks about the heat of fusion

(or heat of melting)or molar heat of fusion and it has the symbol of H(fusion). Liquid -----> vapor (boiling or vaporization): In this process, one speaks about the molar heat of vaporization H(vaporization). In both processes, molar heats are needed to bring about the change in state (phase). [i.e. from solid state to liquid state or from liquid state to vapor state). In closed system almost all liquids will stand in equilibrium with their vapor. This vapor has certain amount of pressure called the vapor pressure. This equilibrium is called the "dynamic equilibrium". In this dynamic equilibrium, the condensation rate of the vapor equals the vaporization rate of the liquid. The equilibrium vapor pressure is measured when dynamic equilibrium exists. The vapor pressure can be calculated using "Clausis - Clapeyron Equation": ln (P) = [- H(vap)/ R][1/T] + C Where: P = vapor pressure H(vap) = molar heat of vaporization in KJ/mol T = temperature in Kelvin C = Constant R = Gas constant = 8.314 J/mol.K With help of "Clausis - Clapeyron Equation", vapor pressure at different conditions (i.e. initial or final state). The following modified Clausis - Clapeyron Equation can be used: ln (P1/P2) = [- H(vap)/ R][(1/T1) - (1/T2)] where: P1 = vapor pressure at temperature T1 P2 = vapor pressure at temperature T2 T1, T2 = temperatures in Kelvin This equation can be simplified as follows: ln (P1/P2) = [H(vap)/ R][(T1 - T2) / (T1T2)] Example: Find P2 using the modified Clausis Claperon Equation, if P1 = 100 mm Hg at 34.9 C and H(vap) = 39.3 KJ/mol. T2 = 63.5 C. Answer: T1 = 34.9 C + 273.15 = 308.05 K

T2 = 63.5 + 273.15 = 336.65 K ln (100 mm Hg / P2) = [(39.3X1000 J)/(8.314 J/mol.K)][(308.05 - 336.65)/(336.65X308.05)] ln (100/ P2) = - 1.3036 100/P2 = e-1.3036 = 0.27155 P2 = (100/0.27155) = 368.25 mm Hg ------------------------------------- Phase diagram: - Is a plot of the pressure against temperature. - In the phase diagram, three phases are put together in a single graph. - Freezing point and boiling point can be obtained at different pressures. - Triple point: is the point where the three phases (solid, liquid, vapor) coexist. - In a phase diagram of water (includes all three states): At higher pressure than 1 atm, one obtains two cases: Boiling point of water increases (elevation)[over 100 degrees]. Freezing point decreases (depression)[lower than zero degrees]. ------------------------------------------------------------------ Selected problems from the textbook: Chapter 11 Please work on these problems. If you have any question, contact your instructor! 11.10;11.12;11.15;11.17;11.61;11.62;11.64;11.94;11.128;11.134 ------------------------------------------------------------------ Selected problems and answers: Chapter 11 1. Define the following: a. intermolecular forces b. intramolecular forces c. dipole - dipole forces d. dispersion forces e. ion - induced dipole forces f. hydrogen bonding g. vapor pressure h. triple point i. phase diagram j. temporary dipole k. permanent dipole Answer: Check notes + textbook! 2. What are the types of intermolecular forces existing between following

pairs? a. C6H6 and CH4 ----> none polar - none polar -----> only dispersion forces. b. CH3Cl and CS2 -----> polar - none polar ----> dipole - induced dipole. c. NaCl and HF -----> ionic - polar -----> ion - dipole d. NH3 and Fe ------> polar - element ---> dipole - induced dipole. 3. Compare the hydrogen bonding strength in the following molecule pairs: a. C6H12O6 and HF b. C6H12O6 and NH3 c. C6H12O6 and H2O Strength of the hydrogen bonding will depend on the electronegativity of the atom carrying the hydrogen atom: F>O>N Therefore, HF>H2O>NH3, HF being the highest and NH3 being the lowest. 4. Arrange the following compounds in the order of increasing melting points: SiH4, SiCl4, SiBr4 and SiI4 The higher the molar mass the higher the melting point: SiH4 < SiCl4 < SiBr4 < SiI4 SiI4 has the highest melting point (higher molar mass) and the iodine atom size is large and contains more electrons. 5. Which one of the following molecules has the highest melting point and why? a. SiI4 and SiH3F Although SiH3F is polar and SiI4 is none polar, SiI4 has the highest melting points because of iodine size and it contains more electrons and larger molar mass. b. CF4 and CFI3 CFI3 has the highest melting point for the reasons mentioned above. 6. The vapor pressure of benzene is 110.5 mm Hg at 20 C. Find its vapor pressure at 150 C. [H(vap) = 31.0 KJ/mol] T1 = 20 + 273.15 = 293.15 K T2 = 150 + 273.15 = 423.15 K ln (110.5 mm Hg / P2) = [(31.0X1000 J)/(8.314 J/mol.K)][(293.15 - 423.15)/(293.15X423.15)] ln (110.5/ P2) = - 3.908 100/P2 = e-3.908 = 0.0201 P2 = (110.5/0.0201) = 5497.51 mm Hg 7. Name the following processes:

a. solid ------> liquid ====> melting b. gas --------> solid =====> deposition c. liquid -----> gas ======> evaporation d. solid ------> gas ======> sublimation e. liquid -----> solid =====> freezing _______________________________________________ Chapter 12: Properties of solutions. A solution is a homogeneous mixture made of a solvent (in excess) and a solute (small amount dissolved in the solvent). A solution = solute + solvent Types of solutions: 1. Unsaturated solution: Contains less amount (least amount) of solute that dissolved at specific temperature. 2. Saturated solution: Contains maximum amount of solute that dissolved at specific temperature. 3. Supersaturated solution: Contains more than the maximum amount of solute that dissolved at specific temperature. Supersaturated solutions are not stable. The solute tends to settle down out of the solution as crystals. Crystallization: is a process when solute comes out of the solution and forms crystals. Concentration: the amount of solute in moles dissolved in liters of the solution. Concentration types: 1. molarity = (# moles of solute) / (liter of solution). Units: mol/L or M or molar 2. molality = (# moles of solute) / (Kg of solvent). Units: mol/Kg or m or molal 3. mass % = [(mass of solute) / (mass of solute + mass of solvent)] X 100. No units 4. Mole fraction X(A) = [( # moles of component A) / (sum of moles of all components)] 5. # moles = (mass of solute in g) / ( its molar mass in g/mol) Example: a sample of 6.44 g of naphthalene C10H8 is dissolved in 80.1 g of benzene C6H6. Calculate the percent by mass of naphthalene in this solution and calculate the molarity of naphthalene if the volume of the solution is 500 mL. Molar mass of naphthalene C10H8 = (10X12) + (8X1) = 128 g/mol # moles of naphthalene = [(6.44 g) / (128 g/mol)]= 0.0503 mol

Molarity of naphthalene = (0.0503 moles) / (0.5 Liters)= 0.1006 M [Note that 500 mL = 0.5 L] Naphthalene mass% = [(6.44 g) / (6.44 g + 80.1 g)]X 100 = 7.44% Example: Express 0.1006 M (molarity)of naphthalene in molality. mol/L ======> mol/Kg 0.1006 mol/1 Liter =====> 0.1006 mol/ Kg solvent Needs to know the density of solution (solute: naphthalene, solvent: benzene) in g/mL. Suppose this density is 0.685 g/mL of the solution. Since the solute is solid, then this density is the density of the solvent as well. Kg of solvent = (1000 mL of solution or solvent)X(0.685 g solution/mL solution or solvent)X(1 Kg /1000 g) = 0.685 Kg solvent. The molality = (0.1006 moles Naphthalene) / (0.685 Kg solvent(benzene)) = 0.147 mol/Kg or m or molal Example: Calculate the molality of 44.6% (by mass) aqueous solution of NaCl. mass % NaCl = [(mass NaCl)]/[mass NaCl + mass of solvent (H2O)] Assume that the sample is 100 g. Then NaCl = 44.6 g NaCl and mass of solvent (H2O)=100 -44.6 g = 55.4 g = 0.0554 Kg Molality = (moles NaCl) / Kg solvent H2O moles NaCl = [(44.6 g NaCl)] / [(23 + 35.5)g/mol]= 0.7623 moles NaCl Molality = [(0.7623 moles NaCl)]/[(0.0554 Kg H2O)]= 13.76 mol/Kg or molal or m. ------------------------------ Important relationship of solubility: 1. Solubility of the solids and liquids increases with increasing temperature: solubility of solids and liquids is direct proportional to the temperature. 2. Solubility of the gases decreases with increasing temperature: solubility of gases is indirect proportional to the temperature. 3. Solubility of the gases increases with increasing pressure: solubility of gases is direct proportional to the pressure. Henry Law: Solubility of gases is direct proportional to the pressure. C = KP Where C the molar concentration of the gas (molarity or solubility). Some textbooks express C with S instead. K = Henry constant.

P = partial pressure of the gas in atm Example: Calculate the molar concentration (solubility) of oxygen in water at 25 degrees for a partial pressure of 0.22 atm. The Henry's law constant for oxygen is 3.5 X 10-4 mol/L.atm Solution: C = KP C = (3.5 X 10-4 mol/L.atm) X (0.22 atm) = 7.7 X10-5 mol/L ------------------------------------------------------- Colligative properties of non electrolyte solutions: Colligative properties: Properties that depend only on the number (amount) of the solute particles in solution and not on the nature of the solute particles. Such properties are: 1. Vapor pressure lowering: - non volatile solutes does not have measurable vapor pressure. - the vapor pressure of the solute is less than vapor pressure of the solvent - vapor pressure lowering depends on the concentration of the solute. Rout's law: P1 = X1 Pi Where P1 = partial pressure of the solvent over the solution. X1 = mole fraction of the solvent in the solution Pi = vapor pressure of the solvent (pure not in any solution) In a solution containing one solute: X1 = 1 - X2 Where X2 = mole fraction of solute P1 = (1 - X2) Pi P1 = Pi - X2Pi X2Pi = Pi - P1 Vapor pressure decrease (lowering) of pure solvent (Pi - P1) = (mole fraction of the solute)X(vapor pressure of pure solvent) Vapor pressure decrease (lowering) of pure solvent is direct proportional to the concentration of the solute given in mol/L Example: Pi of glucose solution is 17.01 mm Hg at 20 degrees, while that of the solvent (H2O) is 17.25 mm Hg at this temperature. Calculate the molality of the solution. Pi - P1 = X2Pi

17.25 - 17.01 = 0.24 mm Hg = X2 (17.25) X2 = 0.24 mm Hg / 17.25 mm Hg = 0.0139 Example: Calculate the molality in the above example (with mole fraction of the solute 0.0139 and the solvent is water with a mass of 1 Kg) solution: 0.0139 = [(moles of solvent)]/[(moles of solvent + moles of solute)] moles of H2O = mass H2O / molar mass H2O moles of H2O = (1000 g) / (18 g/mol)= 55.56 moles mole fraction of H2O(solvent) = 1 - 0.0139 =0.9861 0.9861 = (moles H2O)/ (moles H2O + moles of solute) 0.9861 = (55.56 moles)/(55.56 moles + moles of solute) 0.9861(55.56 moles + moles of solute) = 55.56 54.79 + 0.9861(moles of solute) = 55.56 moles of solute = [(55.56 - 54.79)]/[(0.9861)] = 0.7809 moles Molality = moles of solute / Kg solvent Molality = 0.7809 moles /1 Kg solvent = 0.7809 molal 2. Boiling point Elevation 3. Freezing point Depression Boiling temperature at which the vapor pressure equals the atmospheric pressure. since the solute is none volatile, the vapor pressure of the solution is lowered. This affects the freezing and boiling point of the pure solvent: The boiling point will be increased and the freezing point will be decreased. In both processes, the amount of the increase and the decrease is direct proportional to the amount of the solute (molality) Boiling point increase (Delta T) = K(b) X molality Freezing point decrease (Delta T) = K(f) X molality 4. Osmotic Pressure: Osmosis: The selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. Semipermeable membrane allows solvent molecules but blocks the solute molecules. -At the beginning water level is equal in both compartment of the osmotic cell - after some time, water molecules transfer through the membrane to the more concentrated solution. - process continues till no further change is observed. Osmotic Pressure (pi) = is the pressure required to stop osmosis. Reason of the Osmosis: Vapor pressure of the pure solvent is higher than the

vapor pressure of the solution (solute + solvent). Pure solvent vapor will transfer because its vapor pressure is high. Osmotic pressure (pi) = MRT Where M = molarity in mol/L R = gas constant in 0.082 atm.L/mol.K T = temperature in Kelvin Example: What is the osmotic pressure (in atm) of a solution of a 0.884 M sucrose solution at 16 degrees? T = 16 + 273.15 = 289.15 K Osmotic pressure (pi) = MRT Osmotic pressure (pi) = (0.884 M)(0.082 atm.L/mol.K)(289.15 K) = 20.96 atm ------------------------------------------------------------------ Selected problems from the textbook: Chapter 12 Please work on these problems. If you have any question, contact your instructor! 12.15;12.17;12.51;12.53;12.56;12.61;12.62;12.64;12.68 ------------------------------------------------------------------ Selected problems and answers: Chapter 12 1. Define the following: a. saturated solution b. unsaturated solution c. supersaturated solution d. molarity e. molality f. mole fraction g. mass% h. colligative properties i. osmosis j. osmotic pressure Check notes + textbook! 2. Calculate the molarity of a solution of 30 mL of solute A and 500 mL of solvent B. [Hint: the molecular formula of solute A is C7H8 and its density is less than one, say; 0.86 g/mL] Solution: molar mass C7H8 = (12X7) + (8X1) = 92 g/mol

# moles C7H8 = [(30 mL)X(0.86 g/mL)]/[(92 g/mol)]= 0.2804 moles molarity = [0.2804 moles]/[(500 mL + 30 mL)(1 Liter/1000 mL)] = 0.529 M Note that the volume of the solution is equal the total volume of the solute and solvent = 500 mL + 30 mL = 530 mL 3. Calculate the molality of the above solution if the density of the solvent B is 0.856 g/mL molality = (# moles A) / (Kg B) Kg B = (500 mL)(0.856 g/mL)(1 Kg/1000 g) = 0.428 Kg B molality = (0.2804 mol A) / (0.428 Kg B) = 0.655 molal 4. Calculate the mass% of the above solution. mass% = (mass A)/(mass A + mass B) mass A = 30 mL X 0.86 g/mL = 25.8 g mass B = 500 mL X 0.856 g/mL = 428 g mass% = [(25.8 g) / (25.8 g + 428 g)]X 100 = 5.68% = 5.7% 5. Calculate the mole fraction of the solute A. Molar mass of B is 150 g/mol. mol fraction A = [(mol A)]/[(mol A + mol B)] mol A = 0.2804 mol mol B = [(428 g)/(150 g/mol)]= 2.85 mol B mol fraction A = [(0.2804 mol A)] / [(0.2804 mol A + 2.85 mol B)] = 0.0895 (No Unit!) 6. State the relationship between the following: a. solubility (solid) versus temperature direct proportional b. solubility (gas) versus temperature indirect proportional c. solubility (gas) versus pressure direct proportional 7. State the Henry law: C = KP where c = solubility of a gas in a solution K = Henry constant P = partial pressure. 8. What are the colligative properties discussed? 1. vapor pressure decrease 2. boiling point increase

3. freezing point decrease (depression) 4. osmotic pressure decrease 9. calculate the mole fraction of a solute A dissolved in a solvent B if the vapor pressure of the pure solvent B is 150 mm Hg is depressed by 50 mm Hg at specified temperature. vapor pressure depression = X2Pi 50 mm Hg = X2(150 mm Hg) X2 = (50 mm Hg)/(150 mm Hg) = 0.333 (No Unit!) 10. Calculate the molality of a solution in question # 9, if its freezing point was depressed by the presence of the solute A by 3.5 degrees. The actual freezing point of the pure solvent B is 22 degrees. [K(f) = 31 C/molal] Solution: Freezing point depression = K(f)X molality 3.5 degrees = (31 C/molal)X (molality) Molality = (3.5 C) / (31 C/molal) = 0.1129 molal 11. Calculate the molality of the above solution (question # 10) if its boiling point was elevated by the presence of solute by 10.5 C. The actual boiling point of the pure solvent B is 180 C [ use K(b) = 31 C/molal]. Determine the boiling point of the solution. Solution: Boiling point of the solution: Boiling point elevation = boiling point (solution) - boiling point (pure solvent) 10.5 C = boiling point of solution - 180 boiling point of the solution = 180 + 10.5 = 190.5 C elevation in boiling point = K(b)(molality) Molality = (10.5 C)/(31 C/molal) = 0.339 molal 12. Calculate the osmotic pressure in atm of a naphthalene solution in cyclohexane that has the molarity of 0.973 M at 25 C. Osmotic pressure (pi) = MRT = (0.973 mol/L)(0.082 atm.L/mol.K)(25 + 273.15K) = 23.79 atm _______________________________________________ Chapter 13: Chemical Kinetics Chemical Kinetics is the area concerned with the rate of a chemical reaction. The rate of the reaction:

Consider the following reaction: A -------> B By definition: the rate of the reaction is the consumption of the reactants per time or the production of the products per time. Rate: - Delta[A] / Delta t or + Delta[B] / Delta t where Delta means the change or the difference [actually is represented by a triangle and it means the difference between the final concentration and initial concentration] Delta t means: final time - initial time Note that a minus sign is used when the rate of the reaction is expressed in term of reactants and a positive sign is used when the rate of the reaction is expressed in term of products. Rate of reaction is a positive quantity! The minus sign has no physical meaning! It does show only that the reactants are being consumed. Rate of the reaction is determined by experiment ONLY!! it is an average rate over period of time. Example Consider the reaction of bromine with formic acid: Br2(l) + HCOOH(aq) ----> 2Br-(aq) + 2H+(aq) + CO2(g) Br2(liquid) is a red - purple solution. Therefore, the bromine concentration consumption per time is easy to follow experimentally. Why? because it has color while formic acid HCOOH is colorless. The concentration of bromine will be followed via color change per time UV-Visible spectroscopy techniques. Rate is direct proportional to the concentration of Br2 [Br2]. By experiment: Rate = K[Br2] Where K = the rate of constant. The higher the value of K, the faster the reaction goes, the higher the rate. The rate expression: Rate = K[Br2] is called "Rate Law" Therefore, the rate can be expressed in two terms: 1. In terms of stoichiometric coefficients: Generally written: aA + bB ------> cC + dD Rate in term of stoichiometric coefficients: Rate = (-1/a)[Delta A]/[Delta t] = (-1/b)[Delta B]/[Delta t] = (+1/c)[Delta C]/[Delta t] = (+1/d)[Delta D]/[Delta t] Example:

CH4(g) + 2 O2(g) -------> CO2(g) + 2H2O(g) Rate = (-1/1)[Delta CH4]/[Delta t] = (-1/2)[Delta O2]/[Delta t] = (+1/1)[Delta CO2]/[Delta t] = (+1/2)[Delta H2O]/[Delta t] 2. In term of Rate Law: It is a law that expresses the relationship of the rate of reaction to the rate constant and the concentrations of the reactants (ONLY!!) raised to some powers. Generally written: Rate = K[A]^x [B]^y x and y are reaction orders which express the behavior of the concentration of reactants per time. x or y = 1 -----> the reaction is said to be 1st order x or y = 2 -----> the reaction is said to be 2nd order x or y = 3 -----> the reaction is said to be 3rd order x and y can be positive, negative, fractional or integer numbers Important Note: x and y (reaction orders)are determined by EXPERIMENT ONLY!!! and not from the stoichiometric coefficients! Example: CH4(g) + 2 O2(g) -------> CO2(g) + 2H2O(g) Rate = K[CH4]^x[O2]^y x and y have to determined by experiment ONLY! x and y are ALWAYS defined in terms of reactants ONLY! Example: 2NO(g) + 2H2(g) ------> N2(g) + 2H2O(g) Experimental data: exp. # 1 [NO]: 5.0 X 10^-3 M [H2]: 2.0 X 10^-3 M Initial rate: 1.3 X 10^-5 M/s -------------------------------- exp. # 2 [NO]: 10.0 X 10^-3 M [H2]: 2.0 X 10^-3 M Initial rate: 5.0 X 10^-5 M/s ------------------------------- exp. # 3 [NO]: 10.0 X 10^-3 M

[H2]: 4.0 X 10^-3 M Initial rate: 10.0 X 10^-5 M/s ------------------------------- General rule: 1. As the reactant concentration doubles ----> if the rate doubles (increases by factor of 2) =====> x or y is 1st order. 2. As the reactant concentration doubles ----> if the rate quadruples (increases by factor of 4) =====> x or y is 2nd order. 3. As the reactant concentration doubles ----> if the rate increases by factor of 8) =====> x or y is 3rd order. Back to the example: The general rate law: Rate = K[NO]^x[H2]^y Comparing experiment 1 with 2: as [NO] doubles, the rate increase by factor of 4 [actually by factor of (5.0X10^-5)/(1.3X10^-5) = 3.846] Hence the x = 2 (according to rule # 2) Comparing experiment 2 with 3: as [H2] doubles, the rate increase by factor of 2 (doubles) Hence the y = 1 (according to rule # 1) The rate of the reaction is Rate = Actually 1 is understood and not written! Rate = K [NO]^2[H2] The above method is called the "Inspection Method" There is another method to determine the reaction orders. This method is called "The Method of Initial Rates". Determination of x and y reaction orders by method of initial rates: Dividing rate(1) by rate(2): (1.3X10^-5/5.0X10^-5) = [K(5.0X10^-3)^x{2.0X10^-3)^y] / [(10.0X10^-3)^x{2.0X10^-3)^y] (0.26) = [(5.0X10^-3)(10.0X10^-3)]^x 0.26 = 0.5^x Take the natural logarithm of both sides ln(0.26) = x ln(0.5) (-1.347) = x (-0.693) x = (-1.347) / (-0.693) = 1.94 ~ 2.0 To obtain y, one has to divide rate(2) by rate(3): (5.0X10^-5/10.0X10^-5) = [K(10.0X10^-3)^x{2.0X10^-3)^y] / [(10.0X10^-3)^x{4.0X10^-3)^y]

(0.5) = [(2.0X10^-3)(4.0X10^-3)]^y 0.5 = 0.5^y Take the natural logarithm of both sides ln(0.5) = y ln(0.5) (-0.693) = y (-0.693) y = 1 The rate is then: Rate = K [NO]^2[H2] Note that the results obtained by the inspection method and by the method of initial rates are the same. In both methods one reactant concentration has to be kept constant while the second is changed. Note that the overall reaction order is the sum of all reaction orders involved. In this example, the overall reaction order is the sum of x and y = 2 + 1 = 3 First Order: Consider the following reaction: A ----> product the rate of the reaction is given in two terms: 1. Rate Law: Rate = K[A] 2. Stoichiometric Coefficients: Rate = (-1/1) [(Delta [A])/(Delta t)] The unit of the rate of first order is M/s where M = molarity s = seconds The rate constant K has the unit 1/s Putting the rates obtained in both terms. one obtains: K[A} = (-1/1) [(Delta [A])/(Delta t)] Rearranging: K[(Delta t)] = (-1/1) [(Delta [A])/[A] Further one can obtain: Kt = ln {[A]o}/{[A]} =====> This expression is used for the rate of 1st order Where k = the rate constant t = time in seconds [A]o = initial concentration of the reactant A [A} = final concentration of the reactant A The rate of 1st order can be re-written as: ln[A] = -Kt + ln[A]o

This is a linear relationship of the type Y = mX + C where Y = ln[A] X = t m = slope = - K ln[A]o = intercept Half - time reaction: [A] = 1/2[A]o = t(1/2) Half - time t(1/2) is the time required for the concentration of the reactant to drop (decrease) to half of its initial concentration. The half - time reaction of the first order reaction is given as: t(1/2) = (0.693)/K Where K is the rate constant. Second Order: Consider the following reaction: A ----> product the rate of the reaction is as follows: Rate Law: Rate = K[A]^2 The rate of 2nd order can be re-written as: (1/[A]) - (1/[A]o) = Kt or (1/[A]) = Kt + (1/[A]o) This is a linear relationship of the type Y = mX + C where Y = (1/[A]) X = t m = slope = K (1/[A]o) = intercept The half - time reaction of the second order reaction is given as: t(1/2) = 1/K[A]o Where K is the rate constant. Application of the rate of the reaction: 1. Activation energy: is the minimum energy required to initiate a chemical reaction. The activation energy is related to the rate constant as follows: ln [K1/K2] = (Eo/R)[(T1 - T2)/(T1T2)] ====> 1st order Where: K1 = rate constant at condition 1 (initial condition)

K2 = rate constant at condition 2 (final condition) Eo = the activation energy R = gas constant = 8.314 J/mol.K T1 = temperature at condition 1 (initial condition) T2 = temperature at condition 2 (final condition) In the endothermic processes, the activation energy tends to be high (in the energy graph, the activation energy is drawn from the reactants' energy begin to the energy maximum). The products' energy is higher than the reactants' energy. In the exothermic processes, the activation energy tends to be low. The products' energy is lower than the reactants' energy. 2. Catalysis: A catalyst is a chemical compound when added into the chemical rxn causes the chemical reaction to speed up. The catalyst does not undergo any change in composition. Reason for speeding up the rate of reaction is that the catalyst tends to lower the activation energy. 3. Inhibitor: An inhibitor is a chemical compound when added into the chemical reaction causes the chemical reaction to slow down. The inhibitor does not undergo any change in composition. Reason for slowing down the rate of reaction is that the inhibitor tends to increase the activation energy. ------------------------------------------------------------------ Selected problems from the textbook: Chapter 13 Please work on these problems. If you have any question, contact your instructor! 13.5;13.16;13.17;13.18;13.19;13.28;13.30;13.40;13.44;13.49;13.52;13.70 ------------------------------------------------------------------ Selected problems and answers: Chapter 13 1. Consider the following reaction: C3H8(gas) + 5O2(gas) ------> 2CO2(gas) + 4H2O(gas) Determine the rate of the reaction in terms of: a. stoichiometric coefficients: Rate = (-1/1)[Delta C3H8]/[Delta t] = (-1/5)[Delta O2]/[Delta t] = (+1/3)[Delta CO2]/[Delta t] = (+1/4)[Delta H2O]/[Delta t]

b. in terms of rate law: Rate = K[C3H8]^x[O2]^y 2. The reaction of peroxydisulfate ion (S2O8^2-)with iodide ion (I^-1) is: S2O8^2-(aq) + 3I^-1(aq) ---> 2SO4^2-(aq) + I3^-1(aq) From the following data collected at a certain temperature, determine the rate law and calculate the rate constant and the over all reaction order: Experimental data: exp. # 1 [S2O8^2-]: 0.080 M [I^-]: 0.034 M Initial rate: 2.2 X 10^-4 M/s -------------------------------- exp. # 2 [S2O8^2-]: 0.080 M [I^-]: 0.017 M Initial rate: 1.1 X 10^-4 M/s ------------------------------- exp. # 3 [S2O8^2-]: 0.16 M [I^-]: 0.017 M Initial rate: 2.2 X 10^-4 M/s ------------------------------- Solution: Compare exp 1 with exp 2 ===> [S2O8^2-]is kept constant: As [I^-] doubles, the rate doubles -----> 1st order. Compare exp 2 with exp 3 ===> [I^-] is kept constant: As [S2O8^2-] doubles, the rate doubles -----> 1st order Hence, the rate of reaction is Rate = K[S2O8^2-][I^-] The over all rate of reaction is 1 + 1 = 2 To calculate K: K = Rate / [S2O8^2-][I^-] Using any rate for calculation. (the 2nd rate in this example) K = (1.1 X 10^-4 M/s) / [0.080 M][0.017 M] K = 8.088 X 10^-2 1/Ms 3. The reaction 2a ----> B is first order in A with a rate constant of 2.8 X 10^-2 s^-1 at 80 C. How long (in seconds) it will take for A to decrease from 0.88 M to 0.14 M? Rate = K[A]

ln ([A]o / [A]) = Kt ln [0.88]/[0.14] = (2.8 X 10^-2 s^-1)t 1.838 = (2.8 X 10^-2 s^-1)t t = (1.838) / (2.8 X 10^-2 s^-1) =65.65 s ~ 65.7 s 4. The decomposition of butane (C4H10) to ethyl radicals is a first - order reaction with a rate constant of 5.36 X 10^-4 s^-1 at 700 C: C4H10(g) -----> 2C2H5.(g) Calculate the half - time of the reaction in minutes. 1st order ====> t(1/2) = (0.693) / K t(1/2) = (0.693) / 5.36 X 10^-4 s^-1 = 1296.9 s /60 = 21.55 minutes. 5. Fluorine atoms combine to form molecular fluorine in the gas phase: F(gas) + F(gas) ------> F2(gas) This reaction follows second - order kinetics and has the rate constant of 7.0X10^9/Ms at 23 C. a. If the initial concentration of F(gas) was 0.086 M, calculate the concentration after 2.0 minutes. b. Calculate the half - life of the reaction if the initial concentration of F(gas) is 0.60 M and if it is 0.42 M. a. the rxn 2nd order: (1/[A]) - (1/[A]o) = Kt (1/[A]) - (1/0.086M) = (7X10^9/Ms)(2X60 s) (1/[A]) = (7X10^9/Ms)(2X60 s) + (1/0.086M) (1/[A]) = (8.4X10^11) + 11.628 = 8.4X10^11 [A] = 1 / 8.4X10^11 = 1.19X10^-12 b. t(1/2) = 1/K[A]o = 1/(7X10^9/Ms)[A]o If [A]o = 0.60 M, then: t(1/2) = 1/(7X10^9/Ms)[0.60] = 2.38X10^-10 s. If [A]o = 0.42 M, then: t(1/2) = 1/(7X10^9/Ms)[0.42] = 3.40X10^-10 s. 6. The rate constant of a first - order reaction is 5X10^-2 1/s at 298 K. What is the rate constant at 500 K if the activation energy for the reaction is 60.0 KJ/mol? ln [K1/K2] = (Eo/R)[(T1 - T2)/(T1T2)] ln[(5.0X10^-2)/(K2)] = [(60X1000 J/mol)/(8.314 J/mol.K)][(298K - 500K)/(298X500)] ln[(5.0X10^-2)/(K2)] = - 9.7837

5.0X10^-2 / K2 = e^(-9.7837) = 5.636X10^-5 K2 = 5.0X10-2 / 5.636X10^-5 = 887.154 1/s. 7. Define the following: a. catalysis b. activation energy c. reaction order d. rate of the reaction e. half - time of the reaction 8. Consider the following reaction: Rate = K[A][B]^2 If the reactant [B] doubles, the rate will (Circle one): a. double b. increase by factor of 8 c. decrease by factor of 10 d. none of the above e. all of the above Answer is d because there is missing information about [A]. 9. Consider the following reaction: Rate = K[A]^3[B]^2 If the reactant [A] doubles and the reactant [B] triples, the rate will (Circle one): a. double b. quadruple c. decrease by factor of 100 d. increase by factor of 72 e. none of the above f. all of the above Answer is d. The rate will increase by factor of 72 because: Rate = K [2A]^3[3B]^2 = K(2)^3(3)^2[A]^3[B]^3 = 72K[A]^3[B]^2 _______________________________________________ Chapter 14: Chemical Equilibrium An equilibrium requires that the forward rate of the reaction equals the rate of reverse reaction. Two types of equilibrium:

1. Physical equilibrium: this equilibrium involves the equilibrium between two phases of the same compound. Example: H2O(s) <----------> H2O(l) CO2(s) <---------> CO2(g) (dry ice) An equilibrium involves different or the same phases of different compounds. New compounds are formed in the process. Example: N2O4(g) <-------> 2NO(g) N2(g) + 3H2(g) <-----------> 2NH3(g) Equilibrium constant Kc and Kp: Kc and Kp are the ratio of the products concentrations or partial pressure to the reactants concentrations or partial pressures. Example: N2O4(g) <-------> 2NO(g) Kc = [NO]^2 / [N2O4] Kp = P^2(NO) / P(N2O4) Example: N2(g) + 3H2(g) <-----------> 2NH3(g) Kc = [NH3]^2 / [N2][H2]^3 Kp = P^2(NH3) / P(N2)P^3(H2) Note that the stoichiometric factors are raised up as powers. Note that Kc and Kp have units and actually ignored. Relationship between Kc and Kp: Kp / Kc = [RT]^delta n Where: Kp = equilibrium constant related to partial pressures Kc = equilibrium constant related to concentrations R = gas constant = 0.082 atm.L/mol.K T = temperature in Kelvin delta n = (the sum of all stoichiometric coefficients of the products) - (the sum of all stiochimetric coefficients of the reactants) Example: Write the Kc and Kp expression for the decomposition of N2O5: 2N2O5(g) <------------> 4NO2(g) + O2(g) Calculate Kc and Kp if the following data are given at 25 degrees: [N2O5] = 5.0 M

[NO2] = 2.2 M [O2] = 1.5 M Kc = {[NO2]^4[O2]}/[N2O5]^2 Kp = {P^4(NO2)P(O2)}/P^2(N2O5) Kc = {[2.2M]^4[1.5M]}/[5.0M]^2 Kc = 935.1384)/(25.0)= 1.406 Kp/Kc = [0.082 X (25+273.15)]^(4+1-2) Kp/1.406 = (24.4483)^3 = 14613.223 Kp = (1.406)(14613.223) = 20546.191 = 2.0546 X 10^4 Heterogeneous equilibrium: Such chemical equilibrium includes pure liquids and pure solids. These pure liquids and pure solids are ignored when writing Kc and Kp expressions. Example: NH4HS(s) <-------> NH3(g) + H2S(g) Kc = [NH3][H2S] Kp = P(NH3)P(H2S) Note that [NH4HS(s)] and P(NH4HS) are ignored because there is no change to their values because NH4HS is solid and the pressure and the concentration of a solid and a liquid are constant. Example: Consider the above equilibrium: NH4HS(s) <-------> NH3(g) + H2S(g) If the partial pressure of each gas evolved is 0.265 atm at 298.15 K, calculate Kc and Kp. Kp = P(NH3)P(H2S) = (0.265 atm)(0.265 atm) = 0.070225 Kp/Kc = [0.082 X 298.15]^(1+1-0) = [24.4483]^2 = 597.7194 0.070225/Kc = 597.7194 Kc = 0.070225/597.7194 = 1.1749 X 10^-4 Importance of chemical equilibrium: Predicting the direction of a reaction: By predicting the direction of the equilibrium, one can maximize the yield and hence the profit. To predict the direction of the equilibrium, one can use two methods: 1. First method: Reaction quotient Q(c,p): - Instead of using the actual equilibrium concentrations or equilibrium

pressures, one uses the initial concentrations and pressures and hence calculate Q(c,p) - Calculate Q(c,p) - Compare the value of Q(c,p) with the actual value of K(c,p). Three cases can be obtained and hence the direction of the equilibrium can be obtained: a. If Q(c,p) value > K(c,p) value [ greater than] It means that the initial concentrations of the products are high. Therefore to establish equilibrium the reaction has to re-adjust itself by shifting to the reactants' side [SHIFT TO THE LEFT]. The products have to decompose to produce back the reactants. b. If Q(c,p) = K(c,p) In this case the initial concentration of products equal to the initial concentration of the reactants. The system is in equilibrium. [NO SHIFT] c. If Q(c,p) value < K(c,p) value [ less than] It means that the initial concentrations of the products are low. Therefore to establish equilibrium the reaction has to re-adjust itself by shifting to the products' side [SHIFT TO THE RIGHT]. The reactants have to further react to produce extra products. Example: The equilibrium constant Kc for the formation of nitrosyl chloride, an orange - yellow compound from nitric oxide and molecular chlorine is 6.5X10^4 at 35 C. In a certain experiment, 2.0X10^-2 moles of NO, 8.3X10^-3 moles of Cl2 and 6.8 moles of NOCl are mixed in a 2.0 - Liter flask. In which direction will the system proceed to reach equilibrium? 2NO(g) + Cl2(g) <------> 2NOCl(g) First one has to calculate the initial concentration of the reactants and products: [NO]o = 2.0X10^-2 mol/2.0 L = 1.0X10^-2M [Cl2]o = 8.3X10^-3 mol/2.0L = 4.15X10^-3M [NOCl]o = 6.8 mol/2.0L = 3.4M Qc = [3.4]^2/[1.0X10^-2][4.15X10^-3] = 2.786 X 10^7 But Kc = 6.5 X 10^4 Thus Qc > Kc ======> SHIFT TO LEFT 2. Second method: Inspecting the equilibrium (the balance method) This method depends on inspecting the equilibrium. Imagine that each reactant, each product and each condition (such as heat , pressure) are weights or loads put on a balance. The reactants are put on the left arm of the balance and the products are put on the right arm of the balance. Other conditions such heat and pressure are

put in the corresponding arm according to their position in the chemical reaction equation. Now any chemical (load or weight) is added, it will itself to the common place where it is found to begin with. [common ion, common condition, common molecule, common compound, etc.] Note that solids and liquids addition to the chemical equilibrium will have NO EFFECT! Example: Let consider the above equilibrium: 2NO(g) + Cl2(g) <------> 2NOCl(g) a. Effect of addition of extra Cl2: Cl2 will add itself to the left (common compound/molecule). This will cause the left arm of the balance to be heavier. In order to re-establish the balance between the two arms of the balance, some of the reactants have to produce some products and hence the shift will be to the right. Factors affecting the chemical equilibrium: The "Le Chatelier's" Principle: " If an external stress is applied to a system at equilibrium, the system adjusts itself in such way that the stress is partially offset." The "Le Chatelier's" Principle is the same idea as the balance idea: Factors affecting the chemical equilibrium are: 1. change in the concentration: Example: Consider the following equilibrium: 2NO(g) + O2(g) <--------> 2NO2(g) Predict the shift of the reaction if: a. NO2 is added: NO2 will add itself to the products' side (to the right because of the common molecule/compound effect). SHIFT TO THE LEFT. b. NO is added: NO will add itself to the reactants' side (to the left because of the common molecule/compound effect). SHIFT TO THE RIGHT. c. O2 is added: O2 will add itself to the reactants' side (to the left because of the common molecule/compound effect). SHIFT TO THE RIGHT. 2. Change in the pressure:

Based on the ideal gas equation: PV = nRT Pressure is always indirect proportional to the volume. Therefore, the general rule is: If the pressure of the chemical equilibrium is increased then the shift of the reaction will be to the direction of least number of moles. [P increases, shift to place where number of moles is the least]. The opposite is also true, P decreases, shift to place where number of moles is the highest. Example: 2PbS(s) + 3O2(g) <-------> 2PbO(s) + 2SO2(g) Suppose than one increases the pressure the system (equilibrium), then the shift will be to side with the least number of moles. reactants' side number of moles = 3 moles (note the moles of the solid or liquid reactants are ignored). products' side number of moles = 2 moles. Hence the shift of the reaction will be toward the products (least number of moles: 2 moles). Shift to the RIGHT. If the pressure of the system (equilibrium) is decreased then the shift will be to the LEFT (to the highest number of moles). 3. Change in the volume: The change in the volume of the system is exactly the opposite of the pressure: If the volume of the system is increased, then the shift of the reaction will be to the place with the highest number of moles and vice versa. If the volume of the system is decreased, then the shift will be to the direction of lowest number of moles. 4. Effect of the heat: addition of external heat or removing the heat will depend on the type of the chemical equilibrium: exothermic or endothermic. In case of endothermic equilibrium, the heat will add itself to the reactants' side to the left (because heat is found at the reactants' side). In case of exothermic equilibrium, the heat will add itself to the products' side to the right (because heat is found at the products' side). Example: 2NO(g) + O2(g) <---------> 2NOCl(g) + heat Reaction is exothermic. What will happen if the system is heated? Answer: More heat is added to the right side (products' side) ======> shift to the LEFT

What will happen if the system is cooled down (heat is removed)? Removing heat from the right side of the reaction will lead to lighter arm of the balance at the right. The stress (over weight) will put on the left arm of the balance, shift will be to the RIGHT. 4. Effect of catalyst: No effect. Catalyst speeds up the reaction without any effect on the reaction. 2NO(g) + O2(g) <--------> 2NOCl(g) If a catalyst is added such as Fe3O4, there will be no effect on the equilibrium and hence no shift. ------------------------------------------------------------------ Selected problems from the textbook: Chapter 14 Please work on these problems. If you have any question, contact your instructor! 14.7;14.14;14.16;14.21;14.23;14.24;14.25;14.28;14.30;14.51;14.53;14.54;14.69;14.71;14.76;14.80 ------------------------------------------------------------------ Selected problems and answers: Chapter 14 1. Define the following: a. physical equilibrium b. chemical equilibrium c. rate of forward reaction d. rate of reverse reaction Check notes + textbook. 2. Why does the chemical equilibrium not involve limiting reactant concept? Limiting reactant concepts are dealing with disappearance (100% consumption)of one of the reactants, while chemical equilibrium assumes the presence of both reactants and products at the same time. 3. Which of the following reaction is in equilibrium? (Circle one) a. A + B ------> C + D b. A + B ======> C + D c. A + B <-----> C + D d. A + B ------ C + D answer: c. Equilibrium is always double arrows!

4. Write the equilibrium expressions Kc and Kp for the following reactions: a. 3H2(g) + N2(g) <-----> 2NH3(g) Kc = [NH3]^2/[H2]^3[N2] Kp = P^2(NH3) /P^3(H2)P(N2) b. 2ZnS(s) + 3O2(g) <----> 2 ZnO(s) + 2SO2(g) Kc = [SO2]^2/[O2]^3 Kp = P^2(SO2)/P^3(O2) Note that ZnS and ZnO are solids and therefore they are ignored. c. C6H5COOH(aq) <----> C6H5COO-(aq) + H+(aq) Kc = [C6H5COO-][H+]/[C6H5COOH] Kp = P(C6H5COO-)P(H+)/P(C6H5COOH) d. 2HgO(s) <-------> Hg(l) + O2(g) Kc = [O2] Kp = P(O2) Note that HgO is solid and Hg is liquid therefore they are ignored. 5. Calculate Kc and Kp at 74 C for the following chemical equilibrium: CO(g) + Cl2(g) <----> COCl2(g) [CO] = 1.2X10^-2M [Cl2] 0.054M [COCl2] = 0.14M Kc = [COCl2]/[CO][Cl2] Kc = [0.14M]/[1.2X10^-2M][0.054M] = 2.16X10^2 Kp/Kc = [RT]^delta n Kp/2.16X10^2 = [0.082X(74+273.15)K]^(1-1-1) Kp/2.16X10^2 = [28.4663]^-1 = 0.03513 Kp = (0.03513)(2.16X10^2) = 7.588 6. Write Kc and Kp at 100 C for the following reaction: Ni(s) + 4CO(g) <---------> Ni(CO)4(g) The partial pressure of both gases is 0.250 atm. Kp = P{Ni(CO)4}/P^4(CO) Kp = (0.250 atm)/(0.250)^4 = 64 Kp/Kc = [0.082X(100+273.15)]^(1-4)= 3.4907X10^-5 64/Kc = 3.4907X10^-5 Kc = 64/3.4907X10^-5 = 1.8335X10^6 7. Using reaction quotient Qp concept, predict the direction of the chemical

equilibrium if the following data are given: 2SO2(g) + O2(g) <-------> 2SO3(g) The initial pressures are 0.350 atm for SO2, 0.762 atm for O2 and 1.112 atm for SO3. Kp = 5.60X10^4 Qp = P^2(SO3)/P^2(SO2)P(O2) Qp = (1.112atm)^2/(0.350atm)^2(0.762atm) = 13.247 Kp = 5.60X10^4 Kp is greater than Qp, therefore the initial products' concentration is lower than the corresponding equilibrium concentration. Therefore, the reactants' concentrations are higher than the products' concentrations. Shift =====> Right. 8. Predict the shift in the equilibrium for the following reaction: SO2(g) + Cl2(g) <------> SO2Cl2(g) + Heat a. If SO2 is added: Shift to right b. If SO2Cl2 is removed: less product, shift to the right c. If Cl2 is added: Shift to the right d. If the system is cooled: Cooling the system means removing the heat, shift to the right e. If the system is heated: More heat is added to products' side, shift to the left. f. If the system's pressure is increased (volume is decreased): The pressure is increased (volume is decreased). Then the shift will be to the lowest number of moles. Reactants' side: 1 + 2 = 3 moles Products' side: 1 mole Hence shift will be to the products' side (1 mole). Shift to the right. g. If the system's pressure is decreased (volume is increased): The pressure is decreased (volume is increased). Then the shift will be to the highest number of moles. Reactants' side: 1 + 2 = 3 moles

Products' side: 1 mole Hence shift will be to the reactants' side (3 mole). Shift to the left. h. If a catalyst is added to the system: Catalyst has no effect on the position (shift) of the equilibrium. _______________________________________________ Chapter 15: Acids and Bases Bronsted acid: a compound that donates H+ (H+ - donor) when dissolved in water. Bronsted base: a compound that accepts H+ (H+ - accepts) when dissolved in water. Bronsted base generally tends to donate OH- when dissolved in water. Example: Bronsted acids: HCl, HNO3, H2SO4, H3PO4 Bronsted bases: KOH, NaOH, Ca(OH)2, NH4OH Conjugated acid - base pair: (Conjugated means joined together) 1. Acid <-----------> conjugated base [sometimes called the base form of the acid] 2. Base <-----------> conjugated acid [sometimes called the acid form of the base] Example: HF(aq) <--------> H+(aq) + F-(aq) HF is an acid F-: is conjugated base NH3(aq) + H+(aq) <--------> NH4+(aq) NH3: is a base NH4+: is conjugated acid General Rules: 1. Conjugated acid: a. is a base that has gained H+ b. has a positive charge2 2. Conjugated base:

a. is an acid that has lost H+ b. has a negative charge 3. Acids and bases are generally neutral. To figure our the corresponding conjugated acids and bases, one has to start with the neutral compounds assigning them as an acid and a base. 4. Standard system is used to figure out the conjugated acids and bases is water (H2O). Water can react as an acid or as a base as follows: 2H2O(l) <-------> H3O+(aq) + OH-(aq) H2O: acid or base depending on the reaction itself. H3O+: (Hydronium ion), conjugated acid (by definition) OH-: (Hydroxide ion), conjugated base (by definition) For writing simplicity, one tends to write H3O+ as H+ (proton). One should bear in mind that H+ does not exist in nature!! Use the H2O standard system after assigning the neutral acid and base. Example: Assign the acid, the base, the conjugated acid and the conjugated base! CH3COOH(aq) + H2O(l) <----------> CH3COO-(aq) + H+(aq) First look for neutral compounds: CH3COOH(aq): an acid H2O(l): a base Suppose that one does not know that CH3COOH (acetic acid) is an acid, then use H2O as standard system: H2O gives H+ (or H3O+). H+ is a conjugated acid by definition and hence water has to react as a base. CH3COO-: conjugated base (an acid that lost H+, and is negatively charged0 H+: conjugated acid (by definition) Example: CN-(aq) + H2O(l) <-------> HCN(aq) + OH-(aq) Start assigning the neutral compounds: HCN: an acid H2O: an acid Suppose that one does not know that HCN (hydrocyanic acid) is an acid, then looking at CN- is enough to give the clue that this ion has lost proton (H+). H2O gives OH-. OH- is a conjugated base by definition and hence water has to react as an acid CN-: conjugated base (negatively charged, lost H+) OH-: conjugated base (by definition) The above example is put in a way that difficult to be recognized!! The best

way is to put the acid and the base as neutral compounds as reactants in the chemical equation: HCN(aq) + H2O(l) <----> H3O+(aq) + CN-(aq) HCN: an acid H2O: a base (because it gives H3O+) H3O+: conjugated acid (by definition) CN-: conjugated base (negatively charged, lost H+) The acid - base properties of water: Auto-ionization of water (also called self - ionization of water): H2O(l) + H2O(l) <------> H3O+(aq) + OH-(aq) or: H2O(l) <-------> H+(aq) + OH-(aq) Kw = ion - product of water or water dissociation constant Kw = [H+][OH-]/[H2O] Since H2O is the solvent and is in abundant. Its concentration is constant and does not change. Hence, Kw = [H+][OH-] It was found that at 25 C [H+] = 1X10^-7M and [OH-] = 1X10^-7M Therefore, Kw = [1X10^-7M][1X10^-7M] = 1X10^-14 (ignoring the units) Example: Calculate the concentration of OH- ions in a HCl solution whose hydrogen ion concentration is 1.3M. HCl -----> H+ + Cl- (strong acid) [H+] = 1.3 M [OH-] = Kw/[H+] = (1X10^-14)/(1.3) = 7.69X10^-15 M pH - A measurement of the acidity and basicity: pH = - log[H+] pH = a negative logarithm of the hydrogen ion concentration in mol/L. Acidic solution: [H+] > 1X10^-7M, pH < 7 Neutral solution: [H+]=[OH-]=1X10^-7M, pH =7 Basic solution: [H+] < 1X10^-7M, pH > 7 Example: The pH of rain water collected in a certain region of the northeastern United States on a particular day was 4.82. Calculate the [H+] ion concentration of the rain water and calculate the [OH-] for this rain water.

pH = 4.82 = - log[H+] [H+] = 10^-pH = 10^-4.82 =1.514X10^-5 [OH-] = 1X10^-14/[H+] = 1X10^-14/1.514X10^-5 = 6.605X10^-10 Stength of Acids and Bases: strong acids are strong electrolytes (i.e. they dissociate 100% completely). Strong acids dissociation involves one arrow only in the chemical equation. Weak acids are weak electrolytes (i.e. they do not dissociate 100%). Weak acids dissociation involves two arrows (double arrows) in the chemical equation. Strong and weak bases: they follow the same definition as the strong and weak acids above. Example: Strong acids: (ONE arrow) HCl ------> H+ + Cl- HNO3 -----> H+ + NO3- H2SO4 -----> 2H+ + SO4^2- Weak acids: (DOUBLE arrow) HF <------> H+ + F- H3PO4 <----> 3H+ + PO4^3- CH3COOH <------> CH3COO- + H+ Strong bases: (ONE arrow) KOH -----> K+ + OH- Ca(OH)2 ------> Ca^2+ + 2OH- NaOH -------> Na+ + OH- Weak bases: (DOUBLE arrow) NH4OH <------> NH4+ + OH- Fe(OH)3 <-------> Fe^3+ + 3OH- Al(OH)3 <-------> Al^3+ + 3OH- O^2- is the strongest base, stronger that OH-. However, it does not exist in water. O^2- + H2O ----> 2OH- Weak acid/base dissociation constants (or ionization constants): CH3COOH(aq) <---> H+(aq) + CH3COO-(aq) Ka = [H+][CH3COO-]/[CH3COOH] Ka = acid dissociation constant = the ratio of the products to the reactants.

NH4OH(aq) <----> NH4^+(aq) + OH-(aq) Kb = [NH4^+][OH-]/[NH4OH] Question: Why do Ka and Kb Not appear in most textbooks and literature? Answer: Consider the following acid dissociation: HCl(aq) -------> H+(aq) + Cl-(aq) Ka = [H+][Cl-]/[HCl] Since HCl is strong acid, then [HCl] ~ zero after the dissociation. Ka = [H+][Cl-]/[zero] ~ infinite Ka of strong acids is very large and hence has no physical meaning! Calculation regarding Ka and Kb of weak acids and weak bases: Two cases: 1. First case: Molarity and Ka (or Kb) are given; pH to be calculated. Example: Calculate the pH of a 0.122M monoprotic weak acid whose Ka is 5.7X10^-4. To solve such problems, follow the steps below: A. Write the chemical equation; assuming that the weak acid has the general formula HY HY(aq) <-----> H+(aq) + Y-(aq) B. Identify the species contributing to Ka Ka = [H+][Y-]/[HY] C. Establish the following table: Initial Molarity: [HY] = 0.122M [H+] = 0M [Y-] = 0M ---------------------- Change in the Molarity: [HY] = - X [H+] = + X [Y-] = + X ---------------------- Molarity at equilibrium: [HY] = 0.122M - X [H+] = 0M + X = X [Y-] = 0M + X = X ----------------------

D. Substituting for Ka by the corresponding molarities at equilibrium, one obtains: Ka = [X][X]/[0.122 –X] = [X}^2 / [0.122 – X] 5.7X10^-4 = X^2 / 0.122 – X Assuming that 0.122 >>> X, then: 5.7X10^-4 = X^2 / 0.122 X^2 = 6.954X10^-5 X = [= 6.954X10^-5]^1/2 = 8.339X10^-3 and hence the pH = -log[X] = -log[8.339X10^-3 = 2.08 2. Second case: Molarity and the pH are given; Ka (or Kb) are to be calculated Example: The pH of 0.060 M weak monoprotic acid is 3.44. Calculate Ka of this weak acid. Follow the above procedure: A. Write the chemical equation; assuming that the weak acid has the general formula HY HY(aq) <-----> H+(aq) + Y-(aq) B. Identify the species contributing to Ka Ka = [H+][Y-]/[HY] C. Establish the following table: Initial Molarity: [HY] = 0.060M [H+] = 0M [Y-] = 0M ---------------------- Change in the Molarity: [HY] = - X [H+] = + X [Y-] = + X ---------------------- Molarity at equilibrium: [HY] = 0.060M - X [H+] = 0M + X = X [Y-] = 0M + X = X ---------------------- D. Substituting for Ka by the corresponding molarities at equilibrium, one obtains:

Ka = [X][X]/[0.060 –X] = [X}^2 / [0.060 – X] Assuming that 0.060 >>> X, then: Ka = X^2 / 0.060 X can be calculated as follows: pH = -log[X] = 3.44 === > [X] = 10^-pH = 10^-3.44 = 3.631X10^-4 Ka = [3.631X10^-4 ]^2 / 0.060 = 2.197X10^-6 ------------------------------------------------------------------ Strength of Acids: The strength of the acids depends on the following: 1. the properties of the solvent 2. temperature 3. molecular structure of the acid itself Comparing the strength of the acids in the same solvent at the same temperature will depends on its molecular structures: 1. In case of monoprotic acids:: two factors affect the strength of the acids: Consider the following acid HX ------ > H+ + X- A. The strength of the bond between H+ and X- The stronger the bond, the less dissociation, the weaker the acid. B. The less polar HX, the stronger the acid because the more polar the acid the tighter the bonding, the less dissociation, the weaker the acid. Polarity depends on the electronegativity of an atom with the structure of the acid . The more electronegative the atom, the more pulling of electrons of the adjacent atom, the harder to break the bond, the weak the acid. Therefore the acid strength will follow the following order: HI>HBr>HCl>HF 2. In case of Oxoacids: Oxoacids are acids that have oxygen, hydrogen and a central atom to which the name of the acid is named. Two cases: A. Oxoacids having different central atoms that are from the same periodic GROUP and that have the same oxidation number: Example : |O:: H – :O: - Cl: -:O:: Versus: Bromic acid: |O:: H __ : O : __ Br: -:O:::

Both chlorine and bromine atoms are central atoms and have the same oxidation number (+5). However, chlorine atom is more electronegative than bromine atom. The strength of the acidity increases with increasing electronegativity of the central atom. Hence, HClO3 is stronger acid than HBrO3. [Actually, hydrogen – oxygen bonding is the key of the strength. The stronger the acid the weaker the bond between hydrogen – oxygen. With the connection of oxygen to more electronegative atom, the bonding between hydrogen – oxygen is weakened, the stronger the dissociation of the bond and of the acid, the stronger the acid]. B. Oxoacids having the same central atom but different oxidation number: The higher the oxidation number of the central atom, the stronger the acid. Example: HClO4 (oxidation number +7) > HClO3(oxidation number +5) > HClO2(oxidation number +3) > HClO(oxidation number +1). Salts: Salt is the product of the reaction of acids with bases. There are four types of salts: 1. Salt that is produced from the reaction of a strong acid and a strong base: Example: HCl + NaOH ----- > NaCl + H2O NaCl is producing a neutral solution with a pH of 7. Why? Because both Na+ and Cl- are not hydrolyzed (i.e. do not react with H2O) Hydrolysis: is the reaction with salt with water to produce a weaker acid or a weaker base! 2.Salt that is produced from the reaction of a strong acid and a weak base: Example:: HCl + NH4OH ---> NH4Cl + H2O NH4Cl is producing acidic solution with the pH about 5. Why? Cl- is not hydrolyzed but NH4+ does react with H2O to produce NH4OH (a weak base) and H+: NH4+ + H2O --------> NH4OH + H+ (acidic because of H+) Calculate the pH of 0.24M ammonium chloride solution (NH4Cl). What is the percent hydrolysis? NH4Cl < ------ > NH4+ + Cl- NH4+ + H2O < ------> NH4OH + H+ [Ka = 5.6X10^-10] Initial Molarity: [NH4Cl] = 0.24M [H2O] = No change [NH4OH] = 0M [H+] = 0M

---------------------- Change in the Molarity: [NH4Cl] = - X [H2O] = No change [NH4OH] = + X [H+] = + X ---------------------- Molarity at equilibrium: [NH4Cl] = 0.24M – X [H2O] = No change [NH4OH] = 0M + X = X [H+] = 0M + X = X ---------------------- Ka = 5.6X10^-10 = [NH4OH][H+] / [NH4Cl] = [X]^2 / [0.24 – X] Assuming 0.24 >>> X, then: 5.6X10^-10 = [X]^2 / [0.24] ===> X = {[5.6X10^-10][0.24]}^1/2 = 1.159X10^-5 = [H+] The % hydrolysis = {[H+] / [0.24]}X100% = {[1.159X10^-5] / [0.24]}X100% = 0.004829% [H+} = 1.159X10^-5 pH = - log[H+] = - log[1.159X10^-5] = 4.94 3. Salt that is produced from the reaction of a weak acid and strong base. Example: CH3COOH + NaOH ---- > NaCH3COO + H2O NaCH3COO is producing a basic solution of a pH about 8~9. Why? Na+ is not hydrolyzed but CH3COO- does react with H2O to produce CH3COOH (a weak acid) and OH-: CH3COO- + H2O ------ > CH3COOH + OH- Example: Calculate the pH of 0.24M sodium formate solution (NaHCOO). What is the percent hydrolysis? HCOONa < ------ > Na+ + HCOO- HCOO- + H2O < ------> HCOOH + OH- [Kb = 5.6X10^-10] Initial Molarity: [HCOO-] = 0.24M [H2O] = No change [HCOOH] = 0M [OH-] = 0M ----------------------

Change in the Molarity: [HCOO-] = - X [H2O] = No change [HCOOH] = + X [OH-] = + X ---------------------- Molarity at equilibrium: [HCOO-] = 0.24M – X [H2O] = No change [HCOOH] = 0M + X = X [OH-] = 0M + X = X ---------------------- Kb = 5.6X10^-10 = [HCOOH][OH-] / [[HCOO-] = [X]^2 / [0.24 – X] Assuming 0.24 >>> X, then: 5.6X10^-10 = [X]^2 / [0.24] ===> X = {[5.6X10^-10][0.24]}^1/2 = 1.159X10^-5 = [OH-] The % hydrolysis = {[OH-] / [0.24]}X100% = {[1.159X10^-5] / [0.24]}X100% = 0.004829% [H+} = Kw/[OH-] = 1X10^-14 / 1.159X10^-5 = 8.625X10^-10 pH = - log[H+] = - log[8.625X10^-10] = 9.06 ------------------------------------------------------------------ Selected problems from the textbook: Chapter 15 Please work on these problems. If you have any question, contact your instructor! 15.5;15.6;15.8;15.31;15.32;15.41;15.43;15.44;15.52;15.72;15.77;15.78 ------------------------------------------------------------------ Selected problems and answers: Chapter 15 1.define the following: a. acid – conjugated base pair b. base – conjugated acid pair c. weak acid/base d. strong acid/base e. acidic solution f. basic solution

g. neutral solution h. pH i. Ka j. Kb k. Kw Check Notes + textbook! 2. Why are Ka and Kb values of strong acids and bases rarely recorded in literature? Because their value are very large and it is assumed that no equilibrium does exist. 3. Identify the acid – base conjugated pairs in each of the following:: a. HClO3 + H2O < ------- > H3O+ + ClO3- HClO3 = acid H2O = base (because it gives H3O+ which is conjugated acid by definition) H3O+ = conjugated acid ClO3- = conjugated base b. HSeO4^- + NH3 < -------> NH4+ + SeO4^2- NH3 = base HSeO4^- = acid (although it is a negative charged ion, it does have H+ and it donates it) NH4+ = conjugated acid of the bae NH3 SeO4^-2 = conjugated base of the acid HSeO4^- c. HCO3^- + OH- < ----- > CO3^2- + H2O This example is rather difficult to solve. The best way to go around this is by applying the standard system: H2O. H2O = acid (because it gives OH- which is a conjugated base by definition) OH- = conjugated base by definition. HCO3^- and CO3^2-: which one of these is a base and which is a conjugated acid? HCO3^- has hydrogen (although is negatively charged). By definition a conjugated acid is a base that gained H+, hence HCO3^- is conjugated acid and CO3^2- is a base. d. C5H5NH^+ + H2O < ------ > C5H5N + H3O^+ H2O = base because it gives H3O^+ which is conjugated acid by definition C5H5NH^+ and C5H5N: which one of these is a base and which is a conjugated acid? C5H5NH^+ has hydrogen and positively charged. By definition a conjugated acid is a base that gained H+, hence C5H5NH^+ is conjugated acid and

C5H5N is a base. 4. Calculate the concentration of H+ ions in NaOH solution whose hydroxide ion is 1.3 M. NaOH is strong base. [H+] = Kw/[OH-] = 1X10^-14 / 1.3 = 7.6X10^-15M 5. Calculate the pH and [H+] of 0.0175 M KOH. KOH is strong base. [H+] = Kw/[OH-] = 1X10^-14 / 0.0175 = 5.714X10^-13M pH = - log [5.714X10^-13] = 12.24 6. Calculate the pH and [OH-] of 0.0155M HNO3. HNO3 is strong acid. [H+] = 0.0155 pH = - log[H+] = - log[0.0155] = 1.81 [OH-] = Kw/[H+] = 1X10^-14 / 0.0155 = 6.45X10^-13M 7. Determine if the acids/bases are strong or weak by inspecting the chemical equations and Ka and Kb values. Rank them according to increasing strength: Acids: a. HI ------ > H+ + I- (Ka = 1X10^+9) strong acid # 1 b. H2SO4 ------- > 2H+ + SO4^2- (Ka = 1X10^+3) strong acid # 2 c. HNO2 < ------- > H+ + NO2^- (Ka = 7.2X10^-4) weak acid # 3 Bases: a. CO3^2- + H+ < ------- > HCO3^- (Kb = 2.1X10^-4) weak base # 2 b. CsOH --------- > Cs^+ + OH^- (Kb is very large approx. 10^+9) strong base # 1 c. NH3 + H+ < -------- > NH4+ (Kb = 1.8X10^-4) weak base # 3 8. Calculate the pH of 0.255 M of monoprotic acid whose Ka is 1.8X10^-5 Solution: You might follow the procedure explained in the lecture notes or you can follow the short cut as follows: Ka = [X]^2 / [0.255 – X]; where [X] = [H+] Assuming that 0.255>>> X 1.8X10^-5 = [X]^2 / 0.255 [X] = {[1.8X10^-5][0.255]}^1/2 = 2.14X10^-3M = [H+] pH = - log[X] = - log[H+] = - log[2.14X10^-3] = 2.67 9. The pH of a 0.055 M of monoprotic weak acid is 4.02. Calculate the Ka of the acid.

Shortcut solution: Ka = [X]^2 / [0.055 – X]; where [X] = [H+] Assuming that 0.055>>>X Ka = [X]^2 / 0.055 Since pH = 4.02, [H+] = 10^-pH = 10^-4.02 = 9.55X10^-5 = [X] Ka = [9.55X10^-5]^2 / 0.055 = 1.66X10^-7 10. Explain the acidity trend in the following sets of acids: a. HI>HBr>HCl>HF HI is the least polar and hence weak bonding and higher dissociation. HF is the highest polarity and hence strong bonding and lower dissociation. b. HClO4>HClO3>HClO2>HclO HClO4 has higher oxidation number (+7) HClO has lower oxidation number (+1) 11. Calculate the pH of a 0.255 M sodium propionate (C2H5COONa). Its Kb is 5.6X10^-10. What is the percent hydrolysis? Shortcut solution: C2H5COO- + H2O < ------- > C2H5COOH + OH- Kb = [X]^2 / [0.255 – X] Assuming that 0.255 >>> X Kb = 5.6X10^-10 = [X]^2 / [0.255]; [X] = [OH-] [X] = {[5.6X10^-10][0.255]}^1/2 = 1.195X10^-5M = [OH-] % hydrolysis = {[OH-] / [0.255]}X100% = {[1.195X10^-5] / [0.255]}X100% = 4.69X10^-3% [H+] = Kw/[OH-] = 1X10-14/1.195X10^-5 = 8.368X10^-10 pH = - log [H+] = - log [8.368X10^-10] = 9.077 Chapter 16: Solubility. Common – Ion Effect: Consider the following equilibrium: CH3COOH < ----- > H+ + CH3COO- Adding NaCH3COO solution to the equilibrium, will lead to a shift in the equilibrium to the left. Some incoming CH3COO- will react with H+ to produce acetic acid CH3COOH. This effect is called the common ion effect. One can use the ―Henderson – Hasselbalch‖ Equation to calculate the pH of such weak acid solution if adding the common ion effect:

pH = pKa + log [Conjugated base] / [acid] Example: What is the pH of a solution containing 0.3M HCOOH and 0.52M HCOOK. Ka = 1.7X10^-4 [HCOOH] = 0.3M, [HCOO-] = 0.52M pH = - log (1.7X10^-4) + log [0.52] / [0.3] = 3.769 + 0.2389 = 4.0079 ~ 4.01 One solve this problem without using the so-called ―Henderson – Hasselbalch‖ Equation as follows: Ka = [H+][HCOO-] / [HCOOH] 1.7X10^-4 = [H+][.52] / [0.3] [H+] = {1.7X10^-4 [0.3]} / [5.2] = 9.808X10^-5 pH – log (9.808X10^-5) = 4.0084 ~ 4.01 Buffer solution: A solution that always resists any change in the pH. It does so by neutralizing any incoming acids or bases. A buffer solution is made of: a. weak acid + its salt (conjugated base) or: b. weak base + its salt (conjugated acid). Example: CH3COOH / NaCH3COO CH3COOH = weak acid NaCH3COO ----- > Na+ + CH3COO-; the CH3COO- is the conjugated base of the acid CH3COOH Example: Which of the following solution are buffer solutions. Explain you answer. a. KH2PO4/H3PO4 ====> buffer solution (weak acid H3PO4 + conjugated base H2PO4-) b. HCl/NaCl ===== > Not a buffer (strong acid HCl) c. NaClO4/HclO4 ==== > Not a buffer (strong acid HClO4) d. C5H5N/C5H5NHCl ==== > buffer solution (weak base C5H5N + conjugated acid C5H5NH+) e. KOH/NaOH ====== > Not a buffer (strong base KOH) f. NH3/NH4Cl ===== > buffer solution (weak base NH3 + conjugated acid NH4+) The calculation of the pH of the Buffer solutions:

Calculate the pH of a buffer system containing 1.0M CH3COOH and 1.0M CH3COONa. What is the pH of buffer solution when enough HCl is added so that its molarity is 0.1M? What is the pH of buffer solution when enough NaOH is added so that its molarity is 0.1M? Ka = 1.8X10^-5 Answer: Use the so-called ―Henderson – Hasselbalch‖ Equation: pH = pka + log [conjugated base] / [acid] pH = -log(1.8X10-5) + log [0.1] / [0.1] = 4.745 + log 1 = 4.74 + 0 = 4.745 ~ 4.75 When 0.1 M HCl is added: HCl ----- > H+ + Cl-; the H+ will add itself to the already present H+ of the acetic acid CH3COOH according to: CH3COOH < ----------- > H+ + CH3COO- It means that the incoming H+ (from HCl) will react with some of the CH3COO- to produce some CH3COOH and hence the equilibrium will be shifted to the left (common – ion effect). Therefore, [H+] concentration will go up pH will go down (solution becomes more acidic) [CH3COO-] concentration will go down (some of it will be consumed by the incoming H+ of the HCl acid) [CH3COOH] concentration will go up (some extra CH3COOH will be produced due to the reaction of the incoming H+ and CH3COO-) [CH3COOH] will increase by 0.1 M ==== > 1.0 + 0.1 = 1.1M [CH3COO-] will decrease by 0.1M === > 1.0 – 0.1 = 0.9M pH = pka + log [conjugated base] / [acid] pH = -log(1.8X10-5) + log [0.9] / [1.1] = 4.745 + log (0.818) = 4.745 – 0.0872 = 4.658 ~ 4.66 When 0.1 M NaOH is added: NaOH ----- > Na+ + OH-; the OH- will scavenge the already present H+ of the acetic acid CH3COOH according to: CH3COOH < ----------- > H+ + CH3COO- . OH- + H+ ------- > H2O It means that the incoming OH- (from NaOH) will react with some of the H+ to produce some water and hence the equilibrium will be shifted to the right. Shift to the right means that some [CH3COOH] has to dissociate to produce and to compensate for the loss of H+ (scavenged by the incoming OH-) Therefore, [H+] concentration will go down pH will go up (solution becomes more basic)

[CH3COO-] concentration will go up (some of it will be produced by extra dissociation of CH3COOH) [CH3COOH] concentration will go down (some extra CH3COOH will be dissociated to produce and to compensate for the loss of H+ (scavenged by the incoming OH-) [CH3COOH] will decrease by 0.1 M ==== > 1.0 - 0.1 = 0.9M [CH3COO-] will increase by 0.1M === > 1.0 + 0.1 = 1.1M pH = pka + log [conjugated base] / [acid] pH = -log(1.8X10-5) + log [1.1] / [0.9] = 4.745 + log (1.222) = 4.745 + 0.0871 = 4.8321 ~ 4.83 Titration Curve: It is the plot of the pH against the volume of the titrant (the compound acid or base that put in the buret). One can obtained two important information from the titration curve: a. The pH at the end point (or called the equivalence point). b. The amount of the titrant needed to reach the end point. End point (or called the equivalence point): is the point in the titration curve by which all H+ or OH- are consumed totally. At which the complete neutralization is achieved. Hence, at the end point, number of moles of H+ = number of moles of OH- [or number of equivalent moles of H+ = number of equivalent moles of OH-] Types of titration curve: There are four types: 1. Titration curve is formed from the neutralization of a strong acid with a strong base: Example: HCl + NaOH ------- > NaCl + H2O The pH at the end point will depend on the hydrolysis of the salt produced during the neutralization reaction. Since NaCl is Not hydrolyzed (Na+ and Cl- do not produce a weak acid or a weak base). Hence, the pH at the end point is 7 (neutral solution, no hydrolysis). The pH at the end point is the pH of pure water: H+ + OH- ---- > H2O 2. Titration curve is formed from the neutralization of a strong acid with a weak base: Example: HCl + NH4OH ------- > NH4Cl + H2O The pH at the end point will depend on the hydrolysis of the salt produced during the neutralization reaction. Since NH4Cl is hydrolyzed (NH4+ is hydrolyzed to produce H+ but Cl- is not) Hence, the pH at the end point is acidic: NH4+ + H2O ------ > NH4OH + H+.

The H+ produced by the hydrolysis at the end point is responsible for the acidic pH. 3. Titration curve is formed from the neutralization of a weak acid with a strong base: Example: CH3COOH + NaOH ------- > CH3COONa + H2O The pH at the end point will depend on the hydrolysis of the salt produced during the neutralization reaction. Since CH3COONa is hydrolyzed (CH3COO- is hydrolyzed to produce OH- but Na+ is not) Hence, the pH at the end point is basic: CH3COO- + H2O ------ > CH3COOH + OH-. The OH- produced by the hydrolysis at the end point is responsible for the basic pH. 4. Titration curve is formed from the neutralization of a weak acid with a weak base: Example: CH3COOH + NH4OH ------- > CH3COONH4 + H2O The pH at the end point will depend on the hydrolysis of the salt produced during the neutralization reaction. Since CH3COONH4 is hydrolyzed (CH3COO- is hydrolyzed to produce OH- but NH4+ is hydrolyzed to produced H+): CH3COO- + H2O ------ > CH3COOH + OH- (Kb) NH4+ + H2O ----- > NH4OH + H+ (Ka) Which one (OH- or H+) will determine the outcome of the pH at the end point? The pH value will depend on the value of the acid dissociation constant Ka and on the base dissociation constant Kb. The larger value among Ka and Kb will determine the value of the pH at the end point: Three cases: a. Ka > Kb ===== > pH is acidic b. Ka = Kb ===== > pH is neutral c. Ka < Kb ===== > pH is basic In the example above: CH3COO- + H2O ------ > CH3COOH + OH- (Kb = 1.8X10^-5) NH4+ + H2O ----- > NH4OH + H+ (Ka = 1.8X10^-5) Ka = Kb = 1.8X10^-5. Therefore the pH at the end point is neutral. Predicting the pH of the titration curve theoretically! (by calculation and without conducting any experiment). ------------------------------------------------------------------ Consider the third type: weak acid is neutralized with strong base.

[Note that this type of calculation can be used to all four types of the titration curve without any exception!] Calculate the pH in the titration of 25.0 mL of 0.2 M acetic acid by 0.2 M sodium hydroxide when the following conditions are carried out: a. No NaOH is added. There is only 25.0 mL of CH3COOH in the Erlenmeyer titration flask. b. After addition of 10.0 mL of NaOH c. After addition of 25.0 mL of NaOH d. After addition of 35 mL of NaOH The following supporting data is given: CH3COOH(aq) + NaOH(aq) -------- > CH3COO-(aq) + Na+(aq) + H2O CH3COOH(aq) < -------- > H+(aq) + CH3COO-(aq) Ka = 1.8X10^-5 [Acid dissociation, at the beginning, when No NaOH is added] CH3COO-(aq) + H2O(l) < -------- > CH3COOH(aq) + OH-(aq) Kb = 5.6X10^-10 [ [CH3COO- hydrolysis at the end point. a. When No NaOH is added: CH3COOH dissociates according to: CH3COOH(aq) < -------- > H+(aq) + CH3COO-(aq) Ka = 1.8X10^-5 Ka = [H+]^2 / [CH3COOH] Note here [H+] = [CH3COO-] 1.8X10^-5 = [H+]^2 / [0.2] [H+] = {(1.8X10^-5)[0.2]}^1/2 = 0.001897 pH = - log (0.001897) = 2.722 ~ 2.72 b. After addition of 10.0 mL of NaOH First one has to calculate number of moles of NaOH added to the acetic acid solution CH3COOH. Second one has to calculate the number of moles of CH3COOH (compound being titrated) using the chemical equation of the titration. Third the pH will be calculate use the Ka equation of the acid or using the so – called ―Henderson – Hasselbalch‖ equation: 1. Number of moles of NaOH = (10.0 mL NaOH)(0.2 mol NaOH/1 Liter NaOH) (1 Liter NaOH/1000 mL NaOH) = 0.002 moles NaOH 2. Number of moles of CH3COOH originally present = (25.0 mL CH3COOH)(0.2 molCH3COOH/ 1 Liter CH3COOH)(1 liter CH3COOH/1000 mL CH3COOH) = 0.005 moles CH3COOH 3.Number of CH3COOH left over after 0.002 moles NaOH is added = 0.005 –

0.002 = 0.003 moles CH3COOH left over after neutralization all 0.002 moles of NaOH. Actual [CH3COOH] concentration is 0.003 mole CH3COOH / total volume in liters Total volume of solution mixture (CH3COOH + NaOH) = (25 + 10) / 1000 = 0.035 Liter Thus [CH3COOH] = 0.003 moles / 0.035 L = 0.0857M 4. Number of moles CH3COONa formed is obtained the chemical equation: CH3COOH(0.002 moles) + NaOH (0.002 moles) ------ > CH3COONa (0.002 moles)+ H2O 1:1 mole ratio [CH3COO-] concentration = (0.002 moles) / (0.035 L) = 0.0571 M At this stage, a buffer system is obtained formed by [CH3COOH] / [CH3COO-] pair combination. Using the so – called ―Henderson – Hasselbalch‖ equation: pH = - log(1.8X10^-5) + log[CH3COO-]/[CH3COOH] = - log(1.8X10^-5) + log[0.0571]/[0.0857] pH = 4.745 – 0.1764 = 4.5686 ~ 4.57 c. After addition of 25.0 mL of NaOH This amount of NaOH is corresponding to the equivalence point (because 25 mLof 0.2M NaOH is neutralized by 25 mL of 0.2M CH3COOH and the mole ratio is 1:1). Number of moles of CH3COOH originally present = (25.0 mL CH3COOH)(0.2 molCH3COOH/ 1 Liter CH3COOH)(1 liter CH3COOH/1000 mL CH3COOH) = 0.005 moles CH3COOH CH3COOH(aq)(0.005 moles) + NaOH(aq)(0.005 moles) -------- > CH3COO-(aq)(0.005 moles) + Na+(0.005 moles)(aq) + H2O [CH3COO-] = [Na+] = [0.005] / total volume in liters Total volume of solution mixture (CH3COOH + NaOH) = (25 + 25) / 1000 = 0.05 Liter Thus [CH3COO-] = [Na+] = [0.005 moles] / 0.05 L = 0.1 M At the end point or equivalence point, hydrolysis will take place as follows: CH3COO-(aq) + H2O(l) < ------ > CH3COOH(aq) + OH-(aq) (with Kb = 5.6X10^-10) Using the shortcut solution: Kb = [CH3COOH][OH-] / [CH3COO-]; where [CH3COOH] = [OH-] = [X] 5.6X10^-10 = [X][X] / [0.1]

[X] = [OH-] = {(5.6X10^-10)(0.1)}^1/2 = 7.483X10^-6 [H+] = Kw / [OH-] = (1X10^-14) / (7.483X10^-3) = 1.336X10^-9 pH = - log(1.336X10^-9) = 8.874 ~ 8.87 d. After addition of 35.0 mL of NaOH After the addition of 35.0 mL of NaOH, the solution will be basic (beyond the equivalence point). The pH of solution will be the pH of the strong base NaOH and therefore CH3COO- can be safely ignored. Number of moles of NaOH left over after the complete neutralization (after the end point) = (35 mL – 25 mL)(0.2 moles/L)(1L/1000 mL) = 0.002 moles [NaOH] = (moles NaOH) / total volume Total volume = 35mL NaOH + 25 mL CH3COOH = 60 mL 0.06 Liters Thus [NaOH] = 0.002 moles / 0.06 Liters = 0.0333 M [H+] = Kw / [OH-] = (1X10^-14) / (0.0333) = 3.003X10^-13 pH = - log(3.003X10^-13) = 12.522 ~ 12.52 ------------------------------------------------------------------ Solubility equilibrium: Solubility product: Ksp AgCl(s) < --------- > Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] Note that [AgCl(s)] is ignored because it is constant and does not change! Note that [Ag+] = [Cl-] Hence Ksp = [Ag+]^2 = [Cl-]^2 Example: Ag2Cr2O7(s) < -------- > 2Ag+(aq) + Cr2O7^2-(aq) Ksp = [2 Ag+]^2[Cr2O7^2-] Table 16.2 exhibits a list of most Ksp. Ksp always has a unit of the molarity (mol/L) Example: Find Ksp, if the solubility of lead chromate PbCrO4 is 4.5 X 10^-5 g/L. Solution: Ksp has the unit of the molarity. Therefore if the solubility is given in gram/

Liter (g/L), one should convert the solubility from g/L to mole/Liter (mol/L). The solubility given in mol/L is called the molar solubility. The conversion from g/L to mol/L needs the usage of the molar mass of the compound in consideration. PbCrO4(s) < ---- > Pb^2+(aq) + CrO4^2-(aq) [Pb^2+] = [CrO4^2-] = 4.5 X 10^-5 g/L. Molar mass of PbCrO4 = 207 + 52 + (4X16) = 323 g/mol [Pb^2+] = [CrO4^2-] = (4.5 X 10^-5 g/L) / (323 g/mol) =1.393X10^-7 mol/L = molar solubility. Ksp = [Pb^2+][CrO4^2-] = Pb^2+]^2 = [CrO4^2-]^2 = [1.393X10^-7]^2 =1.941X10^-14 (comparable with literature value of 2.0X10^-14). Example: Calculate the solubility of Ag2CO3 in g/L if Ksp (solubility) is 8.1X10^-12 Ag2CO3(s) < ------- > 2Ag+(aq) + CO3^2-(aq) Ksp = 8.1X10^-12 Ksp = [Ag+]^2[CO3^2-] Establish a table as follows: Initial Molarity: [Ag2CO3] = No change, solid [Ag+] = 0M [CO3^2-] = 0M ---------------------- Change in the Molarity: [Ag2CO3] = No change, solid [Ag+] = +2 X [CO3^2-] = + X ---------------------- Molarity at equilibrium: [Ag2CO3] = No change, solid [Ag+] = 0M + 2X = 2X [CO3^2-] = 0M + X = X Ksp = [Ag+]^2[CO3^2-] = [2X]^2[X] = (4X^2)(X) = 4X^3 8.1X10^-12 = 4X^3 [X] = {(8.1X10^-12)/(4)}^1/3 = 1.277X10^-4M [X] = 1.277X10^-4 mol/L =====> needed to be converted to g/L Molar mass of Ag2CO3 = 2(107.9) + 12 + (3X16) = 275.8 g/mol

[X] = (1.277X10^-4 mol/L)(275.8 g/mol) = 0.03522 g/L = 3.522X10^-2 g/L ------------------------------------------------------------------ Selected problems from the textbook: Chapter 16 Please work on these problems. If you have any question, contact your instructor! 16.9;16.12;16.14;16.28;16.43;16.44;16.49;16.56;16.62 ------------------------------------------------------------------ Selected problems and answers: Chapter 16 1. Define the following: a. molar solubility b. common – ion effect c. buffer solution d. titration curve e. ion product 2. Calculate the pH of a buffer solution containing 0.55 M CH3COOH and 0.85M CH3COONa. What would be the pH of the buffer solution if no salt (CH3COONa) is added? CH3COOH(0.55M) < -------- > H+(aq) + CH3COO-(0.85M)(aq) Ka = 1.8X10^-5 Using the Henderson – Hasselbalch equation: pH = pka + log [conjugated base] / [acid] pH = -log(1.8X10^-5) + log [0.85]/[0.55] = 4.745 + 0.1890 = 4.934 Without addition of the salt CH3COONa: Ka = [X]^2 / [CH3COOH] where [X] = [H+] = [CH3COO-} 1.8X10^-5 = [X]^2 / [0.55M] [X]^2 = (1.8X10^-5)[0.55] = 9.9X10^-6 [X} = (9.9X10^-6)^1/2 = 0.00315 pH = - log (0.00315) = 2.502 ~ 2.50 3. Which of the following compound pair does form a buffer solution? Explain! a. H3PO4 / H2PO4^-: buffer solution; weak acid H3PO4 + its salt (conjugated

base) H2PO4^- b. HNO3 / NaNO3: No buffer; HNO3 is strong acid c. Ca(OH)2 / LiOH: No buffer; both Ca(OH)2 and LiOH are strong bases d. HCOOH / HCOOK: buffer solution; weak acid HCOOH + its salt(conjugated base) HCOO- e. NH3 / NH4Cl: buffer solution; weak base NH3 + its salt(conjugated acid) NH4^- f. H2PO4^- / HPO4^2-: buffer solution; weak acid H2PO4^- + its salt (conjugated base) HPO4^2- 4. Calculate the pH of a buffer solution made of 2.5 M HCOOH (formic acid) and 3.5 M HCOOK (potassium formate).Further, determine the following: a. the pH of this buffer, when enough HCl is added so that its concentration is 1.0 M b. the pH of this buffer, when enough NaOH is added so that its concentration is 1.0 M Pure buffer: HCOOH(aq) < ------- > H+(aq) + HCOO-(aq) Ka = 1.7X10^-4 pH = pka + log [conjugated base] / [acid] pH = -log(1.7X10^-4) + log [3.5M] / [2.5M] = 3.769 + 0.146 = 3.915 ~ 3.92 a. When 1.0 M HCl is added to buffer: pH = pka + log [conjugated base – 1.0M] / [acid + 1.0M] pH = -log(1.7X10^-4) + log [2.5M] / [3.5M] = 3.769 - 0.146 = 3.623 ~ 3.62 b. When 1.0 M NaOH is added to buffer: pH = pka + log [conjugated base + 1.0M] / [acid - 1.0M] pH = -log(1.7X10^-4) + log [4.5M] / [1.5M] = 3.769 + 0.477 = 4.246 ~ 4.25 5. Determine the pH equivalence point for the following systems: a. KOH + HNO3 ===== > No hydrolysis; strong acid/strong base; pH at end point is 7 b. HCOOH + NaOH ==== > HCOONa is hydrolyzed; pH at the end point is basic c. NH3 + HClO4 ==== > NH4ClO4 is hydrolyzed; pH at the end point is acidic d. HCOOH + NH3 ==== > NH4(HCOO) is hydrolyzed; pH at the end point is acidic. Why? Both NH4+ and HCOO- are hydrolyzed: NH4+ + H2O < ------ > NH4OH + H+ [Kb = 1.8X10^-5] HCOO- + H2O < ------ > HCOOH + OH- [Ka = 1.7X10^-4]

Since Ka > Kb ======= > solution is acidic 6. Find Ksp of Bi(OH)3 if the solubility is 5.6X10^-7 g/L Bi(OH)3(s) < ------- > Bi^3+(aq) + 3OH-(aq) Molar mass Bi(OH)3 = Bi +3O + 3H = 209 +3(16) + 3(1) = 260 g/mol [Bi^2+] = [OH-] = [X] Ksp = [X] [3X]^3 = 27 [X]^4 Molar solubility in mol /L = (5.6X10^-7 g/L) / (260 g/mol) = 2.1538X10^-9 mol/L Ksp = 27[2.1538X10^-9 mol/L]^4 = 5.810X10^-34 Note: A table can be use to figure the values X and 3X as it is done in the next question. 7. Calculate the solubility in g/L of Ba3N2. Assume Ksp is 3.0 X10^-30 Ba3N2(s) < ------- > 3Ba^2+(aq) + 2N^3-(aq) Ksp = [Ba^2+]^3[N^3-]^2 Establish a table as follows: Initial Molarity: [Ba3N2] = No change, solid [Ba^2+] = 0M [N^3-] = 0M ---------------------- Change in the Molarity: [Ba3N2] = No change, solid [Ba^2+] = +3X [N^3-] = + 2X ---------------------- Molarity at equilibrium: [Ba3N2] = No change, solid [Ba^2+] = 0M + 3X = 3X [N^3-] = 0M + 2X = 2X Ksp = [Ba^2+]^3[N^3-]^2 = [3X]^3[2X]^2 = 108[X]^5 3.0X10^-30 =108[X]^5 [X] = {(3.0X10^-30) / (108)}^1/5 = 4.8836X10^-7M [mol/L] Molar mass of Ba3N2 = 3Ba + 2N = 3(137.34) + 2(14) = 440.02 g/mol [X] in g/L = (4.8836X10^-7 mol/L)(440.02 g/mol) = 2.1489 g/L

8. Calculate the pH in the titration of 50.0 mL of 0.1 M formic acid by 0.5 M sodium hydroxide when the following conditions are carried out: a. No NaOH is added. There is only 50.0 mL of HCOOH in the Erlenmeyer titration flask. b. After addition of 10.0 mL of NaOH c. After addition of 35.0 mL of NaOH The following supporting data is given: HCOOH(aq) + NaOH(aq) -------- > HCOO-(aq) + Na+(aq) + H2O HCOOH(aq) < -------- > H+(aq) + HCOO-(aq) Ka = 1.7X10^-4 [Acid dissociation, at the beginning, when No NaOH is added] HCOO-(aq) + H2O(l) < -------- > HCOOH(aq) + OH-(aq) Kb = 5.88X10^-11 [HCOO- hydrolysis at the end point]. 1. When No NaOH is added: HCOOH dissociates according to: HCOOH(aq) < -------- > H+(aq) + HCOO-(aq) Ka = 1.7X10^-4 Ka = [H+]^2 / [HCOOH] Note here [H+] = [HCOO-] 1.7X10^-4 = [H+]^2 / [0.1] [H+] = {(1.7X10^-4)[0.5]}^1/2 = 0.00412 M pH = - log (0.00412) = 2.385 ~ 2.39 2. After addition of 10.0 mL of NaOH First one has to calculate number of moles of NaOH added to the acetic acid solution HCOOH. Second one has to calculate the number of moles of HCOOH (compound being titrated) using the chemical equation of the titration. Third the pH will be calculate use the Ka equation of the acid or using the so – called ―Henderson – Hasselbalch‖ equation: a. Number of moles of NaOH = (10.0 mL NaOH)(0.5 mol NaOH/1 Liter NaOH) (1 Liter NaOH/1000 mL NaOH) = 0.005 moles NaOH b. Number of moles of HCOOH originally present = (50.0 mL HCOOH)(0.1 mol HCOOH/ 1 Liter HCOOH)(1 liter HCOOH/1000 mL HCOOH) = 0.005 moles HCOOH c. Number of HCOOH left over after 0.005 moles NaOH is added = 0.005 – 0.005 = 0.00 moles HCOOH Nothing is left over of the formic acid HCOOH after adding 10.0 mL of NaOH.

This amount of NaOH is corresponding to the equivalence point (because 10 mL o f 0.5M NaOH is neutralizing the 50 mL of 0.1M HCOOH and the mole ratio is 1:1). Number of moles of HCOOH originally present = (50.0 mL HCOOH)(0.1 mol HCOOH/ 1 Liter HCOOH)(1 liter HCOOH/1000 mL HCOOH) = 0.005 moles HCOOH HCOOH(aq)(0.005 moles) + NaOH(aq)(0.005 moles) -------- > HCOO-(aq)(0.005 moles) + Na+(0.005 moles)(aq) + H2O [HCOO-] = [Na+] = [0.005] / total volume in liters Total volume of solution mixture (HCOOH + NaOH) = (50 + 10) / 1000 = 0.06 Liter Thus [HCOO-] = [Na+] = [0.005 moles] / 0.06 L = 0.0833M At the end point or equivalence point, hydrolysis will take place as follows: HCOO-(aq) + H2O(l) < ------ > HCOOH(aq) + OH-(aq) (with Kb = 5.88X10^-11) Using the shortcut solution: Kb = [HCOOH][OH-] / [HCOO-]; where [HCOOH] = [OH-] = [X] 5.88X10^-11 = [X][X] / [0.0833] [X] = [OH-] = {(5.88X10^-11)(0.0833)}^1/2 = 2.213X10^-6 [H+] = Kw / [OH-] = (1X10^-14) / (2.213X10^-6) = 4.519X10^-9 pH = - log(4.519X10^-9) = 8.345 ~ 8.35 3. After addition of 35.0 mL of NaOH After the addition of 35.0 mL of NaOH, the solution will be basic (beyond the equivalence point). The pH of solution will be the pH of the strong base NaOH and therefore HCOO- hydrolysis can be safely ignored. Out of 35 mL NaOH is added, 10 mL only used to neutralize HCOOH. The left over volume of NaOH (35mL – 10 mL = 25 mL) will determine the pH of the solution which is strong basic. Number of moles of NaOH left over after the complete neutralization (after the end point) = (35 mL – 10 mL)(0.5moles/L)(1L/1000 mL) = 0.0125 moles [NaOH] = (moles NaOH) / total volume Total volume = 35mL NaOH + 50 mL HCOOH = 85 mL = 0.085 Liters Thus [NaOH] = 0.0125 moles / 0.085 Liters = 0.1471 M [H+] = Kw / [OH-] = (1X10^-14) / (0.1471M) = 6.798X10^-14 pH = - log(6.798X10^-14) = 13.168 ~ 13.17

_______________________________________________ Chapter 18: Thermodynamics. First law of thermodynamics: It state that ― energy can be converted from one form to another form, but it can not be created or destroyed.‖ Formula of the first law of thermodynamics [Delta]E = q + w Where: [Delta] = the difference between final and initial conditions. E = energy Hence [Delta]E = final energy value – initial energy value q = heat w = work q = + (positive): heat absorbed by the system from the surroundings [system is the chemical reaction; the surroundings is any thing out the side the system] q = - (negative): heat is given to the surroundings by the system w = + (positive) work done on the system by the surroundings w = - (negative) work done by the system on the surroundings Spontaneous process and entropy: Spontaneous process is that is carried out without external help. The spontaneous process is carried out under set of controlled conditions such as temperature, pressure etc. Examples: Exothermic reactions are spontaneous Acid – base neutralization reactions are spontaneous Spontaneous processes tend always to decrease the energy of the system. Entropy: is the measure of randomness, disorder, disorganization, mayhem and chaos etc. of a system. Entropy has a symbol ―S‖ The greater the disorder, the greater the entropy Solids are more organized, therefore S will have the smallest value

Liquids are moderately organized, therefore S will have a moderate value Gases are highly disorganized, therefore S will have the highest value [Delta]S(solids) <[Delta]S(liquids)< [Delta]S(gases) (solids) ===== > (liquid): [Delta]S>0 (solids) ===== > (gases): [Delta]S>0 (liquids) ==== > (gases): [Delta]S>0 (liquids) ==== > (solids): [Delta]S<0 (gases) ==== > (solids): [Delta]S<0 (gases) ==== > (liquids): [Delta]S<0 (solids) ===== > (solids): [Delta]S has to be calculated using special tables (liquids) ==== > (liquids): [Delta]S has to be calculated using special tables (gases) ==== > (gases): [Delta]S has to be calculated using special tables Examples: How does the entropy of a system change for each of the following processes? a. condensing water vapor: [Delta]S <0 (gas ---- > liquid) b. forming sucrose crystals from supersaturated solution: [Delta]S<0 (liquid ---- > solid) c. heating hydrogen gas from 60 degrees to 80 degrees: [Delta]S has to be calculated! d. Subliming dry ice: [Delta]S>0 ( solid ---- > gas) [Delta] S = S(final ) – S(initial) In a chemical reaction: aA +bB ---------- > cC + dD [Delta]S(rxn)o = [cS (C )o + dS (D)o] – [aS (A)o + bS (B)o] Where: [cS (C )o + dS (D)o] = [the sum of the entropy of the all products]X[stoichiometric (coefficients) factors] [aS (A)o + bS (B)o] = [the sum of the entropy of the all reactants]X[stoichiometric (coefficients) factors] The sign o = standard conditions at 25 degrees and 1 atm Example: Find [Delta]S(rxn)o for the following reaction: 2NaHCO3(s) ----> Na2CO3(s) + H2O(l) + CO2(g) Using the appendix 3 at the back of the textbook, one obtains:

S(rxn)o for NaHCO3(s): 102.09 J/mol.K [at 1 atm and 25 degrees] S(rxn)o for Na2CO3(s): 135.98J/mol.K [at 1 atm and 25 degrees] S(rxn)o for H2O(l): 69.9 J/mol.K [at 1 atm and 25 degrees] S(rxn)o for CO2(g): 213.6J/mol.K [at 1 atm and 25 degrees] [Delta]S(rxn)o = [135.98 + 69.9 + 213.6] – [2(102.09] = + 215.3 J/mol.K Where the unit is Joule/mol.Kelvin = J/mol.K The second law of thermodynamics: The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. [Delta]S(universe) >0 ===== > spontaneous process [Delta]S(universe) =0 ===== > equilibrium (no change) [Delta]S(universe) <0 ===== > none spontaneous process [Delta]S(universe) = [Delta]S(system) + [Delta]S(surroundings) >0 ===== > spontaneous process Where: a. [Delta]S(system) = [Delta]S(rxn) = {[the sum of the entropy of the all products]X[stoichiometric (coefficients) factors]} – {[the sum of the entropy of the all reactants]X[stoichiometric (coefficients) factors]} b. [Delta]S(surroundings) = - {[Delta]H} / T where T = temperature in Kelvin H = Enthalpy = amount of heat measured at 1 atm and specified temperature (usually 25 degrees in most literature) c. [Delta]H = H(products) – H(reactants) = {[the sum of the enthalpy of the all products]X[stoichiometric (coefficients) factors]} – {[the sum of the enthalpy of the all reactants]X[stoichiometric (coefficients) factors]} Entropy changes in the surroundings: System is giving off heat ==== > entropy of the surroundings increases System is taking in heat === > entropy of the surroundings decreases Example: Consider the following reaction: N2(g) + 3H2(g) ----- > 2NH3(g) [Delta]H(system) = -92.6 KJ/mol Using appendix 3, determine whether the reaction is spontaneous or not at 25 degrees! From appendix 3: S(rxn)o for N2(g): 191.5 J/mol.K [at 1 atm and 25 degrees]

S(rxn)o for H2(g): 131.0 J/mol.K [at 1 atm and 25 degrees] S(rxn)o for NH3(g): 193.0 J/mol.K [at 1 atm and 25 degrees] [Delta]S(rxn) = [2(193.0)] – [191.5 + 3(131.0)] = -198.5 J/mol.K [Delta]S(surroundings) = - {[Delta]H} / T = -[-92.6 X1000 J/mol] / (25 + 273.15)K [Delta]S(surroundings) = + 310.582 J/mol.K [Delta]S(universe) = [Delta]S(system) + [Delta]S(surroundings) = (-198.5 + 310.582)J/mol.K [Delta]S(universe) = +112.082 J/mol.K > 0 ===== > spontaneous process! The third law of thermodynamics: ― The entropy of a perfect crystalline substance is zero at absolute zero temperature [zero Kelvin]. As the temperature increases, the freedom of motion increases and the entropy increases.‖ [Delta]S(system or rxn) = S(final) – S(initial) S(initial) = zero at zero Kelvin, and hence [Delta]S(system or rxn) = S(final) === > This is the formulation of the third law of thermodynamics. Gibbs Free Energy: G = is the energy availble to do work: [Delta]G(rxn) = G(products) – G(reactants) = {[the sum of the Gibbs free energy of the all products]X[stoichiometric (coefficients) factors]} – {[the sum of the Gibbs free energy of the all reactants]X[stoichiometric (coefficients) factors]} [Delta]G(rxn)o = G(products)o – G(reactants)o = {[the sum of the standard Gibbs free energy of the all products]X[stoichiometric (coefficients) factors]} – {[the sum of the standard Gibbs free energy of the all reactants]X[stoichiometric (coefficients) factors]} [Delta]G(rxn)o = [Delta]H(rxn)o – T[Delta]S(rxn)o Three cases: a. [Delta]G(rxn) > 0 ===== > none spontaneous process b. [Delta]G(rxn) = 0 ===== > equilibrium (no change) c. [Delta]G(rxn) < 0 ===== > spontaneous process

Standard free energy of reaction: [Delta]G(rxn)o is the Gibbs free energy for a rxn when it occurs under standard state conditions, when reactants in their standard states are converted to produce products in their standard state: State of matter: gas ===== > standard state: 1 atm pressure State of matter: liquid: ==== > standard state: pure liquid State of matter: solid ==== > standard state: pure solid State of matter: element ==== > standard state: [Delta]G(formation)o = zero State of matter: solution ==== > standard state: 1 M solution Example: Calculate the standard Gibbs free change for the following reaction at 25 degrees. Predict if the reaction is spontaneous or not. H2(g) + Br2(l) --------- > 2HBr(g) Using appendix 3: [Delta]G(rxn)o = [2(-53.2)] – [0 + 0] = - 105.4 KJ/mol < 0 ==== > spontaneous process. Standard Gibbs free energy and chemical equilibrium: [Delta]G(rxn or system) = [Delta]G(rxn or system)o + RT[ln(Q)] Where: R = gas constant = 8.314 J/mol.K T = temperature in Kelvin Q = reaction quotient [Delta]G(rxn or system) = Gibbs free energy of the system [Delta]G(rxn or system)o = Gibbs free energy of the system at standard conditions (i.e. 1 atm, 1M etc) ln = the natural logarithm At equilibrium: [Delta]G(rxn or system) = 0 (zero) Hence: 0 = [Delta]G(rxn or system)o + RT[ln(Q)] [Delta]G(rxn or system)o = -RT[ln(Q)] or: [Delta]G(rxn or system)o = -RT[ln(Ka)] or: [Delta]G(rxn or system)o = -RT[ln(Kb)] or: [Delta]G(rxn or system)o = -RT[ln(Kc)] or:

[Delta]G(rxn or system)o = -RT[ln(Kp)] or: [Delta]G(rxn or system)o = -RT[ln(Ksp)] etc… Example: Calculate the equilibrium constant Kp for the following reaction at 25 degrees: 2O3(g) < --------- > 3O2(g) [Delta]G(rxn or system)o = [3(0)] – [2(163.4 KJ/mol)] = - 326.8 KJ/mol =326800 J/mol [Delta]G(rxn or system)o = -RT[ln(Kp)] -326800 J/mol = - [(8.314 J/mol.K) (25+273.15)K][ln(Kp)] + 131.84 = ln(Kp) Kp = e^+131.84 = 1.809X10^+57 ------------------------------------------------------------------ Selected problems from the textbook: Chapter 18 Please work on these problems. If you have any question, contact your instructor! 18.5;18.13;18.14;18.20;18.23;18.28;18.29 ------------------------------------------------------------------ Selected problems and answers: Chapter 18 1. Define the following: a. first law of thermodynamics b. second law of thermodynamics c. third law of thermodynamics d. spontaneous process e. entropy f. standard Gibbs free energy 2. If [Delta]E = q + w, explain the status of q and w on the surroundings when their values are positive and when their values are negative. q = + (positive): heat absorbed by the system from the surroundings

[system is the chemical reaction; the surroundings is any thing out the side the system] q = - (negative): heat is given to the surroundings by the system w = + (positive) work done on the system by the surroundings w = - (negative) work done by the system on the surroundings 3. How does the entropy of a system change for each of the following processes: a. freezing liquid water: [Delta]S <0 (liquid ---- > solid) e. subliming iodine: [Delta]S >0 (solid ---- > gas) f. evaporating liquid alcohol: [Delta]S >0 (liquid ---- > gas) g. condensing oxygen gas: [Delta]S <0 (gas ---- > liquid) 4. Find [Delta]S(rxn)o for the following reaction: 2Na+(aq) + CO3^2- ---------> Na2CO3(s) Using appendix 3: [Delta]S(rxn)o = [135.98 J/mol.K] – [2(60.25) + (-53.1)] =68.58 J/mol.K 5. Determine whether the following reaction at 25 degrees is spontaneous or not. [Hint: Use appendix 3] H2(g) + Cl2(g) -------- > 2HCl(g) Two solutions are possible: The shortest and the easiest one is to calculate [Delta]G(rxn) and check its value with zero: [Delta]G(rxn) = [2(-95.27)] – [0 + 0] = -190.54 KJ/mol.K <0 ==== > spontaneous. The second way is longest one but it gives the same solution: [Delta]S(rxn) = [2(187.0)] – [131.0 + 223.0] = + 20.0 J/mol.K [Delta]S(surroundings) = - {[Delta]H(system)} / T = - {[Delta]H} / (25 + 273.15)K [Delta]H(system) = [2(-92.3)] – [0 + 0] = - 184.6 KJ/mol = -184600 J/mol [Delta]S(surroundings) = - (-184600 J/mol) / (25 + 273.15)K = + 619.15 J/mol.K [Delta]S(universe) = [Delta]S(system) + [Delta]S(surroundings) = (20 + 619.15)J/mol.K [Delta]S(universe) = + 639.15 J/mol.K > 0 ===== > spontaneous process!

Note that [Delta]G(rxn) = [Delta]H(rxn) – T[Delta]S(rxn) = -184600J/mol - (298.15K)(20.0)J/mol.K [Delta]G(rxn) = -184600J/mol – 5963 J/mol = -190563 J/mol = - 190.563KJ/mol. This is the same value found above in the shortest and the easiest solution. 6. Calculate [Delta]G(rxn) at 25 degrees of a system if the following data were given: [Delta]H(rxn)o = -32.5 KJ/mol [Delta]S(rxn)o = +356.0 J/mol.K [Delta]G(rxn) = [Delta]H(rxn) – T[Delta]S(rxn) = - (32.5X1000)J/mol - (298.15K)(356.0)J/mol.K [Delta]G(rxn) = -32500J/mol – 106141.4 J/mol = -138641.4J/mol = - 138.64 KJ/mol 7. Calculate the standard free energy [Delta]G(rxn)o for the following reaction at 25 degrees. Calculate Kp of this reaction: 3H2(g) + N2(g) < -------- > 2NH3(g) Using appendix 3: [Delta]G(rxn)o = [2(-16.6)] – [0 + 0] = -33.2KJ/mol = - 33200 J/mol [Delta]G(rxn or system)o = -RT[ln(Kp)] -33200 J/mol = - [(8.314 J/mol.K) (25+273.15)K][ln(Kp)] + 13.393 = ln(Kp) Kp = e^+13.393 = 6.554X10^+5 8. If Ksp for a specific chemical reaction was found to be 1.8X10^-6 at 25 degrees, find [Delta]G(rxn)o. Is this reaction is spontaneous? [Delta]G(rxn or system)o = -RT[ln(Kp)] = -(8.314J/mol.K)(298.15K)[ln(1.8X10^-6)] = +32789.134 J/mol [Delta]G(rxn or system)o = + 32.79X10^+3 KJ/mol ====== > 0 Not spontaneous. Note that there are several rules that determine the spontaneousity of the reaction: a. [Delta]G(rxn or system)o < 0 b. [Delta]S(universe) > 0 c. Kp or Kc or Ka or Kb or Ksp > 1

In this problem Ksp is 1.8X10^-6 which less that 1 and hence the reaction (equilibrium) is not spontaneous!! _______________________________________________ Chapter 19: Electrochemistry Electrochemistry is the study of the (exchange) interconversion of electrical energy chemical energy. Balancing Redox Equations: Redox equation: involves oxidation/reduction reaction. Example: Balance the following equation: MnO4^-(aq) + C2O4^2-(aq) --------- > Mn^2+ + CO2 First: identify the reduction and the oxidation reactions by determining the oxidation numbers of the atoms involved. Second: Treat each of oxidation and reduction equation by itself to determine number of electrons lost or gained. Manganese is Mn^7+ in MnO4^- and manganese is Mn^2+ in Mn^2+. [This is a reduction] Carbon is C^3+ in C2O4^2- and carbon is C^4+ in CO2. [This is an oxidation] Third: balance the number of oxygen for each reduction and oxidation equation by itself by adding H2O to the opposite side where oxygen is found. [i.e. if the oxygen found to be in the reactants’ side then add H2O to the products’ side and vice versa]. Fourth: balance number of hydrogen of the H2O by adding H+ to the opposite side of H2O 8H+(aq) + 5e- + MnO4^-(aq) --------- > Mn^2+(aq) + 4H2O(l) [Reduction] C2O4^2-(aq) --------- > 2CO2(g) + 2e- [Oxidation] Fifth: eliminate the number of electrons by balancing each of the oxidation and reduction equation: [multiply each equation by the swapped number of electrons] a. multiply the reduction equation by 2 b. multiply the oxidation equation by 5 [8H+(aq) + 5e- + MnO4^-(aq) --------- > Mn^2+(aq) + 4H2O(l)]X 2

[C2O4^2-(aq) --------- > 2CO2(g) + 2e-]X 5 16H+(aq) + 10e- + 2MnO4^-(aq) --------- > 2Mn^2+(aq) + 8H2O(l) 5C2O4^2-(aq) --------- > 10CO2(g) + 10e- Sixth: add up the two equations to eliminate the number of electrons. [Make sure to add up H2O and H+] 16H+(aq) +2MnO4^-(aq) + 5C2O4^2-(aq) ---------- > 2Mn^2+(aq) + 8H2O(l) + 10CO2(g) Seventh: check up the overall equation as follows: Left side: Reactants’ side # H = 16 # Mn = 2 # O = 8 + 20 = 28 # C = 10 Charges = (16+) + (2-) + (10-) = 4+ Right side: Products side # H = 16 # Mn = 2 # O = 8 + 20 = 28 # C = 10 Charges = 4+ Both left side (reactants’ side) and right side (products’ side) are equal. One can say with confidence that ―THIS REDOX EQUATION IS BALANCED.‖ The above redox equation is balanced in acid medium. How about having the same redox equation in basic medium? In order to have this redox equation in a basic equation, follow the previous steps as if the equation is in acidic medium and at whole end replace the H+ by OH- as follows: 16H+(aq) +2MnO4^-(aq) + 5C2O4^2-(aq) ---------- > 2Mn^2+(aq) + 8H2O(l) + 10CO2(g) But it is known that: H+(aq) + OH-(aq) = H2O(l) Therefore, H+(aq) = H2O(l) – OH(aq) The above redox equation in basic medium turns to be:

16[H2O(l) – OH-(aq)] +2MnO4^-(aq) + 5C2O4^2-(aq) ---------- > 2Mn^2+(aq) + 8H2O(l) + 10CO2(g) Moving the – OH(aq) to the other side with the opposite sign, one obtains: 16H2O(l) +2MnO4^-(aq) + 5C2O4^2-(aq) ---------- > 2Mn^2+(aq) + 8H2O(l) + 10CO2(g) + 16 OH-(aq) Adding up H2O(l) from both sides: 8H2O(l) +2MnO4^-(aq) + 5C2O4^2-(aq) ---------- > 2Mn^2+(aq) + 10CO2(g) + 16 OH-(aq) Checking up the overall equation as follows: Left side: Reactants’ side # H = 16 # Mn = 2 # O = 8 + 8 + 20 = 36 # C = 10 Charges = (2-) + (10-) = 12- Right side: Products side # H = 16 # Mn = 2 # O = 20 + 16 = 36 # C = 10 Charges = (4+) + (16-) = 12- Both left side (reactants’ side) and right side (products’ side) are equal. One can say with confidence that ―THIS REDOX EQUATION IS BALANCED.‖ In basic medium. ------------------------------------------------------------------ Electrochemical cell (also called: ―the galvanic cell‖ or ―the battery‖) The electrochemical cell is made from two sub cells. In each sub cell, there is a half – cell reaction: a. Anode cell (anode half – cell reaction): In the anode cell, an oxidation reaction is always carried out. The cell is made of an electrode immersed in the solution of the same element as the

electrode. [i.e Zn is an electrode is immersed in Zn(NO3)2 solution] The half – cell reaction is called the anode reaction or the oxidation reaction: Zn --------- > Zn^2+ + 2e- (oxidation) The anode cell represents the positive electrode of the cell. b. Cathode cell (cathode half – cell reaction): In the cathode cell, a reduction reaction is always carried out. The cell is made of an electrode immersed in the solution of the same element as the electrode. [i.e Cu is an electrode is immersed in CuSO4 solution] The half – cell reaction is called the anode reaction or the oxidation reaction: Cu^2+ + 2e- --------- > Cu (reduction) The cathode cell represents the negative electrode of the cell. The anode (+) is connected to the cathode (-) through an alligator wire which is connected to the voltammeter. The anode (+) is connected to the red knob (positive) of voltammeter circuit and the cathode (-) is connected to the black (ground or negative) of the voltammeter circuit. Reason for this is that electrons are moving out of the anode through the alligator wire and if they find a negative charged plate (internal battery of the voltammeter) they will be neutralized and can not move through to the cathode cell. A positive plate of the voltammeter will repel these coming electrons which will push them away to the cathode side. The anode solution is becoming more and more positive as the anode is more and more losing electrons. This creates an artificially positive solution. The cathode is accepting the electrons and the solution of the cathode becomes artificially negative solution. This lead to a separation of charges and hence the current (the electron flow) will seize to exist. In order to get around this a salt bridge is used. The salt bridge is made of an electrolyte such as KNO3, NaCl, KCl, NaNO3 etc. [The main thing of this electrolyte is that is made of ionic compound that is chemically inert and inactive (does not react with any chemicals involved)]. This strong electrolyte is made of cation (+), which transfers to the artificially negative solution (the cathode solution), and anion (-), which transfers to the artificially positive solution (the anode solution). The salt bridge: a. closes the electrical circuit b. closes the loop c. neutralizes any excess positive and negative charges The short - hand electrochemical cell expression: Zn(s) + CuSO4(aq) --------- > ZnSO4(aq) + Cu(s)

can be written as: Zn|Zn^2+ || Cu^2+|Cu Anode(oxidation) || Cathode (reduction) The || is symbol for a salt bridge. For detailed drawing of the electrochemical cell, one should consult the textbook! At the end of chemical reaction the blue CuSO4 solution will turn to colorless indicating that all copper ions are being reduced out of the solution and precipitated on the copper electrode. Therefore, copper electrode will increase in mass. On the other hand the anode (Zn) will decrease in mass because more and more of it will dissolve as Zn^2+ (oxidized) into the ZnSO4 solution. Standard Electrode Potential: Standard conditions: Pressure = 1 atm Concentration = 1 M Temperature = any specified temperature, actually 25 degrees is used Hydrogen is used as a standard electrode potential as follows: 2H+(aq)[1M] + 2e- ----- > H2(g)[1 atm] E(H)o = standard reduction potential = zero Volt (0 V) at 25 C. Standard oxidation cell: Zn(s) ------- > Zn^2+(aq)[1M] + 2e- At 1 atm, 1M solution of Zn^2+ and 25 C. Combining both cells, the following electrochemical cell is produced: Zn(s) + 2H+(aq) ------- > Zn^2+(aq) + H2(g) E(Zn)o = standard oxidation potential = + 0.76 V Short – hand expression of such electrochemical cell: Zn|Zn^2+||H+||H2 The following equation can be use for the above electrochemical cell to calculate the overall potential (anode and cathode cell): E(cell)o = E(anode)o + E(cathode)o Where: E(cell)o = total potential [or total voltage or electromotive force (emf) of the cell] E(anode)o = anode half – cell potential (or voltage). E(anode)o is also expressed by E(ox)o

E(cathode) = cathode half – cell potential (or voltage). E(cathode)o is also expressed by E(red)o Using table 19.1 page 766 in textbook of the electrochemical series, one obtains: E(cell)o = (+ 0.76V) + (0 V) = + 0.76V Electrochemical series: The electrochemical series is made of sets of reduction chemical reactions with the metal of group 1 and 2 being at the top of the series (which are easily oxidized). Hydrogen represents the zero volt and the noble elements such gold, silver, platinum, mercury and copper are at the bottom of the series (which are very easily reduced). Note that all the values in the electrochemical series are reduction values. In order to convert to the corresponding oxidation values, one has to change the sign of the reduction values ONLY. [for example, Zn^2+ + 2e- --------- > Zn is a reduction reaction and the reduction potential is – 0.76 V. The oxidation potential for the reaction Zn ------- > Zn^2+ is + 0.76 V. The same value but opposite sign. The standard potential does not change with the change in stoichiometric factors (coefficient) or the change in the size of the cell. Example: A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO3)2 solution and an Ag electrode in a 1.0 M AgNO3 solution. Calculate the standard emf of this electrochemical cell at 25 C. Using values from the electrochemical series (table19.1) Mg will be oxidized: Mg(s) ------ > Mg^2+(aq) + 2e- [E(ox)o = + 2.377 V] Ag will be reduced: Ag+(aq) + 1e- ------ > Ag(s) [E(red)o = + 0.80 V] E(cell)o = E(ox)o + E(red)o = 2.377 + 0.80 = 3.177 V Relationship between [Delta]Go and E(cell)o [Delta]Go = - RT[ln(K)] [Delta]Go = - n F E(cell)o Where: n = # of the exchanged electrons F = Faraday constant; 1 F = 96500 J/mol.V From above:

E(cell)o = {(0.0257 V) [ln (K)]} / n E(cell)o > 0 ====== > process is spontaneous. The summary of the conditions of spontaneousity is as follows: a. [Delta]G(rxn or system)o < 0 b. [Delta]S(universe) > 0 c. Kp or Kc or Ka or Kb or Ksp > 1 d. E(cell)o > 0 (at standard conditions: 1 atm and 1 M) e. E(cell) > 0 (Not at standard condition) Example: Calculate the equilibrium constant for the following equilibrium at 25 degrees. Predict if the equilibrium reaction is spontaneous or not. Fe(s) + 2Ag+(aq) < ----- > Fe^2+(aq) + 2Ag(s) Fe is being oxidized: Fe(s) ------ > Fe^2+(aq) + 2e- Ag is being reduced: 2Ag+(aq) + 2e- ------- > 2Ag(s) E(cell)o = E(ox)o + E(red)o = (0.44) + (0.80) = + 1.24 V E (cell)o > 0 ====== > process is spontaneous! E(cell)o = {(0.0257 V) [ln (K)]} / n + 1.24 V = (0.0257 V)[ln(K)] / 2 96.498 = ln(K) K = e^+96.498 = 8.101X10^+41 Nernsts’ equation: This equation is used when no standard conditions are used [i.e pressure is not 1 atm, concentration is not 1 M] Consider the general redox chemical equation: [Not at standard conditions] aA + bB -------- > cC + d D E(cell) = E(cell)o - [(RT) / (nF)] [ln(Q)] Using F = 96500 J/mol.V and T = 25 degrees; and R =8.314 J/mol.K one obtains: E(cell) = E(cell)o - [(0.0257 V) / (n] [ln(Q)] Where: E(cell) = total cell potential not at standard conditions of 1 atm and 1 M E(cell)o = total standard cell potential at 1 atm and 1 M n = number of exchanged electrons

Q = reaction quotient = {[C]^c[D]^d} / {[A]^a[B]^b} Example: Will the following reaction occur spontaneously at 25 degrees, given that [Fe^2+] = 0.60 M And [Cd^2+] = 0.010 M. The general redox equation is: Fe(s) + Cd^2+(aq) ------- > Fe^2+(aq) + Cd(s) Fe is being oxidized: Fe(s) ---- > Fe(aq) + 2e- [E(ox)o = + 0.44 V] Cd is being reduced: Cd^2+ + 2e- ------- > Cd(s) [E(red)o = - 0.40 V] E(cell)o = = E(ox)o + E(red)o = (+ 0.44 V) + (- 0.40 V) = + 0.04 V Q = [Fe^2+] / [Cd^2+] Note the solids Fe(s) and Cd(s) are ignored. n = 2 E(cell) = E(cell)o - [(0.0257 V) / (n] [ln(Q)] E(cell) = + 0.04 V – [(0.0257 V) / (2) [ln (0.60/0.01)] = + 0.04 V – 0.0526 V = - 0.0126 V E(cell) < 0 ====== > the process is Not spontaneous! ------------------------------------------------------------------ Selected problems from the textbook: Chapter 19 Please work on these problems. If you have any question, contact your instructor! 19.1;19.2;19.12;19.22;19.24;19.30;19.31;19.63;19.107 ------------------------------------------------------------------ Selected problems and answers: Chapter 19 1. Define the following: a. electrochemistry b. cathode c. anode d. salt bridge

e. half – cell reaction f. standard electrode cell g. oxidation h. reduction Check notes + textbook! 2. Balance the following redox equation in acidic and basic medium: Cr2O7^2-(aq) + C2O4^2-(aq) ------ > Cr^3+(aq) + CO2(g) Following all steps mentioned in the lecture notes: Chromium s Cr^6+ in Cr2O7^2- and chromium is Cr^3+ in Cr^3+. [This is a reduction] Carbon is C^3+ in C2O4^2- and carbon is C^4+ in CO2. [This is an oxidation] 14H+(aq) + 6e- + Cr2O7^2-(aq) ----- > 2Cr^3+(aq) + 7H2O [reduction] C2O4^2-(aq) ------- > 2CO2(g) + 2e- [oxidation] The ratio between the electrons 6e-:2e- can be reduced to 3:1 Multiply the reduction equation by 1 and the oxidation equation by 3 14H+(aq) + 6e- + Cr2O7^2-(aq) ----- > 2Cr^3+(aq) + 7H2O [reduction] 3 C2O4^2-(aq) ------- > 6CO2(g) + 6e- [oxidation] Adding up both equations: 14H+(aq) + Cr2O7^2-(aq) + 3C2O4^2-(aq) ----- > 2Cr^3+(aq) + 7H2O + 6CO2(g) This redox equation is balanced regarding its atoms and charges in acidic medium. To balance the above redox equation in basic medium, replace H+(aq) by [H2O(l) – OH-(aq)] as follows: 14[H2O(l) – OH-(aq)] + Cr2O7^2-(aq) + 3C2O4^2-(aq) ----- > 2Cr^3+(aq) + 7H2O + 6CO2(g) 14H2O(l) + Cr2O7^2-(aq) + 3C2O4^2-(aq) ----- > 2Cr^3+(aq) + 7H2O + 6CO2(g) + 14OH-(aq) 7H2O(l) + Cr2O7^2-(aq) + 3C2O4^2-(aq) ----- > 2Cr^3+(aq) + 6CO2(g) + 14OH-(aq)

This redox equation is balanced regarding its atoms and charges in basic medium. 3. Draw an electrochemical cell (galvanic cell) for the following reaction at 25 C: Al(s) + Au^3+(aq) ------- > Al^3+(aq) + Au(s) Consult textbook for detailed drawing! 4. Determine the half – cell reactions (anode and cathode reactions) for the above reaction (question # 3). Al ------ > Al^3+(aq) + 3e- [oxidation: +1.66 V] Au^3+(aq) + 3e- ------- > Au(s) [reduction: +1.50 V] 5. Write the short – hand description of the cell for the above question. Al|Al^3+ || Au^3+|Au 6. Determine E(cell)o for the above question. Is the reaction spontaneous? E(cell)o = E(ox)o + E(red)o = (+ 1.66 V) + (+ 1.50 V) = +3.16 V E(cell)o > 0 ===== > spontaneous reaction! 7. Determine [Delta]Go for the above reaction. Is the reaction spontaneous? [Delta]Go = - n F E(cell)o = [-3X96500 J/mol.V] [+3.16 V] = - 914820 J/mol = - 914.82 KJ/mol [Delta]Go < 0 ===== > spontaneous reaction! 8. Using E(cell)o determine K for the above reaction. Is the reaction spontaneous? E(cell)o = [(0.0257) / n][ln(K)] +3.16 V = [0.0257/3][ln(K)] 368.872 = ln(K) K = e^+(368.872) = infinite (very large positive number) K > 1 ===== > spontaneous reaction! 9. If [Al^3+] = 0.05 M and [Au^3+] = 0.150 M in the above question, predict if the reaction is spontaneous or not. Using the Nernsts’ equation: Q = [Al^3+(aq)] / [Au^3+(aq)] Note that both solids Al(s) and Au(s) are ignored. E(cell) = E(cell)o - [(0.0257 V) / (n] [ln(Q)]

E(cell) = + 3.16V – [(0.0257 V) / (3) [ln (0.05/0.150)] = + 3.16 V – (- .0.00941) V = +3.16V + 0.00941 = E(cell) = + 3.16941 V E(cell) > 0 ===== > spontaneous reaction!

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