CH 17 Thermochemistry
Thermochemistry- energy change that
occurs during a chemical reaction and
changes of state.
- Stored in chemical bonds and is called
chemical potential energy
Heat- represented by the letter q
- Transfer of energy between two objects
- Always flows from warmer to colder
objects
There are two parts to heat:
System- what you are focusing on
Surroundings – the area around the
system
All thermal reactions must obey the law of
conservation of energy.
Therefore,
- energy released by the surroundings,
must be absorbed by the system
(endothermic)
- energy released by the system is
absorbed by the surroundings.
(exothermic)
Heat is measured in calories (cal) or joules
(j)
1 J = .2390cal and 1 cal = 4.184 J
Joules are smaller than calories.
Foods calories are kilocalories.
Heat capacity and specific heat.
Heat capacity- the amount of heat needed
to raise the temperature of an object 1 oC.
- This is different for all objects and the
size is not standardized.
We use specific heat instead.
Specific Heat is the amount of heat
required to raise 1 g of substance 1 oC.
Formula :
C=
q________
M x ∆ T (Change in temp)
C = specific heat found on a chart.
Q = heat added to system ( J or cal)
M =mass in g or kg
Change in temp.= ending temp – starting
temp
Calorimetry is the process by which we
measure heat transfer
Uses a calorimeter to measure
Two types
- Constant Pressure- Measure enthalpy
(H) of the system.
- Equation q = H these are used
interchangeably
- Q system = H =- q surr = - m x C x T
- H is negative fro exothermic reations
and positive for endothermic.
Constant Volume Calorimeters –
- Sample is burned and the surrounding
water heats up.
- Change in water indicates calories
Problems pg 513
Thermochemical equations
Heat becomes a reactant or a product in
thermochemical reactions.
For endothermic reactions, heat is a
reactant and H is positive.
For exothermic reactions, heat is a
product and H is negative.
Usually with reactions the change in temp
is included beside the completed equation.
Look on page 514 and 515.
The physical state of the reactants and
products must also be listed.
Problems on page 516
Heat of combustion
Heat required to burn 1 mole of a
substance
Heat in Changes of State
Heat of Fusion and Solidification.
The heat needed to melt 1 mole of a
substance is the molar heat of fusion (Hfus)
The heat needed to freeze 1mole of a
substance is the molar heat of
solidification( Hsolid)
Hfus = -Hsolid
The melting of ice to water requires 6.01
kJ/mol
The freezing of water to ice requires –6.01
kJ/mol
Problems p521
Heat of Vaporization and Condensation
Same as the molar heat of fusion and
solidification
Heat needed to vaporize 1 mole of a
substance is the heat of vaporization
Heat needed to condense 1 mole of
substance is the heat of condensation.
For water it is 40.7 kJ/mol
Problems pg 524
Heat of solutionThe amount of heat generated by the
dissolution of 1 mole of substance
NaOH(s)  Na+ + OH- H soln = 445.1kJ/mol
If the heat of solution is negative the
reaction is exothermic
If the heat of solution is positive the
reaction is endothermic
Problems p 526
Calculating Heat of Reactions
Hess’s Law of Heat Summation
- Two thermochemical equations can be
used to determine a final equation, the
two heats of reaction can be added to
give a final heat of reaction.
- Ex C(s) diamonds  C(s) graphite
o A) C(s) graphite + O2  CO2
 H = - 393.5kJ
o B) C(s) diamonds + O2  CO2
 H- 395.5 kJ
o Write equation A in reverse
o CO2  C(s) graphite + O2 H =
393.5kJ (kJ is positive because of
the reverse reaction)
o Add the two equations:
CO2  C(s) graphite + O2 H = 393.5kJ
+ C(s) diamonds + O2  CO2 H- 395.5 kJ
C(s) diamonds  C(s)graphite H = -1.9kJ
Heat of Formation
- Hf0= heat to form 1 mole of substances
- Hf0 = 0 for elements in their natural
state , elements and diatomic elements.
The enthalpy, or change in heat can be
determined by the following equation.
Enthalpy = the heat of formation(Hf0)of
the products – heat of formation(Hf0) of
the reactants.
Problem p. 531
CO (g) + O2 (g)  CO2 (g)
Balance the equation first.
2CO (g) + O2(g)  2 CO2 (g)
Hf0 reactants and products is found on
table 17.4 on page 530.
For CO it is –110.5 kJ
O2 = 0 kJ because it is in it’s natural state.
CO2 = -393.5 kJ
2 moles of CO at –110kJ/ mol = -221.0 kJ
2 mole of CO2 at –393.5 kJ/ mol = -787.0
kJ
So if we plug in the numbers for the
equation:
Enthalpy = (Hf0)of the products – (Hf0) of the
reactants.
Enthalpy = -787.0 kJ-(-221.0 kJ+ 0) =
H = -566kJ