Chapter 11

Thermochemistry

Molar Heat of Fusion (Hfus):

The energy needed to melt 1 mole of a substance.

Water: ΔHfus= 6.01 kJ/mol

ΔH values are positive because substances need added heat in order to melt.Example: How much heat does it take to melt 35.0 g of ice?

35.0 g H2O = 11.7 kJ x

1

HHydrogen

1.01

8

OOxygen16.00

1.01x 2

2.02

H2O

16.00x 1

+16.00 = 18.02 g

g H2O

mol H2O

18.02

1x

Calculator: 35.0 ÷ 18.02 x 6.01 = 11.67314095 kJ

mol H2O

6.01 kJ

1

(Endothermic)

Molar Heat of Solidification (Hsolid):

The heat lost when 1 mole of a liquid freezes.

Water: ΔHsolid= - 6.01 kJ/mol

ΔH values are negative because substances have to lose heat in order to freeze.

(Exothermic)

Molar Heat of Vaporization (Hvap):

The energy needed to vaporize 1 mole of a substance.

Water: ΔHvap= 40.7 kJ/mol

Example: How much heat does it take to vaporize 35.0 g of water?

35.0 g H2O = 79.1 kJ x

1

HHydrogen

1.01

8

OOxygen16.00

1.01x 2

2.02

H2O

16.00x 1

+16.00 = 18.02 g

g H2O

mol H2O

18.02

1x

Calculator: 35.0 ÷ 18.02 x 40.7 = 79.05105438 kJ

mol H2O

40.7 kJ

1

(Endothermic)

Molar Heat of Condensation (Hcond):

The amount of heat released when 1 mole of a substance condenses.

Water: ΔHcond = - 40.7 kJ/mol

(Exothermic)

Heat of Combustion (H):

The heat released during a chemical reaction in which 1 mole of a substance is completely burned.

Also called Molar Heat of Combustion

(Exothermic)

Molar Heat of Solution (Hsoln):

The heat change that results when 1 mole of a substance is dissolved in water.

Molar heat of solution for calcium chloride ΔHsoln= -82.8 kJ/mol

Example: Certain hot packs work by mixing calcium chloride with water.How much heat energy is produced if you mix 6.32 moles of CaCl2 in water?

6.32 mol CaCl2 = -523 kJ xmol CaCl21

Calculator: 6.32 x -82.8 = -523.296 kJ

-82.8 kJ

(Hess’s Law)Use Hess’s Law to calculate (∆H) for the reaction where graphite becomes diamond.

C(graphite) C(diamond)

Use the enthalpy changes for the combustion of graphite and diamond.

∆H = ______ kJ

C(graphite) + O2 (g) CO2 (g) ∆H1 = -394 kJ1)

C(diamond) + O2 (g) CO2 (g) ∆H2 = - 396 kJ2)

Graphite is supposed to be a reactant so leave 1st reaction alone.

C(graphite) + O2 (g) CO2 (g) ∆H1 = -394 kJ1)

Diamond is supposed to be a product so reverse 2nd reaction and change sign for ∆H.

C(diamond) + O2 (g) ∆H2 = + 396 kJ

2)

CO2 (g)

+C(graphite) C(diamond) ∆H = -394 + 396 = + 2 kJ+2

It is called Hess’s Law of Heat SUMMATION

Standard Heat of Formation:

2 CH3OH (l) + 3 O2 (g) 2 CO2(g) + 4 H2O (l)

∆Hf° = ∆Hf°(products) - ∆Hf°(reactants)

-239x 2- 478 kJ

- 1,454 kJ

∆H° = - 1,454 kJ

∆H° = - 1,932 kJ – (- 478 kJ) =

Substance

∆Hf°

(kJ/mol)

CH3OH (g) -239

O2(g) 0.0

CO2 (g) -394

H2O(g) -286

0.0x 3 0 kJ+

- 478 kJ(reactants)

-394x 2 -788 kJ

-286x 4 -1,144 kJ+

- 1,932 kJ(products)

∆H° = ∆Hf°(products) - ∆Hf°(reactants)