EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC? qoverall = qice + qfusion + qwater + qboil + qsteam

q = (10.0g 2.09J/goC 15.0oC)

+ (10.0g 333J/g)

+ (10.0g 4.18J/goC 100.0oC)

+ (10.0g 2260J/g)

+ (10.0g 2.03J/goC 27.0oC)

q = (314 + 3.33×103 + 4.18×103 + 2.26×104 + 548)J

= 30.9 kJ

Heat Flow in Reactions

exothermic –reaction that gives off energy

q < 0

isolated system E=0 heat released by reaction raises the temperature of the solvent

constant T, heat is released to the surroundings

endothermic – reaction that absorbs energy

q > 0

Expansion Type Work

w = -PV system does work

P

P

Vinitial

V

V = Vfinal - Vinitial

qp = +2kJ

Do 250 J of work to compress a gas, 180 J of heat are released by the gas

What is E for the gas?

430

J70

J-7

0 J

-180

J

-250

J

0% 0% 0%0%0%

1. 430 J

2. 70 J

3. -70 J

4. -180 J

5. -250 J

0

0

130

10

Enthalpy HE = q + w

at constant V, wexpansion = 0

E = qv

at constant P, wexpansion = -PV

E = qp - PVDefinePV) = PV at constant P

Hence H = qp

EnthalpyEnthalpy heat at constant pressure or the

heat of reaction

qp = H = Hproducts - Hreactants

Exothermic ReactionH = (Hproducts - Hreactants) < 0

2 H2(g) + O2(g) 2 H2O(l) H < 0

Endothermic ReactionH = (Hproducts - Hreactants) > 0

2 H2O(l) 2 H2(g) + O2(g) H > 0

State FunctionsH and E along with P, T, V (or P, T,

V) and many others are state functions. They are the same no matter what path we take for the change.

• q and w are not state functions, they depend on which path we take between two points.

initial

final

Eq

w

q

wE=Efinal-Einitial

q and w can be anything

Path Independent Energy Changes

Which day would you like OWL quizzes due (4 AM)

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0% 0% 0%0%0%

1. Monday2. Tuesday3. Wednesday4. Thursday5. Friday

100

0

130

Stepwise Energy Changesin Reactions

Laws of Thermochemistry

1. The magnitude of is directly proportional to the amount of reaction.

H is for 1 mole of reaction as written

2 H2(g) + O2(g) 2 H2O(l) H = -571.6 kJ

H2(g) + ½ O2(g) H2O(l) H = -285.8 kJ

Can have ½ mole O2 just not ½ molecule

Laws of Thermochemistry

2. H for a reaction is equal in magnitude but opposite in sign to H for the reverse reaction.

H2(g) + ½ O2(g) H2O(l) H = -285.8 kJ

H2O(l) H2(g) + ½ O2(g) H = +285.8 kJ

Laws of Thermochemistry

3. The value of H for the reaction is the same whether it occurs directly or in a series of steps.

Hoverall = H1 + H2 + H3 + · · ·

also called Hess’ Law

Enthalpy Diagram

H2(g) + ½ O2(g) H2O(l) H = -285.8 kJ

H2O(l) H2O(g) H = +44.0 kJ

H2(g) + ½ O2(g) H2O(g) H = -241.8 kJ

Given 3 CO + 3/2 O2 3 CO2 H = -849 kJ

What is H for CO2 CO + ½ O2 ?

100

0

130

-283

kJ

+283

kJ

+849

kJ

-254

7 kJ

+254

7 kJ

0% 0% 0%0%0%

1. -283 kJ

2. +283 kJ

3. +849 kJ

4. -2547 kJ

5. +2547 kJ

Energy and Stoichiometry

• Since H is per mole of reaction we can relate heat to amount of reaction

• Given C2H6 + 7/2 O2 2 CO2 + 3 H2O H=-1559.7 kJ

• If 632.5 kJ are released to surroundings what mass of H2O is formed?

• 632.5 kJ released means H = -632.5 kJ for this much H2O

OHg92.21OHmol1

OHg016.18

reactionmol1

OHmol3

kJ7.1559

reactionmol1kJ5.632 2

2

22

Bomb Calorimeter measure qv

qrxn + qcal = 0qrxn = -qcal

qrxn = - ccalTErxn = qrxn/moles rxnErxn ≈ Hrxn

H = E + (PV)H = E + RTngas

@298K RT = 2.5 kJ/mol

“Coffee Cup” Calorimeter qp

Photo by George Lisensky

Measuring H

• When 25.0 mL of 1.0 M H2SO4 are added to 50.0 mL of 1.0 M KOH, both initially at 24.6C the temperature rises to 33.9C. What is H for H2SO4 + 2 KOH K2SO4 + 2 H2O ? (Assume d = 1.00 g/mL, c = 4.18 J/g.C)

• qsoln = mcT

• m = (25.0 + 50.0)mL×1.00g/mL = 75.0 g

Measuring H cont.

• q=mcT

• qsoln = 75.0 g × 4.18 J/g.C × (33.9-24.6)C

• qsoln = 2916 J

• qrxn + qsoln = 0

• qrxn = -2916 J

Hrxn = qrxn/moles rxn

Measuring H cont

• How many moles rxn?

• 1 mol rxn / 1 mol H2SO4

• 1 mol rxn / 2 mol KOH

4242 SOHmol025.0

mL1000

L1

L1

SOHmol00.1mL0.25

KOHmol050.0mL1000

L1

L1

KOHmol00.1mL0.50

Stoichiometric mixture so 0.025 mol rxn

Measuring H cont

Hrxn = qrxn/moles rxn

Hrxn = -2916 J / 0.025 mol rxn

Hrxn = -116622 J / mol rxn

Hrxn = -117 kJ

H is per mole of reaction as written

If excess Al is added to 50 mL of 0.250 M H2SO4 how many moles of

the following reaction occur?2 Al + 3 H2SO4 Al2(SO4)3 + 3 H2

10

0

0

130

0% 0% 0%0%0%

1. 0.0125 mol

2. 0.0375 mol

3. 0.025 mol

4. 0.00625 mol

5. 0.00417 mol

Hess’s Law

• Can find H for an unknown, or hard to measure, reaction by summing measured H values of known reactions.

EXAMPLE

H for formation of CO cannot readily be measured since a mixture of CO and CO2 is always formed.

C (s) + ½ O2 (g) CO (g) H = ?

C (s) + O2 (g) CO2 (g) H1 = -393.5 kJ

CO (g) + ½ O2 (g) CO2 (g) H2 = -283.0 kJ

C (s) + ½ O2 (g) CO (g) H = H1 - H2

H = H1 - H2 = -393.5 – (-283.0) = -110.5 kJ

Standard Enthalpy of Formation

the enthalpy associated with the formation of 1 mol of a substance from its constituent elements under standard state conditions at the specified temperature

For an element this is a null reaction

O2 (g) O2 (g) H = 0

Hf = 0 for all elements in their standard states

For which one of these reactions is ΔHºrxn

= ΔHºf?

0% 0% 0%0%0%

1. N2(g) + 3 H2(g) 2 NH3(g)

2. C(graphite) + 2 H2(g) CH4(g)

3. C(diamond) + O2(g) CO2(g)

4. CO(g) + ½ O2(g) CO2(g)

5. H2(g) + Cl2(g) 2 HCl(g)

100

0

130

Calculation of Ho

Ho = mols Hfoproducts – mols Hf

oreactants

We can always convert products and reactants to the elements.

Hess’s law says H is the same whether we go directly from reactants to products or go via elements

Example What is the value of Hrxn for the reaction:2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)

from Appendix J TextC6H6(l) Hf

o = + 49.0 kJ/mol

O2(g) Hfo = 0

CO2(g) Hfo = - 393.5

H2O(g) Hfo = - 241.8

Hrxn mols Hfoproduct

– mols Hforeactants

ExampleWhat is the value of Hrx for the reaction:2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)

from Appendix J TextC6H6(l) Hf

o = + 49.0 kJ/mol; O2(g) Hfo = 0

CO2(g) Hfo = - 393.5; H2O(g) Hf

o = - 241.8 Hrxn molsHf

oproduct - mols Hf

oreactants

Hrxn - 393.5) + 6(- 241.8)product

- 2(+ 49.0 ) + 15(0)reactants kJ/mol

= - 6.2708 103 kJ

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