Chapter 10 Thermochemistry - Georgia Institute of...

OFB Chap. 10 110/28/2004

Bond Enthalpy

• Bond Enthalpy (∆H Dissociation) is the enthalpy change in a reaction in which a chemical bond is broken in the gas phase.

• Bond Energy is the energy needed to break 1 mol of the particular bond.– Energy is released when bonds

are formed (exothermic)– Energy must be supplied when

bonds are broken (endothermic)

OFB Chap. 10 210/28/2004

Bond Enthalpies• The breaking of chemical bonds in

stable substances often generates highly reactive products (or intermediates)CH4 → ·CH3 + ·H ∆H°= + 439 kJmol-1

• Called Bond Enthalpy• OFB Table 10-3 p 462 gives Average

Bond Enthalpies

OFB Chap. 10 310/28/2004

• Average Bond Enthalpies

C2H6 → ·C2H5 + ·H ∆H°= + 410 kJmol-1

CHF3 → ·CF3 + ·H ∆H°= + 429 kJmol-1

CHCl3 → ·CCl3 + ·H ∆H°= + 380 kJmol-1

CHBr3 → ·CBr3 + ·H ∆H°= + 377 kJmol-1

average ∆H°C-H = + 412 kJmol-1

• Applications of Bond Enthalpy– Given a reaction– 1st Step is break all bonds to give free atoms in

the gas phase (Endothermic)– 2nd Step is form new bonds for the products.

(Exothermic)

OFB Chap. 10 410/28/2004

CCl2F2 + 2H2 → CH2Cl2 + 2HF

∆Hr = ? = H products – H reactants

= –114 kJ mol-1

• Applications of Bond Enthalpy– Given a reaction– 1st Step is break all bonds to give free atoms in

the gas phase (Endothermic)– 2nd Step is form new bonds for the products.

(Exothermic)

OFB Chap. 10 510/28/2004

Exercise 10-10• Estimate the Standard Enthalpy of

Reaction for the gas-phase reaction that forms methanol from methane and water and compare it with the ∆H°r obtained from the data in Appendix D.

CH4(g) + H2O(g)→CH3OH(g) + H2(g)

OFB Chap. 10 610/28/2004

CH4(g) + H2O(g)→CH3OH(g) + H2(g)

H

C

H H

H

+O

H

H

H

HH

+ H HC

O H

-2489 kJmol-1

Exothermic+2578 kJmol-1

Endothermic

3 C-H or 3x(-413)1 O-H or 1x(-463)1 C-O or 1x(-351)1 H-H or 1x(-436)

4 C-H or 4x4132 O-H or 2x463

FormedBroken

∆H= ∆H bonds broken + ∆H bonds formed

= +2578 +(-2489)= + 89 kJmol-1 (estimated)

Using Appendix D ∆H = 116 kJmol-1

OFB Chap. 10 710/28/2004

PRS Quiz

Q1 Which of the following bonds has the lowest enthalpy?

1. C−C

2. C=C

3. C=O

4. C=N

OFB Chap. 10 810/28/2004

PRS QuizFor water near its triple point, ∆Hfus = 6.0 kJ mol–1 and ∆Hvap = 45.0 kJ mol–1.

Thus ∆H for the sublimation process, H2O(s) → H2O(g), is

1. 6.0 kJ mol–1

2. 39.0 kJ mol–1

3. 45.0 kJ mol–1

4. 51.0 kJ mol–1

H2O(s) → H2O(l)

H2O(l) → H2O(g)

Add

H2O(s) → H2O(g)

OFB Chap. 10 910/28/2004

PRS QuizEstimate the Enthalpy of Reaction

for the following reaction in the gas phase.

BBr3 + BCl3 → BBr2Cl + BCl2Br

Average Bond Enthalpies (kJ mol-1)B―Br ~ 250B―Cl ~ 300

1. 0 kJ mol-1

2. 50 kJ mol-1

3. 100 kJ mol-1

4. Can not be determined

Breaking and forming the same number of B―Br and B―Clbonds (3 each)

OFB Chap. 10 1010/28/2004

Chapter 10Thermochemistry

• First Law of ThermodynamicsThe change in the internal energy of a system is equal to the work done on it plus the heat transferred to it.

∆ E = q + w• Second Law of Thermodynamics

In a real spontaneous process the Entropyof the universe (meaning the system plus its surroundings) must increase.

∆ Suniverse > 0• Third Law of Thermodynamics

In any thermodynamic process involving only pure phases at equilibrium, the entropy change, ∆ S, approaches zero at absolute zero temperature; also the entropy of a crystalline substance approaches zero.

∆ S = 0 at 0 K

OFB Chap. 10 1110/28/2004

First Law of Thermodynamics

∆E = q + w• q was previously defined as the

Heat Absorbed by a system• If q > 0 , heat is absorbed• If q < 0 , heat is given off• w is the work done on the body

Recall• w = F x d• w = ∆EKinetic=∆ (1/2mv2)• w = ∆Epotential= mg∆h

In chemistry this kind of mechanical work is Pressure-Volume work (P-V)

OFB Chap. 10 1210/28/2004

• Force exerted by heating = P1A– Where P1 is the pressure inside the vessel– Where A is the Area of the piston

• Pext = P1 if balanced• ∆V = A ∆h• w = −Pext ∆V

– Units are atm·L or L atm– Where 1 L atm = 101.325 Joules

OFB Chap. 10 1310/28/2004

w = -Pext ∆V

• Expansion– ∆V > 0 therefore w < 0– The system does work on the

surroundings• Compression

– ∆V < 0 therefore w > 0– The surroundings have done work on

the system• Try Example and Exercise 10-11

– Calculate the work done on a gas and express it in Joules

OFB Chap. 10 1410/28/2004

PRS Quiz Typical exam questions (1-4)

A gas is compressed from 40L to 15L at a constant pressure of 5 atm. In the course of this compression 10 kJ of energy is released

Q1 qQ2 wQ3 ∆EQ4 ∆H

OFB Chap. 10 1510/28/2004


Q1 The heat q for this process is1. 125 kJ2. –125 kJ3. −10 kJ4. 10 kJ Answer 3 “10 kJ of

energy is released”

Recall: If q is negative, then heat is evolved or given off or released

OFB Chap. 10 1610/28/2004


Q2 The work w for this process is

1. 125 L atm2. -125 L atm3. -10 L atm4. 10 L atm

OFB Chap. 10 1710/28/2004


Q2 The work w for this process is 1. 125 L atm2. -125 L atm3. -10 L atm4. 10 L atm

Answer 1. w = – Pext (∆V)Note the compression, ∆V < 0 therefore ∆W > 0

∆V = V final – V initial

w = – Pext (∆V) = – Pext (15-40)

= – 5(– 25) = + 125 L atm

OFB Chap. 10 1810/28/2004


Q3 ∆ E for the process is1. –22.7 kJ2. 2.7 kJ3. 22.7 kJ4. 125 kJ

1st Law of thermodynamics

∆E = q + w

= (-10 kJ) + (125 L atm x 101.325 J/L atm)

= – 10 kJ + 12666 J

= – 10 kJ + 12.7 kJ

= + 2.7 kJ Answer 2

q from Question 1 = –10kJ

w from Question 2 =

125 L atm (note units)

1st Law ∆E = q + w

OFB Chap. 10 1910/28/2004


Q4 ∆H for the process is1. -10 kJ2. -7.3 kJ3. 2.7 kJ4. 10 KJ

Answer 1

Recall qp = ∆ H enthalpy

From Q1 q at constant pressure is minus 10 kJ

OFB Chap. 10 2010/28/2004

Enthalpy and Energy• 1st Law of Thermodynamics

∆E = q + w – At constant volume ∆V =0– Thus w = -Pext ∆V = 0

∆E = qv (constant volume)• Recall as previously stated

∆H = qp (constant pressure)• Enthalpy is defined as

H = E + PV• or Expressed as a change in

Enthalpy∆H = ∆E + ∆(PV)

• Rearrange∆E = ∆H - ∆(PV)

• If PV = nRT is the ideal gas law∆(PV) ≈ ∆(nRT)= RT ∆ng

OFB Chap. 10 2110/28/2004

Chapter 10Thermochemistry

∆(PV) = RT ∆ng

• ∆ng is the change in the total chemical amount of gases in a reaction

• ∆ng = Total moles of product gases minus the Total moles of reactant gases

∆E = ∆H - RT∆ng

∆H = ∆E + ∆(PV)• Rearrange

∆E = ∆H - ∆(PV)

OFB Chap. 10 2210/28/2004

ExampleCalculate the internal energy @ 25°C for the following reaction

kJ 110.5∆H

1CO(g)(g)2O21)C(graphite

−=

→+

Strategy– Step 1 calculate ∆(PV)– Step 2 calculate ∆E at 25°C

∆H = ∆E + ∆(PV)

OFB Chap. 10 2310/28/2004

Example• Calculate the internal energy @ 25°C

for the following reaction

kJ110.5∆H

1CO(g)(g)O21)C(graphite 2

−=

→+

• Solution– Step 1 calculate ∆(PV)

• ∆(PV) ≈ RT ∆ng

• ∆ng = 1 – ½ = ½ mol• ∆(PV) ≈ RT ∆ng= (8.35 kJmol-1) (298 K) (1/2 mol)= 1.24 kJ

– Step 2 calculate ∆E at 25°C• ∆E = ∆H – ∆ (PV)• = – 110.5 – (1.24)• = – 111.74 kJ

OFB Chap. 10 2410/28/2004

PRS Quiz

In which of the following reactions will there be the largest difference between ∆H and ∆E?

1. Pb (s) + I2 (s) → PbI2 (s)2. N2 (g) + O2 (g) → 2 NO (g)3. 4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O(g)4. N2(g) + 3 H2(g) → 2 NH3(g)

∆H = ∆E + ∆(PV)∆H − ∆E = ∆(PV)

OFB Chap. 10 2510/28/2004

PRS QuizIn which of the following reactions will there be the largest difference between ∆H and ∆E?

1. Pb(s) + I2(s) → PbI2(s)2. N2(g) + O2(g) → 2 NO(g)3. 4 NH3(g) + 7 O2(g) → 4 NO2(g) +

6 H2O(g)4. N2(g) + 3 H2(g) → 2 NH3(g)

∆(PV) = RT ∆ng

∆H − ∆E = ∆(PV)

OFB Chap. 10 2610/28/2004

Chapter 10Thermochemistry Summary

if

pressureconstant at CapacityHeat denotes Cp where

p

T-T∆T

∆Tabsorbedheat

∆TqCCapacityHeat

=

===

11-pp molJK are units

nCc

CapacityHeat Molar

−= 1-1-ps gJK are units

mCc

CapacityHeat Specific

=

( ) ( )2if p221if p11

2 p221 p11

21

TTcnTTcn∆Tcn∆Tcn

qqeTemperaturionEquilibrat

−−=−−=−=

OFB Chap. 10 2710/28/2004

• Enthalpy, H.

qp= ∆H(note p denotes constant pressure)

• Enthalpy is a state property, which means it depends on its initial and final state, not the path to get there.

– If ∆H positive = ∆H > 0 = q > 0 Means heat is absorbedAnd is called Endothermic– If ∆H negative = ∆H < 0 = q < 0 Means heat is given offAnd is called Exothermic

•Hess’s Law: If two or more chemical equations are added to give a new equation, then adding the enthalpies of the reactions that they represent gives the enthalpy of the new reaction.

OFB Chap. 10 2810/28/2004

Standard State Enthalpies• ∆H° is the sum of products minus the

sum of the reactants• For a general reaction

a A + b B → c C + d D

(B)H(A)H

(D)H(C)H∆Hof

of

of

of

o

∆∆

∆∆

ba

dc

−−

+=

Bond Enthalpies∆H= ∆H bonds broken + ∆H bonds formed

First Law of ThermodynamicsThe change in the internal energy of a system is equal to the work done on it plus the heat transferred to it.

∆ E = q + w

OFB Chap. 10 2910/28/2004

Enthalpy and Energy• Enthalpy is defined as

H = E + PV• or Expressed as a change in

Enthalpy∆H = ∆E + ∆(PV)

• For gases, If PV = nRT is the ideal gas law

∆(PV) ≈ ∆(nRT)= RT ∆ng

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