Chapter 11
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Chapter 11. Thermochemistry. H 2 O. 1.01 x 2 2.02. 16.00 x 1 +16.00. = 18.02 g. Molar Heat of Fusion (  H fus ) : The energy needed to melt 1 mole of a substance. W ater: ΔH fus = 6.01 kJ/mol

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Chapter 11

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Chapter 11

Chapter 11

Thermochemistry


Chapter 11

H2O

1.01

x 2

2.02

16.00

x 1

+16.00

= 18.02 g

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.

(Endothermic)

1

H

Hydrogen

1.01

8

O

Oxygen

16.00

Example: How much heat does it take to melt 35.0 g of ice?

mol H2O

6.01 kJ

1

35.0 gH2O

= 11.7 kJ

x

x

mol H2O

1

g H2O

18.02

Calculator: 35.0 ÷ 18.02 x 6.01 = 11.67314095 kJ


Chapter 11

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)


Chapter 11

H2O

1.01

x 2

2.02

16.00

x 1

+16.00

= 18.02 g

Molar Heat of Vaporization(Hvap):

The energy needed to vaporize 1 mole of a substance.

Water: ΔHvap= 40.7 kJ/mol

(Endothermic)

1

H

Hydrogen

1.01

8

O

Oxygen

16.00

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

mol H2O

40.7 kJ

1

35.0 gH2O

= 79.1 kJ

x

x

mol H2O

1

g H2O

18.02

Calculator: 35.0 ÷ 18.02 x 40.7 = 79.05105438 kJ


Chapter 11

Molar Heat of Condensation(Hcond):

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

Water: ΔHcond = -40.7 kJ/mol

(Exothermic)


Chapter 11

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)


Chapter 11

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?

-82.8 kJ

6.32 molCaCl2

= -523 kJ

x

mol CaCl2

1

Calculator: 6.32 x -82.8 = -523.296 kJ


Chapter 11

(Hess’s Law)

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

C(graphite) C(diamond)

∆H =______ kJ

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

1)

C(graphite)+ O2 (g) CO2(g)

∆H1 =-394 kJ

CO2(g)

2)

C(diamond)+ O2 (g)

C(diamond)+ O2 (g) CO2(g)

∆H2 =- 396 kJ

∆H2 =+ 396 kJ

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

It is called Hess’s Law of Heat SUMMATION

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

1)

C(graphite)+ O2 (g) CO2(g)

∆H1 =-394 kJ

+

2)

+2

C(graphite) C(diamond)

∆H =-394 + 396 =+ 2 kJ


Chapter 11

Standard Heat of Formation:

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

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

-239

x 2

- 478 kJ

0.0

x 3

0 kJ

-394

x 2

-788 kJ

-286

x 4

-1,144 kJ

+

+

- 478 kJ

- 1,932 kJ

(reactants)

(products)

∆H°=∆Hf°(products) - ∆Hf°(reactants)

∆H° = - 1,932 kJ – (- 478 kJ) =

- 1,454 kJ

∆H° = - 1,454 kJ


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