Calculating Heat Changes
First Law of Thermodynamics:
the internal energy of an isolated system is
constant
Signs (+/-) will tell you if energy is entering or
leaving a system
+ indicates energy enters a system
- indicates energy leaves a system
•A change in internal energy can be identified with the heat
supplied at constant volume
ENTHALPY (H)
(comes from Greek for “heat inside”)
•
The heat supplied is equal to the change in another
thermodynamic property called enthalpy (H)
i.e. H = q
• this relation is only valid at constant pressure
As most reactions in chemistry take place at
constant pressure we can say that:
A change in enthalpy = heat supplied
Heat Transfer
 Specific Heat (Cp)
 amount of energy
required to raise the
temp. of 1 kg of
material by 1 degree
Kelvin
 units: J/(kg·K)
or J/(kg·°C)
Heat Transfer
 Which sample will take
longer to heat to 100°C?
50 g Al
50 g Cu
• Al - It has a higher specific heat.
• Al will also take longer to cool down.
Heat Transfer
Q = m  T  Cp
Q:
m:
T:
Cp :
heat (J)
mass (kg)
change in temperature (K or °C)
specific heat (J/kg·K)
T = Tf - Ti
– Q = heat loss
+ Q = heat gain
Heat Transfer

A 32-g silver spoon cools from 60°C to 20°C.
How much heat is lost by the spoon?
GIVEN:
m = 32 g
Ti = 60°C
Tf = 20°C
Q=?
Cp = 235 J/kg·K
WORK:
Q = m·T·Cp
m = 32 g = 0.032 kg
T = 20°C - 60°C = – 40°C
Q = (0.032kg)(-40°C)(235J/kg·K)
Q = – 301 J
Thermochemical Equations
 A chemical equation that includes the heat
change
 Heat of reaction- heat change for the
equation exactly as written, represented by
ΔH, when pressure is constant.
Hess’s law
 Hess’s Law states that the heat of a whole reaction
is equivalent to the sum of it’s steps.
 For example: C + O2  CO2 (pg. 165)
The book tells us that this can occur as 2 steps
C
+ ½O2  CO
H = – 110.5 kJ
CO + ½O2  CO2
H = – 283.0 kJ
C + CO + O2
 CO + CO2
H = – 393.5 kJ
I.e. C + O2
 CO2
H = – 393.5 kJ
 Hess’s law allows us to add equations.
 We add all reactants, products, & H values.
 We can also show how these steps add together via
an “enthalpy diagram” …
Steps in drawing enthalpy diagrams
1. Balance the equation(s).
2. Sketch a rough draft based on H values.
3. Draw the overall chemical reaction as an enthalpy
4.
5.
6.
7.
diagram (with the reactants on one line, and the
products on the other line).
Draw a reaction representing the intermediate step by
placing the relevant reactants on a line.
Check arrows: Start: two leading away
Finish: two pointing to finish
Intermediate: one to, one away
Look at equations to help complete balancing (all
levels must have the same # of all atoms).
Add axes and H values.
C
+ ½O2  CO
CO + ½O2  CO2
H = – 110.5 kJ
H = – 283.0 kJ
 CO2
H = – 393.5 kJ
C
+ O2
C + O2
Reactants
Products
Enthalpy
Intermediate
H = – 110.5 kJ
CO + ½O2
H =
– 393.5 kJ
H =
– 283.0 kJ
CO2
Note: states such as (s) and (g) have been ignored to reduce clutter on these
slides. You should include these in your work.
Bond Strengths
Bond strengths measured by bond enthalpy HB (+ve values)
• bond breaking requires energy (+ve H)
• bond making releases energy (-ve H)
Lattice Enthalpy
A measure of the attraction between ions (the enthalpy
change when a solid is broken up into a gas of its ions)
• all lattice enthalpies are positive
• I.e. energy is required o break up solids
Another Example of Hess’s Law
Given:
C(s) + ½ O2(g)  CO(g)
H = -110.5 kJ
CO2(g)  CO(g) + ½ O2(g)
H = 283.0 kJ
Calculate H for:
C(s) + O2(g)  CO2(g)
Enthalpies of Formation
• If 1 mol of compound is formed from its constituent
elements, then the enthalpy change for the reaction is
called the enthalpy of formation, Hof .
• Standard conditions (standard state): Most stable form of
the substance at 1 atm and 25 oC (298 K).
• Standard enthalpy, Ho, is the enthalpy measured when
everything is in its standard state.
• Standard enthalpy of formation: 1 mol of compound is
formed from substances in their standard states.
Enthalpies of Formation
• If there is more than one state for a substance under
standard conditions, the more stable one is used.
• Standard enthalpy of formation of the most stable form of
an element is zero.
Enthalpies of Formation
Using Enthalpies of Formation to
Calculate Enthalpies of Reaction
• For a reaction
H rxn   n  H f products    m  H f reactants 
• Note: n & m are stoichiometric coefficients.
• Calculate heat of reaction for the combustion of propane
gas giving carbon dioxide and water.
Foods and Fuels
Foods
• 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal.
• Energy in our bodies comes from carbohydrates and fats
(mostly).
• Intestines: carbohydrates converted into glucose:
C6H12O6 + 6O2  6CO2 + 6H2O, H = -2816 kJ
• Fats break down as follows:
2C57H110O6 + 163O2  114CO2 + 110H2O, H = -75,520 kJ
Fats contain more energy; are not water soluble, so are good for
energy storage.