8.0 thermochemistry (student's copy)

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CHAPTER 8: CHAPTER 8: THERMOCHEMISTRYTHERMOCHEMISTRY

8.1 System8.2 Concept of enthalpy8.3 Calorimetry8.4 Hess’s Law8.5 Born-Haber cycle

8.1 SYSTEM8.1 SYSTEM

LEARNING OUTCOMELEARNING OUTCOME1. Describe:

i. open systemii. closed systemiii. isolated system.

2. Explain endothermic and exothermic reactions using the complete energy profile diagrams.

What happens to the ice cream and the hot coffee after 10 minutes?

FIRST LAW OF THERMODYNAMICS

The law states that energy can be tansformed (changed from one form to another) but cannot be created or destroyed.

There are generally 3 types of systems:

Open System Closed system Isolated system

An open system a system that can exchange mass and energy, usually in the form of heat with its surroundings

closed system, which allows the exchange of energy with its surroundings

isolated system that does not allow the exchange of either mass or energy with its surrounding

TERMS DEFINITIONHeat energy transferred between

two bodies of different temperatures

System any specific part of the universe

Surroundings

everything that lies outside the system

IMPORTANT TERMS

ENERGY the ability to do work Units of energy

• SI Unit: Joule (J)1 J = 1 kgm2s-2

• Older unit: calorie (cal)1 cal = 4.184 J

A study of heat change in chemical reactions.

Two types of chemical reactions:- Exothermic Endothermic

ΔH is (-ve) Enthalpy of products < Enthalpy of reactants Heat is released from the system to the

surroundings. E.g: combustion, neutralization etc.

2H2(g) + O2(g) 2H2O(l) + heatCH4(g) + 2O2(g) CO2(g) + 2H2O(g) + heat

Consider the following reaction:-

A (g) + B (g) → C (g) ΔH = ve (reactants) (product)

Energy profile diagram for exothermic reaction

A(g) + B(g)

enthalpy

Reaction pathway

ΔH = -ve

ΔH is (+ve) Enthalpy of products > enthalpy of reactants Heat is absorbed by the system from the

surrounding E.g: ice melting

Heat + H2O(s) H2O(l)heat + H2O(l) H2O(g)

Energy profile of diagram endothermic reactions

Consider the following reaction:-

A (g) + B (g) → C (g) ΔH = +ve (reactants) (product)

A(g) + B(g)

enthalpy

Reaction pathway

ΔH = +ve

8.2 CONCEPT OF ENTHALPY8.2 CONCEPT OF ENTHALPY

LEARNING OUTCOMELEARNING OUTCOME1. State standard conditions of reaction and define the

following terms :i. enthalpyii. standard enthalpy

2. Define enthalpy of :i. formationii. combustioniii. atomisationiv. neutralisation v. hydrationvi. solution

3. Write thermochemical equation for each of the following enthalpies 15

The heat content or total energy in the system

Commonly measured through heat change.

Examples: system undergoes combustion or ionization.

Heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants Hproducts > Hreactants

9.1-15

Enthalpy of reaction, ∆H:◦The enthalpy change associated with a chemical reaction.

( ΔHreaction = ΣΔHf product – ΣΔHf reactant )

Standard enthalpy, ∆Hº◦The enthalpy change for a particular reaction that occurs at 298K and 1 atm (standard state)

There are many kind of enthalpies such as:

TYPES OF ENTHALPIES

The heat changed when 1 mole of a compound is formed from its elements in their most stable state.

H2 (g) + ½ O2(g) → H2O (l) ∆Hf = 286 kJ

The standard enthalpy of formation of any element in its most stable state form is zero.E.g. :-

∆H f(O2 (g) ) = 0 ∆H f (Cl2 (g)) = 0

K(s) + ½ Br2(l) KBr(s) H = Hf

The heat released when 1 mole of substance is burned completely in excess oxygen.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = Hcomb

C4H10(l) +13/2 O2(g) 4CO2(g) + 5H2O(l) H = Hcomb

The heat absorbed when 1 mole of gaseous atoms is formed from its element

Ha is always positive because it involves only breaking of bonds

E.g:- Na(s) Na(g) Ha = +109 kJ

½Cl2(g) Cl(g) Ha = +123 kJ

The heat released when 1 mole of water, H2O is formed from the neutralization of acid and base.

E.g:-HCl(aq)+ NaOH(aq)→ NaCl(aq) +H2O(l) ΔHn = 58 kJ

The heat released when 1 mole of gaseous ions is hydrated in water.

E.g:-Na+

(g) Na+(aq) Hhyd = 406 kJ

Cl-(g) Cl-(aq) Hhyd = 363 kJ

The heat changed when 1 mole of a substance is dissolves in water.

E.g: KCl(s) K+

(aq) + Cl(aq) Hsoln = +690 kJ

Enthalpy of Solution, Hsoln

Standard enthalpy of reaction

9.1-35

The enthalpy change of a reaction carried out at standard states (1 atm, 25OC)

Shows the enthalpy changes.

E.g : H2O(s) → H2O(l) ΔH = +6.01 kJ

1 mole of H2O(l) is formed from 1 mole of H2O(s) at 0°C, ΔH is +6.01 kJ

However, when 1 mole of H2O(s) is formed from 1 mole of H2O(l), the magnitude of ΔH remains the same but with the opposite sign of it. H2O(l) → H2O(s) ΔH = – 6.01 kJ

The combustion of a sample of aluminium produces 0.250 mol of aluminium oxide and releases 419 kJ of heat at standard conditions.

Al(s) + 3/4O2(g) 1/2Al2O3(s)

(i) Calculate the standard enthalpy of combustion of aluminium.(ii) Determine the enthalpy of formation of Al2O3 and write its thermochemical equation.

ExampleExample

(i) Al(s) + 3/4O2(g) 1/2Al2O3(s)

0.250 mol Al2O3 released 419 kJ heat 0.500 mol Al2O3 released ? = 0.500x419 0.250 = 838 kJ

ΔHo for combustion of Al is -838 kJmol-1

Answer

(ii) ΔHof Al2O3 = 2 x ΔHo

c Al = 2 x (-838) = -1.68 x 103 kJmol-1

Thermochemical equation:2Al(s) + 3/2 O2(g) Al2O3(s) ΔHo

= - 1.68 x 103 kJ

8.3 CALORIMETRY8.3 CALORIMETRY

LEARNING OUTCOMELEARNING OUTCOME1. Define

i. heat capacity , Cii. specific heat capacity, c

2. Calculate heat of reaction in a calorimeter for two possible conditions.

i. Heat of reaction = heat absorb by mediumii. Heat of reaction = heat absorb by calorimeter + heat absorb by medium

CALORIMETRYCALORIMETRY

A method used in the laboratory to measure the heat change of a reaction. Apparatus used is known as the calorimeter Two types of calorimeter: i. Simple calorimeter (constant pressure) ii. Bomb calorimeter (constant volume)

constant–pressure calorimeter

(simple calorimeter)

constant–volume calorimeter

(bomb calorimeter)

ConstantConstant–P–Pressure Calorimeterressure Calorimeter

The outer Styrofoam cup insulate the reaction mixture from the surroundings (it is assumed that no heat is lost to the surroundings)

Heat release by the reaction is absorbed by solution and the calorimeter

Constant–Volume Calorimeter

Heat capacity, CIs the amount of heat required to raise the

temperature of a given quantity of the substance by one degree Celsius (JC1)

Specific heat capacity, c Is the amount of heat required to raise the

temperature of one gram of the substance by one degree Celsius (J g 1C1).

Important Terms in Calorimeter

Heat released by a reaction =

Heat absorbed by surroundings

• Surroundings may refer to the:i. Calorimeter itself or;ii. The medium(e.g. medium) and calorimeter

qreaction= mcΔT or qreaction= CcΔT

Basic Principle in Basic Principle in CalorimeterCalorimeter

Heat released by reaction

= Heat absorbed by calorimeter+water

q = heat released by reaction

mw= mass of water

Cw= specific heat capacity for water

Cc = heat capacity for calorimeter

∆T = temperature change

q = Cc∆T + mwcw∆T

In an experiment, 0.100 g of H2 and excess of O2 were

compressed into a 1.00 L bomb and placed into a calorimeter with heat capacity of 9.08 x 104 J0C1. The initial temperature of the calorimeter was 25.0000C and finally it increased to 25.155 0C. Calculate the amount of heat released in the reaction to form H2O, expressed in kJ per mole.

Example 1

Heat released = Heat absorbed by the calorimeter q = C∆T = (9.08 X 104 J/oc) X (0.1550C) = 14074 J = 14.074 kJ

H2(g) + ½O2(g) → H2O(c)

mole of H2 = 0.100

2.0 = 0.05 mol

Answer

no moles of H2O = no mole of H2

0.05 mol of H2O Ξ 14.074 kJ energy 1 mol H2O released Ξ ? kJ

= 14.074 0.05 = 281.48 kJ

Heat of reaction, ∆H = - 281 kJ mol1

Calculate the amount of heat released in a reaction in an aluminum calorimeter with a mass of 3087.0 g and contains 1700.00 mL of water. The initial temperature of the calorimeter is 25.0°C and it increased to 27.8°C.

Given: - Specific heat capacity of aluminum = 0.553Jg-1 °C-1

Specific heat capacity of water = 4.18 Jg-1 °C-1

Water density = 1.0 g mL-1

ΔT = (27.8 -25.0 )°C = 2.8°C

Example 2

q = mwcwΔT + mcccΔT = (1700.0 g)(4.18 Jg-1 °C-1)(2.8 °C) + (3087.0 g)(0.553 Jg-1 °C-1)(2.8°C) = 2.5 x 104 J = 25.0 kJ

Answer

The heat of neutralization for the following reaction is -56.2 kJmol-1.

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

100.0 ml of 1.50 M HCl is mixed with 100.0 ml of 1.50 M NaOH in a calorimeter having a heat capacity of 15.2 JoC-1. The initial temperature of HCl and NaOH solution are 23.2oC. Calculate the final temperature for this reaction.(ρsolution=1.00gml-1 ; csolution = 4.18 Jg-1oC-1)

Exercise 1

Answer

A calorimeter contains 400 ml of water at 25.0oC. If 600 ml of water at 60.0oC is addedTo it, determine the final temperature.Assume that the heat absorbed bycalorimeter is negligible

Exercise 2

Answer

8.4 HESS’S LAW8.4 HESS’S LAW

LEARNING OUTCOMELEARNING OUTCOME

1. State Hess’s Law2. Apply Hess’s Law to calculate

enthalpy changes using the algebraic method and energy cycle method

Hess’s Law states that when reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in the series of steps.

The enthalpy change depends only on the initial and final states of the reactants and products but is independent of the path taken.

∆H03 ∆Ho

∆Ho1 = ∆Ho

2 + ∆Ho

HESS’S LAWHESS’S LAW

QUESTION :Given the following enthalpies of reaction,

C(s) + O2(g) CO2(g) ∆H = - 393 kJH2(g) + 1/2O2 H2O(g) ∆H= - 286 kJC2H6(g) + 7/2O2(g) 2CO2(g)+ 3H2O(g) ∆H = -1560kJ

Calculate the enthalpy formation for , C2H6(g)

Algebraic MethodAlgebraic Method

C(s) + O2(g) CO2(g) ∆H = - 393 kJ H2(g) + 1/2O2 (g) H2O(g) ∆H= - 286 kJ C2H6(g) + 7/2O2(g) 2CO2(g)+ 3H2O(g) ∆H = -1560kJ

Step 1

☻ List all the thermochemical equations involved

Algebraic MethodAlgebraic Method

)(23)(2 gHCfH

Step 2

☻ Write the enthalpy of formation reaction for C2H6

-84kJ/mol 3H2H1HfH

84kJ- )(62

)(23)(2

_______________________________________________________________

1560kJ3H )(227

)(62)(232(g)2CO (iii)

-286kJ32H )(23)(223

)(23 3)(

kJ -3932 1H )(22)(22)(2 2)(

gOgHCgOHreverse

gOHgOgHii

gCOgOSCi

Step 3

☻Add the given reactions so that the result is the desired reaction.

Given the following enthalpies of reaction,

C(s) + 2F2(g) CF4 ∆H= -680 kJH2(g) + F2(g) 2HF(g) ∆H= -537 kJC2H4(g) + 6F2(g) 2CF4(g) + 4HF(g) ∆H= -2490 kJ

Calculate the enthalpy change for the reaction between carbon and hydrogen to form ethane, C2H4(g) 2C(s) + 2H2(g) C2H4(g)

2C(s) + 4F2(g) 2CF4 ∆H= 2x(-680 kJ)

H2(g) + F2(g) 2HF(g) ∆H= 2x(-537 kJ)

2CF4(g) + 4HF(g) C2H4(g) + 6F2(g) ∆H= +2490 kJ___________________________________________ 2C(s) + 2H2(g) C2H4(g)

∆ H = +2490 + 2(-680) + 2(-537) = +56 kJ/mol

Given :H2(g) + F2(g) 2HF(g) ΔH = -573 kJ

C(s) + 2F2(g) CF4(g) ΔH = -680 kJ

2C(s) + 2H2(g) C2H4(g) ΔH = +52.3 kJ

Using algebraic method, calculate the enthalpy change, ΔH for the following reaction :

C2H4(g) + 6F2(g) 2CF4(g) + 4HF(g)

2 C ( s ) + 3 H2 ( g ) C2 H6 ( g )

2 C O2 ( g ) + 3 H2 O ( g )

2 O2 ( g ) 3 / 2 O2 ( g ) H

7 / 2 O2 ( g ) O

3 = 2 ( - 3 9 3 )

= 3 ( - 2 8 6 )

= - ( - 1 5 6 0 )

Draw the energy cycle and apply Hess’s Law to calculate the unknown value.

Energy Cycle MethodEnergy Cycle Method

∆Hof = 2(∆Ho

1) + 3(∆Ho2) +

∆Ho3

= -786-858+1560= -84 kJ mol-1

Given: S(s) + O2(g) SO2(g) ΔH = 287 kJSO2(g) + ½O2(g) SO3(g) ΔH = 92 kJ

Calculate ΔHf for the following reaction using energy cycle method:

S(s) + 3/2O2(g) SO3(g)

EXERCISE

8.5 8.5 Born-Haber CycleBorn-Haber Cycle

LEARNING OUTCOMELEARNING OUTCOME1. Define

i. lattice energyii. electron affinity

1. Explain the following effects on the magnitude of lattice energyi. ionic chargeii. ionic radii

3. Explain the dissolution process of ionic solids4. Construct Born-Haber cycle for simple ionic

solids using energy cycle diagram and energy level diagram

5. Calculate enthalpy changes using Born-Haber cycle 67

Lattice energy formation is the energy released when one mole of a solid (ionic compound) is formed from its gaseous ions

Na+(g) + Cl-(g) NaCl(s) Hlattice = -771 kJ

(lattice formation)

Lattice Energy, Hlattice

Lattice energy dissociation is the energy required to completely separate one mole of a solid (ionic compound) into its gaseous ions

NaCl(s) Na+(g) + Cl-(g) Hlattice = +771 kJ

(lattice dissociation)

Lattice Energy, Hlattice

The magnitude of lattice energy increases as:-۩ the ionic charges increase ☞ ions attract each other more strongly

۩ the ionic radii decrease ☞ they get closer together

Effect on the magnitude of lattice energy

H for MgO is more negative than H for Na2O because Mg2+ is smaller in size and has bigger charge than Na+

Hºlattice (MgO) > Hºlattice (Na2O)

Electron AffinityElectron AffinityD

Definition Electron Affinity*

The amount of energy change to added 1 mole of electron into 1 mole of gaseous atoms or ions in their ground state.

* These reactions usually exothermic (release energy) because when an electron is added to a neutral atom, it will experience an attraction of nucleus and release an amount of energy.

Atom(g) + e– ion–(g) E = EA1 = -ve

However, affinity does not always release energy. In some cases, affinity requires energy.

Example: Formation of oxide, O2-;

EA1=-ve

O-(g)+ e- O2- (g) EA2=+ve

O(g)+ e- O- (g)

* so the values of EA are generally negative.

* the higher (more negative) the EA, the more easily it accepts an electron.

KEEP IN MINDKEEP IN MINDFirst Ionisation Energy- Is the minimum energy required to

remove an electron from a neutral gaseous atom in its ground state.

X(g) X+(g) + e-

Second Ionisation Energy- Is the energy required to remove an

electron from a gaseous positive ion in its ground state.

X+(g) X2+

(g) + e-

Dissolution is the process by which a solid or liquid forms a solution in a solvent.

Occur when an ionic solid dissolve in water Water molecules are polar

Most ionic crystals are soluble in water Ions in the solid crystal can be separated from each other and

converted to the gaseous ions (Hlattice) The attraction forces between gaseous ions and polar water

molecules cause the ions to be surrounded by water molecules (Hhyd) Hsoln = Hlattice dissociation+

Dissolution Process of Ionic Solid

NaCl(s) Na+(aq) + Cl-(aq)

Lattice Energy

dissociation Heat of Hydration

Heat of solution

Na+(g) and Cl-

NaCl(s) Na+(aq) and Cl- (aq)

Based on the data given below:

ΔHohydration Na+ = -390 kJ mol-1

ΔHohydration Cl- = -380 kJ mol-1

ΔHosolution NaCl = +6 kJ mol-1

(i) Construct energy cycle diagram to represent the dissolution of NaCl

(ii) Calculate the lattice energy of NaCl

(i) Na+(g) + Cl-(g) NaCl(s)

(aq) + Cl-(aq)

(ii) = ΔHohydrationNa+ + ΔHo

hydration Cl- + (-ΔHo

solutionNaCl)

= -390 + (-380) – 6 = -776 kJ mol-1

ΔHolattice

ΔHo= -6 kJ

ΔHo= -380 kJ

ΔHo= -390 kJ

Lattice energy

Based on the data given below:

(i) Construct an energy cycle diagram to represent the dissolution of LiCl

(ii) Calculate the lattice energy of LiCl

enthalpy hydration of Li+

-510 kJ/mol

enthalpy hydration of Cl-

-413 kJ/mol

enthalpy of solution of LiCl

-77kJ/mol

Energy cycle for ionic compounds

Connects enthalpy of formation with lattice energy

Born-Haber Cycle

Energy Cycle Diagram

Find the lattice energy from the followingdata:

Enthalpy of atomisation of potassium : +90 kJ mol1First ionisation energy of potassium : +418kJ mol1Atomisation energy of chlorine : +121kJ mol1Electron affinity of chlorine : –364 kJ mol1Enthalpy of formation of potassium chloride : –436 kJ mol1

K(s) + 1/2Cl2(g) KCl(s)

K(g) Cl(g)

K+ (g) + Cl- (g)

ΔHf= -436 kJ

ΔHa=+ 90 kJ

IE= +418 kJ

ΔHa= +121 kJ

EA = -364 kJ

ΔHf = ΔHa(K)+ IE + ΔHa(Cl) + EA + Δhlatt

-436 = 90+418+121-364+ ΔHlattΔHlatt = -701 kJ/mol

ΔHlattice

Construct a Born-Haber cycle for the formation of magnesium fluoride, MgF2 by using the data below :

Based on the Born-Haber cycle, determine the lattice energy of magnesium fluoride.

Enthalpy atomisation of Mg +148 kJ/molEnthalpy atomisation of F +159 kJ/mol

First ionization of Mg +738 kJ/molSecond ionization of Mg +1450 kJ/mol

Electron affinity of F -328 kJ/molEnthalpy formation of MgF2 -1123 kJ/mol

Energy level diagram of Energy level diagram of Born-Haber cycleBorn-Haber cycle

In the Born-Haber cycle energy diagram, by convention, positive values are denoted as going upwards, negative values as going downwards.

Example :Example :

Given;i. Enthalpy of formation NaCl = - 411 kJmol-1

ii. Enthalpy of sublimation of Na= + 108 kJmol-1

iii. First ionization energy of Na = + 500 kJmol-1

iv. Enthalpy of atomization of Cl = + 122 kJmol-1

v. Electron affinity of Cl = - 364 kJmol-1

vi. Lattice energy of NaCl = ?

Consider the enthalpy changes in the formation of sodium chloride.

A Born-Haber energy cycle diagram for NaCl

Na(s) + ½ Cl2(g)

Na(g) + ½ Cl2(g)

NaCl(s)

energy

Na(g) + Cl(g)

Na+(g) + e + Cl(g)

Na+(g) + Cl- (g)

Ionisation Energy of Na

Electron Affinity of Cl

Lattice energy

Hf NaCl-ve

Calculation:Calculation:

kJ777HkJ364kJ122kJ500kJ108kJ411H

EAHIEHHH

HEAHIEHH

lattice

)Cl(aS0flattice

lattice)Cl(aS0f

From Hess’s Law:Hf NaCl = HaNa + HaCl +IENa + EACl + Lattice Energy

END OF SLIDE SHOW

8.0 thermochemistry (student's copy)

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