Heat in Changes of State

What happens when you place an ice cube on a table in a warm room?

Molar Heat of Fusion (ΔHfus): heat absorbed by one mole of a substance in melting from a solid to a liquid at a constant temperatureMolar heat of Solidification (ΔHsolid): heat lost when one mole of a liquid solidifies at a constant temperature

ΔHfus = - ΔHsolid

Molar Heat of Vaporization (ΔHvap): amount of heat necessary to vaporize one mole of a given liquid

H2O (l) H2O (g) ΔHvap = 40.7 kJ/mol

Molar Heat of Condensation (ΔHcond): amount of heat released when 1 mol of vapor condenses

ΔHvap = - ΔHcond

Vapor

Liquid

Solid

vapo

rizat

ion

cond

ensa

tion

fusio

n

solid

ifica

tion

ΔHvap

-ΔHfus

-ΔHcond

-ΔHsolid

High Enthalpy

Low Enthalpy

ΔHvap

ΔHfus

The Heating Curve of Water

http://netcamp.prn.bc.ca/nuggets/heatingcurve.swf

Molar Heat of Solution (ΔHsoln): heat change caused by dissolution of one mole of substance

Exothermic Reaction: CaCl2 (s) Ca2+(aq) + 2Cl-(aq)ΔHsoln = -445.1 kJ/mol

Endothermic Reaction: NH4NO3 (s) NH4

+ (aq) + NO3- (aq)

ΔHsoln = 25.7 kJ/mol

H2O(l)

H2O(l)

DHsoln = Hsoln - Hcomponents

How much heat (in kJ) is released when 2.500 mol NaOH (s) is dissovled in water? ΔHsoln = -445.1 kJ/mol

List Knowns and UnknownsKnowns: ΔHsoln = -445.1 kJ/molAmount of NaOH(s): 2.500 mol

Unknown:ΔH = ?kJ

Solve:

ΔH = 2.500 mol NaOH (s) x -445.1 kJ = -1113 kJ 1 mol NaOH

Hess’s LawHess’s Law of heat summation: If you add two or

more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

Example: C (diamond) C (graphite)

Use Hess’s Law to find the enthalpy changes for the conversion of diamond to graphite by using the following combustion rxns

C (s, graphite) + O2 (g) CO2 (g) ΔH = -393.5 kJC (s, diamond) + O2 (g) CO2 (g) ΔH = -395.4 kJ

1st Step: Write equation the first equation in reverseCO2 (g) C (s, graphite) + O2 (g) ΔH = 393.5 kJ**when you write an equation in reverse, change sign**

2nd Step: Add the two equations together

CO2 (g) C (s, graphite) + O2 (g) ΔH = 393.5 kJC (s, diamond) + O2 (g) CO2 (g) ΔH = -395.4 kJ

C (s, diamond) C (s, graphite) ΔH = -1.9 kJ

Use Hess’s Law to find the enthalpy change for the formation of carbon monoxide from its elements.

C (s, graphite) + O2 (g) CO2 (g) ΔH = -393.5 kJCO (g) + ½ O2 (g) CO2 (g) ΔH = -283.0

kJ

1st Step: Write the first equation in reverseCO2 (g) CO (g) + ½ O2 (g) ΔH = 283.0

kJ

2nd step: Add the equations togetherCO2 (g) CO (g) + ½ O2 (g) ΔH = 283.0 kJC (s, graphite) + O2 (g) CO2 (g) ΔH = -393.5 kJ

C (s, graphite) + ½ O2 (g) CO (g) ΔH = -110.5 kJ

Standard Heat of Formation (ΔHf0): the change in

enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 25°C

ΔHf0 of a free element in its standard state is

arbitrarily set at 0.ΔHf

0 = 0 for diatomic molecules

Standard Heat of Reaction (ΔH0rxn ):heats of

reaction at standard conditions

ΔH0rxn = ΔHf

0 (products) - ΔHf0 (reactants)