Chapter 6

Thermochemistry

Pressure –Volume WorkPV work is work caused by a volume change against an external pressure.

When gases expand, ΔV is +, but the system is doing work on the surroundings, so wgas is ─.

As long as the external pressure is kept constant

─ Workgas = (External Pressure) x (Change in Volumegas)

w = ─PΔV

To convert the units to joules use 101.3 J = 1 atm·L

ΔE (J) = q + w

ΔE (J) = J + (-P)(ΔV)

Joule (J) = kg-m2

sec2

Newton (N) = kg-msec2

Pascal (Pa) = N m2 = kg

m-sec2

kg-m2

sec2 + kg m-sec2

x m3 Joule (J) =

+kg-m2

sec2 kg-m2

sec2

( )

If a balloon is inflated from 0.100 L to 1.85 L against an external pressure of 1.00 atm, how much work is done?

A certain process results in a gas system releasing 68.3 kJ of

energy. During the process, 15.8 kcal of heat is released by

the system. If the external pressure is kept constant at 1.00

atm and the initial volume of the gas is 10.0 L, what is the

final volume of the gas?

A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is released by the system. How

much work was done by the process?

ΔE = −68.3 kJ, q = −15.8 kcal, w = ?

1 kJ = 1000 J, 1 cal = 4.18 J

q, ∆E w ΔE = q + ww = ΔE - q

w = −PΔV, 101.3 J = 1 atm·L, V1 = 10.0 L, P = 1.00 atm V2 = ?

A certain process results in a gas system releasing 68.3 kJ of energy. During the process, 15.8 kcal of heat is released by the system. If the external pressure is kept constant at 1.00 atm and the initial volume of

the gas is 10.0 L, what is the final volume of the gas?

w, P ∆V ∆V, V1 V2

w = - (2.26 x 103 J) x 1 atm ·L101.3 J

= - 22.7 atm· L

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3

Exchanging Energy BetweenSystem and Surroundings

Exchange of heat energy

q = (mass) x (specific heat) x (ΔTemperature)

Exchange of work

w = −(Pressure) x (ΔVolume)

Measuring ΔE, Calorimetry at Constant Volume

Because ΔE = q + w, we can determine ΔE by measuring q and w.

In practice, it is easiest to do a process in such a way that there is no change in volume, so w = 0. (w = −PΔV)

At constant volume, ΔEsystem = qsystem.

Measuring ΔE, Calorimetry at Constant Volume

In practice, it is not possible to observe the temperature

changes of the individual chemicals involved in a reaction – so instead, we measure the temperature change in the surroundings.

We use insulated, controlled surroundings.

qsystem = −qsurroundings

Bomb Calorimeter

Used to measure ΔE

The heat capacity of the calorimeter is the amount of heat absorbed by the calorimeter for each degree rise in temperature and is called the calorimeter constant.

Ccal, kJ/ºC

The “surroundings” is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water.

qsurroundings = qcalorimeter = ─qsystem

When 1.010 g of sugar is burned in a bomb calorimeter, the temperature rises from 24.92 °C to 28.33 °C. If Ccal = 4.90 kJ/°C,

find ΔE for burning 1 mole (in kJ/mol).

1.010 g C12H22O11, T1 = 24.92 °C, T2 = 28.33 °C, Ccal = 4.90 kJ/°C ΔErxn = ?

Ccal, ΔT qcal qrxn qrxn, mol ΔE

qcal = Ccal x ∆T = -qrxn

When 1.010 g of sugar is burned in a bomb calorimeter, the temperature rises from 24.92 °C to 28.33 °C. If Ccal = 4.90 kJ/°C,

find ΔE for burning 1 mole (in kJ/mol).

1.010 g C12H22O11, T1 = 24.92 °C, T2 = 28.33 °C, Ccal = 4.90 kJ/°C ΔErxn = ?

Ccal, ΔT qcal qrxn qrxn, mol ΔE

qcal = Ccal x ∆T = -qrxn MM C12H22O11 = 342.3 g/mol

qrxn = -16.7 kJ

During the burning of one mole, i.e.,

C11H22 O11(g) + 11 O2(g) → 11 CO2(g) + 11 H2O(g)

qreaction = −5660 kJ/mol,

and 5660 kJ of energy are given off to the surroundings.

What does ∆E = -5660 kJ/mol mean???

C11H22 O11(g) + 11 O2(g) → 11 CO2(g) + 11 H2O(g) + 5660 kJ

Enthalpy

The enthalpy, H, of a system is the sum of the internal energy of the system and the product of pressure and volume.

H is a state function.H = E + PV

The enthalpy change, ΔH, of a reaction is the heat evolved in a reaction at constant pressure.

ΔHreaction = qreaction at constant pressure = qp

Usually ΔH and ΔE are similar in value, the difference is largest for reactions that produce or use large quantities of gas.

The Meaning of Enthalpy (H)

w = - PΔV

ΔH = ΔE + PΔV

qp = ΔE + PΔV = ΔH

H = E + PVwhere H is enthalpy

ΔH ≈ ΔE in1. Reactions that do not involve gases

2. Reactions in which the number of moles of gas does not change.

3. Reactions in which the number of moles of gas does change but q is >>> PΔV.

Endothermic and Exothermic Reactions

When ΔH is ─, heat is being released by the system.

Reactions that release heat are called

exothermic reactions.

When ΔH is +, heat is being absorbed by the system.

Reactions that absorb heat are called

endothermic reactions.

Exothermic and Endothermic Processes

Molecular View of Exothermic Reactions

During the course of a reaction, old bonds are broken and new bonds are made.

The products of the reaction have

less chemical potential energy than the reactants.

For an exothermic reaction, the surrounding’s temperature rises due to release of thermal energy by the reaction.

This extra thermal energy comes from the conversion of some of the chemical potential energy in the reactants into kinetic energy in the form of heat.

Molecular View of Endothermic Reactions

During the course of a reaction, old bonds are broken and new bonds are made.

The products of the reaction have more chemical potential energy than the reactants.

To acquire this extra energy, some of the thermal energy of the surroundings is converted into chemical potential energy stored in the products.

In an endothermic reaction, the surrounding’s temperature drops due to absorption of some of its thermal energy by the reaction.

Enthalpy of Reaction

The enthalpy change in a chemical reaction is an extensive property.

By convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written.

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH = −2044 kJ

How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)?

mol kJ kg g

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH = −2044 kJ

Measuring ΔHCalorimetry at Constant Pressure

Reactions done in aqueous solution are at constant pressure - open to the atmosphere.

The calorimeter is often nested foam cups containing the solution.

qreaction = ─ qsolution = ─(masssolution x Cs, solution x ΔT)

ΔHreaction = qconstant pressure = qreaction

To get ΔHreaction per mol, divide by the number of moles.

The reaction takes place inside an insulated vessel outfitted with a loose-fitting top, a thermometer, and a stirrer.

Measuring the temperature change that accompanies the reaction makes it possible to calculate ΔH

Measuring ΔHCalorimetry at Constant Pressure

What is ΔHrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in

100.0 mL of solution and changes the temperature from 25.6 °C to 32.8 °C?

0.158 g Mg, 100.0 mL, T1 = 25.6 °C, T2 = 32.8 °C, Cs = 4.18 J/g·°C ΔH, kJ/mol = ?

Cs, ΔT, m qsol’n qrxn qrxn, mol ΔH

What is ΔHrxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in

100.0 mL of solution and changes the temperature from 25.6 °C to 32.8 °C?

0.158 g Mg, 100.0 mL, T1 = 25.6 °C, T2 = 32.8 °C, Cs = 4.18 J/g·°C ΔH, kJ/mol = ?

Cs, ΔT, m qsol’n qrxn qrxn, mol ΔH

qrxn = -3.0 x 103 J

Relationships Involving ΔHrxn

When reaction is multiplied by a factor, ΔHrxn is multiplied by that factor. ΔHrxn is extensive

2 C(s) + 2 O2(g) → 2 CO2(g) ΔH = 2(−393.5 kJ) = −787.0 kJ

If a reaction is reversed, then the sign of ΔH is changed.

CO2(g) → C(s) + O2(g) ΔH = +393.5 kJ

C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ

Relationships Involving ΔHrxn - Hess’s Law

If a reaction can be expressed as a series of steps, then the ΔHrxn for the overall reaction is the sum of the heats of reaction for each step.

A +2B ---------> C C ---------> 2DA +2B ---------> 2D

Standard Conditions

The standard state is the state of a material at a defined set of conditions

pure gas at exactly 1 atm pressure

pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest (usually 25 °C)

substance in a solution with concentration 1 M

The standard enthalpy change, ΔH°, is the enthalpy change when all reactants and products are in their standard states.

Practice – Hess’s LawGiven the following information:

Cu(s) + Cl2(g)➜CuCl2(s) ΔH° = −206 kJ2 Cu(s) + Cl2(g) ➜2 CuCl(s) ΔH° = − 36 kJ

Calculate the ΔH° for the reaction below:

! Cu(s) + CuCl2(s)➜ 2 CuCl(s) ΔH° = ? kJ

CuCl2(s) ➜ Cu(s) + Cl2(g) ΔH° = +206 kJ2 Cu(s) + Cl2(g)➜2 CuCl(s) ΔH° = − 36 kJ

Cu(s) + CuCl2(s)➜2 CuCl(s) ΔH° = +170. kJ

Example: Hess’s LawGiven the following information:

2 NO(g) + O2(g) ➜ 2 NO2(g) ΔH° = −116 kJ

2 N2(g) + 5 O2(g) + 2 H2O(l) ➜ 4 HNO3(aq) ΔH° = −256 kJ

N2(g) + O2(g) ➜ 2 NO(g) ΔH° = +183 kJ

Calculate the ΔH° for the reaction below:

3 NO2(g) + H2O(l) ➜ 2 HNO3(aq) + NO(g) ΔH° = ?

3 NO2(g) + H2O(l) ➜ 2 HNO3(aq) + NO(g) ΔH° = ?

2 NO(g) + O2(g) ➜ 2 NO2(g) ΔH° = −116 kJ 2 N2(g) + 5 O2(g) + 2 H2O(l) ➜ 4 HNO3(aq) ΔH° = −256 kJ

N2(g) + O2(g) ➜ 2 NO(g) ΔH° = +183 kJ

3 NO2(g) + H2O(l)➜2 HNO3(aq) + NO(g) ΔH° = − 137 kJ

[3 NO2(g) ➜ 3 NO(g) + 1.5 O2(g)] ΔH° = (+174 kJ)

[1 N2(g) + 2.5 O2(g) + 1 H2O(l)➜ 2 HNO3(aq)] ΔH° = (−128 kJ)

[2 NO(g) ➜ N2(g) + O2(g)] ΔH° = (−183 kJ)

Example: Hess’s Law

Standard ConditionsThe standard state is the state of a material at a defined set of

conditions

pure gas at exactly 1 atm pressurepure solid or liquid in its most stable form at exactly 1 atm pressure and

temperature of interest (usually 25 °C)substance in a solution with concentration 1 M

The standard enthalpy change, ΔH°, is the enthalpy change when all reactants and products are in their standard states.

The standard enthalpy of formation, ΔHf°, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements

the elements must be in their standard statesthe ΔHf° for a pure element in its standard state = 0 kJ/mol

Standard Enthalpies of Formation

Formation Reactions

Reactions of elements in their standard state to form 1 mole of a pure compound.

If you are not sure what the standard state of an element is, find the form that has a ΔHf° = 0

Because the definition requires 1 mole of compound be made, the coefficients of the reactants may be fractions.

Writing Formation ReactionsWrite the Formation Reaction for CO(g)

The formation reaction is the reaction between the elements in the compound, which are C and O.

C + O → CO(g)The elements must be in their standard state

There are several forms of solid C, but the one with ΔHf° = ø is graphite.

Oxygen’s standard state is the diatomic gas.C(s, graphite) + O2(g) → CO(g)

The equation must be balanced, but the coefficient of the product compound must be 1. Use whatever coefficient in front of the reactants is necessary to make the atoms on both sides equal without changing the product coefficient.

C(s, graphite) + ½ O2(g) → CO(g)

Write the formation reactions for the following:

CO2(g)

Al2(SO4)3(s)

C(s, graphite) + O2(g)➜CO2(g)

2 Al(s) + 3/8 S8(s, rhombic) + 6 O2(g)➜Al2(SO4)3(s)

Calculating Standard Enthalpy Change for a Reaction

Calculating Standard Enthalpy Change for a Reaction

Any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products.

The ΔH° for the reaction is then the sum of the ΔHf° for the component reactions.

ΔH°reaction = Σ n ΔHf°(products) − Σ n ΔHf°(reactants)

Σ means sumn is the coefficient of the reaction

ΔH° = [(ΔHf° CO2(g) + 2·ΔHf°H2O(g))− (ΔHf° CH4(g) + 2·ΔHf°O2(g))]

C(s, graphite) + 2 H2(g) → CH4(g) ΔHf°= − 74.6 kJ/mol CH4

ΔH° = [((−393.5 kJ)+ 2(−241.8 kJ)− ((−74.6 kJ)+ 2(0 kJ))] = −802.5 kJ

H2(g) + ½ O2(g) → H2O(g) ΔHf°= −241.8 kJ/mol H2O

CH4(g) → C(s, graphite) + 2 H2(g) ! ΔH° = + 74.6 kJ

C(s, graphite) + O2(g) → CO2(g) ΔHf°= −393.5 kJ/mol CO2

2 H2(g) + O2(g) → 2 H2O(g) ! ! ΔH° = −483.6 kJ

Calculate the Standard Enthalpy Change for CH4(g)+ 2 O2(g)→ CO2(g) + 2 H2O(g)

Calculate the Standard Enthalpy Change for CH4(g)+ 2 O2(g)→ CO2(g) + 2 H2O(g)

ΔH°reaction = Σ n ΔHf°(products) − Σ n ΔHf°(reactants)

Calculate the enthalpy change in the reaction 2 C2H2(g) + 5 O2(g) ➜ 4 CO2(g) + 2 H2O(l)

1. Write formation reactions for each compound and determine the ΔHf° for each

2 C(s, gr) + H2(g) ➜ C2H2(g) ΔHf° = +227.4 kJ/mol

C(s, gr) + O2(g) ➜ CO2(g) ΔHf° = −393.5 kJ/mol

H2(g) + ½ O2(g) ➜ H2O(l) ΔHf° = −285.8 kJ/mol

2. Arrange equations so they add up to desired reaction

2 C2H2(g) ➜ 4 C(s) + 2 H2(g) !ΔH° = 2(−227.4) kJ

4 C(s) + 4 O2(g) ➜ 4CO2(g)! ΔH° = 4(−393.5) kJ

2 H2(g) + O2(g) ➜ 2 H2O(l) ΔH° = 2(−285.8) kJ

2 C2H2(g) + 5 O2(g) ➜ 4 CO2(g) + 2 H2O(l) ΔH = −2600.4 kJ

Calculate the enthalpy change in the reaction 2 C2H2(g) + 5 O2(g) ➜ 4 CO2(g) + 2 H2O(l)

ΔH°reaction = Σ n ΔHf°(products) − Σ n ΔHf°(reactants)

ΔHrxn = [(4•ΔHCO2 + 2•ΔHH2O) – (2•ΔHC2H2 + 5•ΔHO2)]

ΔHrxn = [(4•(−393.5) + 2•(−285.8)) – (2•(+227.4) + 5•(0))]

ΔHrxn = −2600.4 kJ

Calculate the enthalpy change in the reaction 2 C2H2(g) + 5 O2(g) ➜ 4 CO2(g) + 2 H2O(l)

Calculate the ΔH° for decomposing 10.0 g of limestone, CaCO3, under standard conditions.

CaCO3(s) → CaO(s) + O2(g)

Calculate the ΔH° for decomposing 10.0 g of limestone, CaCO3, under standard conditions.

CaCO3(s) → CaO(s) + O2(g)

ΔHºrxn = +572.7 kJ