OFB Chap. 10 110/26/2004

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• Examples / exercises10-1, 10-2, 10-5, 10-6, 10-7, 10-8, 10-9, 10-10, 10-11, 10-12• HW Problems 12, 15a, 22, 34, 40a&b, 64

OFB Chap. 10 210/26/2004

Exercise 10-1Exactly 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. 10 310/26/2004

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

13

3

p

JK33.33x10K15

J500x10K15

500kJ∆TqC

−

=

=°

=°

=

113

13p

p

113

13p

s

mol20.8JK

4g/molg6.42x10

JK33.33x10n

Cc

g5.19JKg6.42x10

JK33.33x10mC

c

−−−

−−−

=

==

===

– b.) Use cs=Cp/m and use cp=Cp/n

OFB Chap. 10 410/26/2004

Enthalpy• Heat is a means by which energy is

transferred among systems• For processes carried out at constant

pressure, the heat absorbed equals a changein Enthalpy, H.

qp= ∆H Enthalpy Change(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.

∆H = Hf – Hi

• Enthalpy is the heat content of a substance

OFB Chap. 10 510/26/2004

OFB Chap. 10 610/26/2004

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 EndothermicReactants + Heat → Products– If ∆H negative = ∆H < 0 = q < 0 Means heat is given offAnd is called ExothermicReactants → Products + Heat

∆Hreaction = Hproducts – Hreactants

OFB Chap. 10 710/26/2004

Endothermic Reactions

reactionn dissolutiowater

3434

2

2

1

q q)()((s) NONH

reaction)on (dissoluti SystemOH

(water) SystemTransferHeat

=−+→ −+ aqNOaqNH

The dissolution reaction is endothermic

The heat is supplied by the water (the water temperature drops)

qp for water is negative (exothermic), qp for the reaction is positive (endothermic)

OFB Chap. 10 810/26/2004

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 EndothermicReactants + Heat → Products– If ∆H negative = ∆H < 0 = q < 0 Means heat is given offAnd is called ExothermicReactants → Products + Heat

∆Hreaction = Hproducts – Hreactants

OFB Chap. 10 910/26/2004

CO (g) + ½ O2 (g) → CO2 (g)

∆H= -283kJ

An exothermic reaction

If Quantities doubled (x2), then double ∆H

2CO (g) + 1O2 (g) → 2CO2 (g) ∆H= -566kJ

If the original reaction is reversed

CO2 (g) → CO (g) + ½ O2 (g)

∆H= +283kJ

An endothermic reactionYou can Try Example 10-5

OFB Chap. 10 1010/26/2004

Phase ChangesPhase changes are not chemical reactions

but involve enthalpy change

H2O(s) → H2O(l) ∆Hfusion= + 6 kJ at 0C

∆Hfusion= + means endothermic (add heat)

if reversedH2O(l)→ H2O(s) ∆Hfreeze= – 6 kJ

∆Hfreeze= – ∆Hfusion at melting point

∆Hvaporization at Boiling point

= – ∆HCondensation

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

OFB Chap. 10 1110/26/2004

OFB Chap. 10 1210/26/2004

How 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 decompose 2 moles of HBr

is +72.8kJ/mol

OFB Chap. 10 1310/26/2004

How much heat energy is needed to decompose 9.74g of HBr (g) (M =80.9 g/mol) into its elements?

H2 + Br2 → 2HBr (g)

∆Hr = -72.8 kJmol-1

Solution:2HBr (g) → H2 + Br2

∆H = +72.8 kJmol-1

kJ 4.38moleHBr g 80.91

HBr g 9.76HBr moles 2

kJ 72.8∆Hq

=

==

OFB Chap. 10 1410/26/2004

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 1510/26/2004

Standard State Enthalpies• Designated using a superscript °

(pronounced naught)• This is needed because Enthalpy varies

with conditions• 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.

• Standard State conditions– All concentrations are 1M or 1 mol/L– All partial pressures are 1 atm

οοK298.15

οC25 ∆Hor ∆Hor ∆H oo

OFB Chap. 10 1610/26/2004

OFB Chap. 10 1710/26/2004

Standard State Enthalpies

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

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

• ∆Hr°= ΣH°products- ΣH°reactants

• from Hess’s Law

(B)Hb(A)Ha

(D)Hd(C)Hc∆Hof

of

of

of

or

∆∆

∆∆

−−

+=

OFB Chap. 10 1810/26/2004

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 1910/26/2004

10-7 Exercise2N2H4 (l) + 7O2 (g) →

2N2O5 (s) + 4H2O (g)∆Hr° = 2X∆Hf°N2O5 (s) + 4X∆Hf°H2O (g)-2X∆Hf° N2H4 (l) -7X∆Hf°O2 (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

=2 x (–43.1) + 4 x (–241.82) – 2 x (50.63) – 7 x (0)

= – 86.2 – 967.78 – 101.26∆Hr° = –1154.74 kJmol-1

OFB Chap. 10 2010/26/2004

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)

1-r

22283

1-r

222

1-r

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 2110/26/2004

BHbAHaDHdCHcHdDcCbBaA

General

ooooo ∆∆∆∆∆ −−+=

+→+

283

223

22

283

5143

)(4)(3)(5)(.3

OHHCHOHHCOHH

lOHgCOgOgHC

oo

ooo

∆∆

∆∆∆

−−

+=

+→+

1-r

22283

1-2f

222

1-2f

22

kJmol (-2452)∆H

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

O(l)H(g)1/2O(g)H 2.kJmol (-393))(CO∆H

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

=

+→+=

→+=

→+

o

o

o

(AnswerB) kJmol 129HC∆HX

5(0)X286)4(393)3(24520)(O∆H recall and s∆H' 3 are Given

183f

C3H8

2f

−==

−−−+−=−

=

o

o