Chemical Energetics Learning Objectives To predict the energy changes associated with chemical processes

EnergyDefined as capacity of matter to do work. Can exist in many forms, such as chemical, electrical, heat, nuclear or light energyMatter has potential and kinetic energy

What is potential energy?Known as stored energy and is called the heat content or enthalpy of a substanceGiven as symbol HIt is the energy in the object due to its positionExample:Gasoline is a source of chemical potential energy.It burns when combined with oxygen and the heat released decreases potential energy

What is kinetic energy?Energy that matter possesses due to its motionMoving bodies possess kinetic energy

How molecules take up thermal energy?Temperature is a measure of the average kinetic energy due to translational motions of moleculesVibrational and rotational motions of molecules can also accept thermal energy and raise the heat capacity of a substanceMonoatomic species (noble gas elements) have the lowest heat capacitiesSuch species do not have vibrational and rotational motions

Polyatomic molecules have the other motions which contribute to an increase heat capacity

Enthalpy (heat capacity)In chemical reactions the enthalpy depends on temperatureEvery substance has a characteristic heat capacity and is also known as specific heat (H)Specific heat is defined as the quantity of heat (lost or gained) required to change the temperature of 1g of that substance by 1C

The greater the heat capacity of a substance, the smaller will be the effect of a given absorption or loss of heat on its temperatureThe relation between mass, specific heat, temperature change (t) and quantity of heat lost or gained by a system is expressed by the equation

(mass of ) (specific heat) t = heat substance substance

ExampleWhat is the amount of heat needed to raise the temperature of 200g of water by 10C? (specific heat of water is 4.18 J/gC)

= 200 x 4.18 x 10C = 8.36 x 103 J

Common assumptions for reaction mixtures made up of aqueous solutions:

density of aqueous solution assumed to be the same as for water, 1g/mL eg, 100mL of solution is said to have a mass of 100g

additivity of volumes of reactants is assumed eg, 100mL of reactant a + 200mL of reactant b = 300mL of reaction mixture

specific heat capacity of the reaction mixture assumed to be the same as water, ie, specific heat capacity = 4.184 JK-1g-1

Energetics of chemical reactionsA zero in enthalpy is when the elements exist in a stable form at 298K and 1atm pressureIf enthalpy decreases during a chemical reaction, a corresponding amount of energy must be released to the surroundings.This can be observed in combustion reactionsCH4 (g) + 2O2 (g) CO2(g) + 2H2O(g) + energy

The release of energy for reactions like this is represented by the figure below

The enthalpy difference between the reactants and products is equal to the amount of energy released to the surroundingsSuch reactions where energy is released to the surroundings is known as an exothermic reactionThe stored potential energy has become lower for the products than the reactants

Reactions which absorb energy result products to have greater enthalpy than reactantsThis can be seen in the reaction below4C (s) + 8S(g) + energy 4CS2 (g)

The enthalpy difference between the reactants and products is equal to the amount of energy absorbed from the surroundingsSuch reactions are known as endothermic reactionsThe enthalpy of products is greater then the reactantsExothermic reactions release energy causing the surroundings to heat upEndothermic reactions absorb energy from the surroundings, causing the surroundings to cool down

Describing energy changes in chemical reactionsH values are negative for exothermic reactions and positive for endothermic reactionsIn exothermic reactions, final enthalpy is less than initial enthalpyIn endothermic reactions the final enthalpy is greater than the initial enthalpy

The sign H reflects the change in enthalpy of the systemFor example H2O (g) H2(g) + 1/2O2 (g)

H is +242kJ indicating that 242kJ of heat are absorbed in the decomposition of one mole of steam

Sometimes the equation is written as below:

H2O (g) H2(g) + 1/2O2 (g) + 242kJ

Note that the quantity of heat is written as part of the equation.

In the combustion of sulfur, S (s) + O2 (g) SO2 (g)

H is -297kJ which indicates that 297kJ of heat are released when one mole of sulfur reacts with oxygen to form sulfur dioxide

Enthalpy of formation (Hf o )The standard enthalpy of formation, Hf o of a substance is defined as the heat change required to produce one mol of the substance from its elements in their standard states.e.g. Ag (s) + Cl2 (g) AgCl (s); Hf o = -127 kJ mol-1In general, the standard enthalpy change for any reaction is given by the following expression

Hfo is the sum of the standard enthalpies of formation of all products and reactants.By definition, the standard enthalpy (heat) of formation of an element in its standard state is zero, Hfo = 0.

The heat of formation of CH4(g), CO2(g) and H2O (l) are -74.8 kJmol-1, -393.5 kJmol-1 and 285.8 kJmol-1 respectively. Calculate the heat change for the reaction. Calculation:

Solution: H for the reaction, is: H = Hfo(products) -Hfo(reactants) {Hfo[CO2(g)] + 2 Hfo[H2O(l)]} - {Hfo[CH4(g)] + 2 Hfo[O2(g)]} Hfo[CO2(g)] = -393.5kJmol-1, Hfo[H2O(l)] = - 285.8kJmol-1 Hfo[CH4(g)] = - 74.8kJmol-1, Hfo[O2(g)] = 0 H = {-393.5 + 2(-285.8)} - {-74.8 + 2 x 0)} = - 965.1 + 74.8 = - 890.3kJmol-1

Hesss LawThe standard enthalpy of reaction depends only on the difference between the total standard enthalpy of the products and the total standard enthalpy of the reactants. It does not depend on the route by which the reaction occurs.

In other word, if a reaction can take place by more than one route, the overall change in the enthalpyis the same, whichever route is followed.

Question 1:

Given the following information, calculate the enthalpy of formation of C2H4Cl2(g) in kJ/mol:

2 C(s) + 2 H2(g) + Cl2(g)

Given:

2 C(s) + 2 H2(g)

C2H4Cl2(g)

C2H4(g);Ho = +52.3 kJ/mol Cl2(g) + C2H4(g);Ho = +116 kJ/molC2H4Cl2 (g)

Question 2.Given the following equations and Ho values, determine the heat of reaction (kJ/mol) at 298 K for the reaction: N2(g) + 1/2 O2(g) N2O(g)

Given:

2 NH3(g) + 3 N2O(g) 4 N2(g) + 3 H2O(l)Ho = -1010 kJ/mol 2 N2(g) + 6 H2O(l)Ho = -1531 kJ/mol4 NH3(g) + 3 O2 (g)

Question 3.

Calculate Ho/kJmol-1 for the following reaction using the listed standard enthalpy of reaction data:

3 C2H2 (g) C6H6 (g)

Given:

2 C(s) + H2(g) C2H2(g) Ho = +226.7 kJ/mol 6 C(s) + 3 H2(g) C6H6(g)Ho = +82.93 kJ/mol

Question 4.

Given the following equations and Ho values, determine the heat of reaction at 298 K for the reaction which occurs in a welder's acetylene torch:

2 C2H2 (g) + 5 O2 (g) 4 CO2 (g) + 2 H2O (l)

Given:

H2(g) + 1/2 O2(g) H2O(l) Ho = -285.8 kJ/mol

2 C(s) + H2(g) C2H2(g) Ho = +226.7 kJ/mol

C(s) + O2(g) CO2(g) Ho = -393.5 kJ/mol

Enthalpy of combustion (Hc o)The standard enthalpy of combustion, Hco of a substance is defined as the heat released when one mole of a substance is completely burnt in oxygen in their standard states.e.g. C2H4 (g) + 3O2 (g) 2CO2 (g) + 2H2O (l); Hco = -1411 kJ mol-1Note that the Hco applies to the complete combustion of the carbon to CO2, not CO. Hco are always ve as heat is always released in a combustion process.

The heats of reaction for the formation of some compounds at 25C at 1 atm

Say a chemist is interested to determine the heat of combustion of propane.

C3H8 (g) + 5O2 (g) 3CO2(g) + 4H2O(g)

From the table, locate equations involving C3H8, CO2, H2O (g)3C (s) + 4H2(g) C3H8 (g) H = -104kJ/molC(s) + O2(g) CO2 (g) H = -394kJ/mol H2(g) + 1/2O2 (g) H2O(g) H = -242kJ/mol

b) The equations are rearranged so that C3H8 appear on left hand side and CO2, H2O (g) on right hand side.The equations are multiplied so that the coefficients are the same as those in the overall equation

C3H8 (g) 3C (s) + 4H2(g) H = +104 kJ/mol

3C(s) + 3O2(g) 3CO2 (g) H = -1182 kJ/mol

4H2(g) + 2O2 (g) 4H2O(g) H = -968 kJ/mol

c) Add the equations and the H valuesC3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g)

H = -2046 kJ/mol

When you add the equations, the C (s) and H2 (g) terms cancel out. Thus the sum of the H values give the heat of reaction as -2046kJ/mol for the combustion of one mole of C3H8

This principal of additivity of heats of reaction is known as Hess Law Questions:Calculate the heats of reaction for the following reactions.NO (g) + 1/2O2(g) NO2 (g) CH4 (g) + 2O2(g) CO2(g) + 2H2O(g) SO2(g) + 1/2O2(g) + H2O(l) H2SO4 (l)

Born-Haber Cycle

An approach in analyzing reaction energiesThe cycle involves the formation of an ionic compound from the reaction of a metal (often a Group I or Group II element) with a non-metal. Used in calculating lattice enthalpies

The lattice enthalpy cannot be measured directly and so we make use of other known enthalpies and link them together with an enthalpy cycle.This enthalpy cycle is the Born-Haber cycle.What do we mean by lattice enthalpy?For an ionic compound the lattice enthalpy (U) is the enthalpy change when one mole of solid in its standard state is formed from its ions in the gaseous state.

Born Haber CycleBorn Haber cycle calculates this lattice enthalpy through a comparison of the standard enthalpy of formation of the ionic compound to the enthalpy required to make gaseous ions from the elements.

It is slightly complex as it is required to atomise elements before it is possible to make gaseous ions

If the element is a molecule, we have to consider the bond dissociation enthalpy, sublimation enthalpy, ionization energy, lattice energy and electron affinity must also be considered.

Example Look at the Born-Haber cycle provided below. Calculate the enthalpy of formation for sodium chloride, NaCl

The net energy changes during the formation of sodium chloride from metallic sodium and chlorine gas can be represented by Hf

The overall process is thought to take place in following intermediate steps:

Intermediate steps:(i) Metallic sodium into gaseous sodium atomThe energy required per mole of sodium is 'enthalpy of sublimation' which is represented by Hs . This step is energy consuming process.

(ii) Dissociation of chlorine molecule into chlorine atoms

The energy required per mole of chlorine is 'enthalpy of dissociation' represented by Hd.

(iii) Gaseous sodium atom into gaseous cation

The energy required in this process is called Ionization energy (IE).

(iv) Gaseous chlorine atom into gaseous anion

This step involves the release of energy referred as Electron Affinity (EA).

(v) Combination of oppositely charged gaseous ions to form solid crystal

This involves the release of energy referred as lattice energy (U).

The various energy changes in different steps are as shown below:

PracticeGiven below are the reaction energies in the formation of Zinc sulfide (ZnS). Calculate the lattice enthalpy for ZnS &Construct the Born Haber cycle by placing and labeling the reactions in the cycle.

Zn(g) Zn2+(g) + 2e-H =+2680 kJZn(s) Zn(g)H =+116 kJS(s) S(g)H =+264 kJS(g) + 2e- S2-(g)H =+246 kJZn2+(g)+S2-(g) ZnS(s)H =? kJZn(s)+ S(s) ZnS(s)H =-182 kJ

PracticeGiven below are the reaction energies in the formation of Magnesium Fluoride. Calculate the lattice enthalpy for the compound & construct the Born Haber cycle by placing and labeling the reactions in the cycle.

Hs of Mg = 146.4 kJ IE1 and IE2 values of Mg = 737 and 1449 kJ Hd of F2 = 158.8 kJEA of F = - 328 kJHf of Magnesium Fluoride = - 1096.5 kJ

PracticeGiven below are the reaction energies in the formation of hypothetical compound ArCl. Calculate the Hf for the compound & construct the Born Haber cycle by placing and labeling the reactions in the cycle.

IE1 of Ar = 526.3 kJ Hd of Cl2 = 243 kJEA of Cl = - 349 kJLattice energy (U) of ArCl (s) = - 703 kJ

PracticeGiven below are the reaction energies in the formation of Lithium Fluoride (LiF). Construct the Born Haber cycle by placing and labeling the reactions in the cycle.

Li (g) Li+ (g) + 1e-Ho =+520 kJ Li (s) Li (g)Ho =+161 kJ F2 (g) 2F (g)Ho =+159 kJ F (g) + 1e- F- (g)Ho =-328 kJ Li+ (g) + F- (g) LiF (s)Ho =? kJ Li (s) + F2 (g) LiF (s)H =-617 kJ