OFB Chap. 10 13/6/2003

Chapter 10Thermochemistry

• 10-1 Heat• 10-2 Calorimetry• 10-3 Enthalpy• 10-4 Standard-State Enthalpies• 10-5 Bond Enthalpies• 10-6 The First Law of

Thermodynamics

OFB Chap. 10 23/6/2003

Chapter 10Thermochemistry

•Heat Capacity, Cp , is the amount of heat required to raise the temperature of a substance by one degree at constant pressure.

•Cp is always a positive number•Q and ∆T must be both negative or positive•If q is negative, then heat is evolved or given off and the temperature decreases•If q is positive, then heat is absorbed and the temperature increases

OFB Chap. 10 33/6/2003

Heat Capacity (continued)• Cp units are energy per temp. change• Cp units are Joule/oK or JK-1

• Units are J/K/mole or JK-1mol-1

• Molar heat capacity

OFB Chap. 10 43/6/2003

Specific Heat Capacity• The word Specific before the name

of a physical quantity very often means divided by the mass

4.18H2O0.719O2

0.739SiO2

0.140Hg(l)Cs

• Specific Heat Capacity is the amount of heat required to raise the temperature of one gram of material by one degree Kelvin (at constant pressure)

OFB Chap. 10 53/6/2003

OFB Chap. 10 63/6/2003

Chapter 10Thermochemistry

• Example (not in book)a.) Calculate the molar heat capacity of quartz (SiO2)b) Calculate the amount of heat required to raise 45.0 Kg of rock (quartz) by 15°C.

• Strategya) Use relationship cp=Mcs

b) Use q=mcs∆T

OFB Chap. 10 73/6/2003

Example (not in book)a.) Calculate the molar heat capacity of quartz (SiO2), if csSiO2=0.739 J/K/gb) Calculate the amount of heat required to raise 45.0 Kg of rock (quartz) by 15°C.

Solutiona) Find the molar mass of quartz

and Use relationship cp=Mcs

b) Use q= mcs∆T

OFB Chap. 10 83/6/2003

Exercise 10-1• Exactly 500.0 kJ of heat is absorbed

at a constant pressure by a sample of gaseous helium. The temperature increases by 15.0K.a.) Compute the heat capacity of the

sampleb.) The mass of the helium sample is 6.42

kg. Compute the specific heat capacity and molar heat capacity of helium.

Strategy– a.) Use the relationship Cp=q/∆T– B.) Use cs=Cp/m and use cp=Cp/n

OFB Chap. 10 93/6/2003

Exercise 10-1• Exactly 500.0 kJ of heat is absorbed at a

constant pressure by a sample of gaseous helium. The temperature increases by 15.0K.a.) Compute the heat capacity of the sampleb.) The mass of the helium sample is 6.42 kg.

Compute the specific heat capacity and molar heat capacity of helium.

Solution– a.) Use the relationship Cp=q/∆T

– b.) Use cs=Cp/m and use cp=Cp/n

OFB Chap. 10 103/6/2003

Equilibration Temperature

• When 2 bodies are placed in contact, heat is exchanged until they reach a common final temperature

• Heat gained by the cooler body equals the heat lost by the warmer body

• If q is positive, heat is gained• If q is negative, heat is lost

• Can also write similar equations for cs

OFB Chap. 10 113/6/2003

Chapter 10Thermochemistry

• Example (not in book)A 43.9g piece of Copper (cs=0.385 J g-1 K - 1) at 135 °C is plunged into 254g of water (cs=4.18 J g-1 K - 1) at 39 °C. Assuming no heat is lost to the surroundings, what will be the final temperature?

a) 100.0 Cb) 87.0 Cc) 53.1 Cd) 40.5 Ce) None of these

OFB Chap. 10 123/6/2003

Also Try 10-2 Example and Exercise

OFB Chap. 10 133/6/2003

Chapter 10Thermochemistry

• For processes carried out at constant pressure, the heat absorbed equals a change in 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.

OFB Chap. 10 143/6/2003

Enthalpy of Reaction• If a chemical reaction occurs at

constant pressure then• (Heat absorbed) = qp = ∆H• ∆H may be either positive or

negative– 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

OFB Chap. 10 153/6/2003

CO (g) + ½ O2 (g) → CO2 (g) ∆H= -283kJ

An exothermic reaction

If doubled (x2), double ∆H2CO (g) + 1O2 (g) → 2CO2 (g) ∆H= -566kJ

If the original reaction is reversedCO2 (g) → CO (g) + ½ O2 (g) ∆H= +283kJ

An endothermic reaction

You can Try Example 10-5

OFB Chap. 10 163/6/2003

Phase Changes• Phase changes are not chemical

reactions but involve enthalpy change

• Note that there are no T units, these occur at constant temperatures for freezing, fusion, vaporization or condensation

OFB Chap. 10 173/6/2003

Typical Exam questionHow much heat energy is needed to

decompose 9.74g of HBr (g) (M =80.9 g/mol) into its elements?

H2 + Br2 → 2HBr (g)

∆H = -72.8 kJmol-1

Strategy:• Decomposition is actually the

reverse reaction• ∆H to burn 2 moles of HBr is

72.8kJ/mol

OFB Chap. 10 183/6/2003

Typical Exam questionHow much heat energy is needed to

decompose 9.74g of HBr (g) (M =80.9 g/mol) into its elements?

H2 + Br2 → 2HBr (g)

∆H = -72.8 kJmol-1

Solution:2HBr (g) → H2 + Br2

∆H = +72.8 kJmol-1

OFB Chap. 10 193/6/2003

Hess’s Law• Sometimes its difficult or even

impossible to conduct some experiments in the lab.

• Because Enthalpy (H) is a State Property, it doesn’t matter what path is taken as long as the initial reactants lead to the final product.

• Hess’s Law is about adding or subtracting equations and enthalpies.

• 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. 10 203/6/2003

Hess’s Law?H (g) CO (g) O 1/2 (graphite) CUnknown 2 =→+ ∆

This is Hess’s Law. It always works no matter how complicated the path.

OFB Chap. 10 213/6/2003

Standard State Enthalpies

• Designated using a superscript °(pronounced naught)

• This is needed because Enthalpy varies with temperature

• Is written as ∆H°

• If no temperature is indicated in the sub-script, assume 25°C.

• OFB text appendix D lists standard enthalpies of formation for a variety of chemical species at 1 atm and 25°C.

οοK298.15

οC25 ∆Hor ∆Hor ∆H oo

OFB Chap. 10 223/6/2003

Standard State Enthalpies

• ∆H° is the sum of products minus the sum of the reactants

• For a general reactiona A + b B → c C + d D

OFB Chap. 10 233/6/2003

Chapter 10Thermochemistry

• Exercise 10-7 Suppose hydrazine and oxygen react to give dinitrogen pentaoxide and water vapor:

2N2H4 (l) + 7O2 (g) ⇒2N2O5 (s) + 4H2O (g)

Calculate the ∆ H of this reaction, given that the reaction:N2 (g) + 5/2O2 (g) ⇒N2O5 (s)has a ∆ H of -43 kJ.

OFB Chap. 10 243/6/2003

10-7 Exercise2N2H4 (l) + 7O2 (g) ⇒

2N2O5 (s) + 4H2O (g)

• What values to use?– ∆Hf°N2O5 (s) given ∆H°=-43.1 kJmol-1

– ∆Hf°H2O (g) look up in App. D– ∆Hf° N2H4 (l) look up in App. D– ∆Hf°O2 (g) look up in App. D

• ∆Hf° = -1154.74 kJmol-1

OFB Chap. 10 253/6/2003

Typical Exam questionGiven the following thermo chemical

equations, calculate the standard enthalpy of formation for propane, C3H8 (g)

AnswersA. (3131) kJmol-1

B. 129C. 4775D. (1773)E. None of these

1-

22283

1-

222

1-

22

kJmol (-2452)∆H

O(l)4H(g)3CO(g)5O(g)HC 3.kJmol (-286)∆H

O(l)H(g)1/2O(g)H 2.kJmol (-393)∆H

(g)CO(g)OC(s) 1.

=

+→+=

→+=

→+

o

o

o

OFB Chap. 10 263/6/2003

BHbAHaDHdCHcHdDcCbBaA

General

ooooo ∆∆∆∆∆ −−+=

+→+

OFB Chap. 10 273/6/2003

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. 10 283/6/2003

• 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. 10 293/6/2003

• 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. 10 303/6/2003

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. 10 313/6/2003

H

C

H H

H

+O

H

H

H

HH

+ H HC

O H

kJmol-1

ExothermickJmol-1

Endothermic

3 C-H 1 O-H 1 C-O 1 H-H

4 C-H2 O-H

FormedBroken

∆H= ∆H bonds broken + ∆H bonds formed

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

Using Appendix D ∆H = 116 kJmol-1

CH4(g) + H2O(g)→CH3OH(g) + H2(g)

OFB Chap. 10 323/6/2003

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. 10 333/6/2003

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. 10 343/6/2003

• 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. 10 353/6/2003

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 wok on the

system• Try Example and Exercise 10-11

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

OFB Chap. 10 363/6/2003

Typical exam questions (1-4)A gas is compressed from 39.92L to 12.97L at a constant pressure of 5.00 atm. In the course of this compression 9.82 kJ of energy is released

Q1 The heat q for this process isa 135 kJb -135 kJc -9.82 kJd 9.82 kJe can not be determined

OFB Chap. 10 373/6/2003

A gas is compressed from 39.92L to 12.97L at a constant pressure of 5.00 atm. In the course of this compression 9.82 kJ of energy is released

Q2 The work w for this process is a 135 L atmb -135 L atmc -9.82 L atmd 9.82 L atme can not be determined

OFB Chap. 10 383/6/2003

A gas is compressed from 39.92L to 12.97L at a constant pressure of 5.00 atm. In the course of this compression 9.82 kJ of energy is released

Q3 ∆ E for the process isa -23.47 kJb 3.86 kJc 23.47 kJd 125 kJe none of these

OFB Chap. 10 393/6/2003

A gas is compressed from 39.92L to 12.97L at a constant pressure of 5.00 atm. In the course of this compression 9.82 kJ of energy is released

Q4 ∆ H for the process isa -9.82 kJb -7.09 kJc 3.83 kJd 9.82 KJe none of these

OFB Chap. 10 403/6/2003

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. 10 413/6/2003

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

OFB Chap. 10 423/6/2003

kJ110.5∆H

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

−=

→+

Example• Calculate the internal energy @ 25°C

for the following reaction

OFB Chap. 10 433/6/2003

Example• Calculate the internal energy @ 25°C

for the following reaction

kJ110.5∆H

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

−=

→+

OFB Chap. 10 443/6/2003

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. 10 453/6/2003

– 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

• 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.

•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. 10 463/6/2003

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. 10 473/6/2003

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

OFB Chap. 10 483/6/2003

Chapter 10Thermochemistry

• Examples / exercises10-1, 10-2, 10-5, 10-6, 10-7, 10-8, 10-9, 10-10, 10-11, 10-

12• HW Problems11, 13, 19, 23, 33, 37, 40, 43,

49, 53, 59, 63