Enthalpies of Reaction
Enthalpy change (H) for a chemical reaction is called
Hrxn and is given by
Hrxn = H(products) - H(reactants)
Notice: H(prod) and H(react) can’t be measured
individually, but Hrxn is measured by measuring heat flow
at constant P!!!

Notice:
Hrxn > 0: reaction endothermic; system absorbs heat
Hrxn < 0: reaction exothermic; system releases heat
to surroundings
E.g.,
N2(g) + 3H2(g)  2NH3(g) Hrxn = -46.19 kJ/mol
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Important features of H
H is an extensive property - its magnitude is directly
proportional to amount of reactant consumed, e.g.,
2SO2(g) + O2(g)  2SO3(g)
Hrxn = -196 kJ
What is Hrxn for the reaction of 4 mol of SO2(g) with 2 mol
of O2(g)?
What is Hrxn for the reaction of 2.4 g of SO2(g)?
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Hrxn for a reaction is equal in magnitude but opposite in
sign to Hrxn for the reverse reaction, e.g.,
2SO3(g)  2SO2(g) + O2(g)
Hrxn = 196 kJ
H for a reaction depends on the state of the reactants
and products (i.e., solid, liq., gas)
The states of the reacting species must be specified - we
will assume that all are at 25o C
How do we determine H values experimentally?
Calorimetry
Measurement of heat flow
Heat capacity C: energy required to raise the temperature
of an object by 1oC
C determines the temperature change a body experiences
when it absorbs heat
The greater the heat capacity, the greater the heat required
to produce a temperature increase
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For pure substances, the heat capacity is usually given for
a specified amount of substance
Heat capacity of 1 mol of a substance: molar heat capacity
Heat capacity of 1 g of a substance: specific heat
E.g., the specific heat of water is 4.184 J/g-K
Calculate the molar heat capacity of water
Notice these units of the specific heat: if we multiply a
specific heat x grams x T, we get units of heat
Therefore, heat flow q can be calculated from
q = (specific heat) x (mass) x T
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E.g., How many kJ of heat are required to raise the
temperature of 2.56 kg of water from 44.8oC to 92.0oC?
Calorimetric measurement of heats of reaction are usually
done in one of two ways: constant pressure or constant
volume
Constant Pressure Calorimetry
Works well for reactions in solution
Easy to hold pressure constant – perform reaction in
an insulated vessel open to atmosphere
Assume that reaction vessel (calorimeter) neither
absorbs nor loses heat – all heat produced (or
absorbed) comes from the reaction being studied
How to calculate Hrxn from a constant-pressure
calorimetric measurement?
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E.g., When 4.25 g of NH4NO3 is dissolved in 60.0 g H2O in a
'coffee cup' calorimeter, the temperature drops from 22.0
oC to 16.9 oC. Find H for the solution process. Assume
that the heat capacity of the solution is 4.184 J/g-K.
Hess's Law
H is defined in terms of state functions, so H is a state
function
H depends only on initial and
final states.......
Suppose we want to calculate Hrxn for the reaction
A(g) + B(g)  C(g)
The initial state is A(g, P, 25oC) + B(g, P, 25oC)
The final state is C(g, P, 25oC)
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The enthalpy change Hrxn depends only on the
enthalpy difference between the final state and the initial
state, and not how we get from one to the other............
Suppose that we know Hrxn for the following reactions.....
A(g) + B(g)  E(g)
Hrxn = 52 kJ/mol
C(g)  E(g)
Hrxn = -105 kJ/mol
Can these two known reactions be combined to give the
reaction that we want? (i.e., A(g) + B(g)  C(g)) (assume
that all reactants and products are at pressure P and 25oC)
Question: are all reactants and products in the
reaction we want contained in the reactions we know?
Question: are there any species in the known reactions
that aren’t present in the reaction we want?
How to make these species go away?
What happens to the sign of Hrxn if we reverse a reaction?
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Hess’s Law tell us that if a reaction can be carried out in a
series of steps, then....
The sum of the H for the individual steps must = H for
the overall process
In the above example, instead of going directly from A&B
to C, we convert A&B to E, and then E to C
As long as we start with A&B and end up with C, it doesn’t
matter how we get there!!!
E.g., from the following heats of reaction,
N2(g) + 2O2(g) 2NO2(g)
Hrxn = 67.6 kJ
2NO(g) + O2(g) 2NO2(g)
Find the heat of reaction Hrxn for
N2(g) + O2(g)2NO(g)
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Hrxn = -56.6 kJ
Can we tabulate Hrxn values that would allow us to use
Hess's Law to calculate enthalpy changes for common
reactions?
We can’t tabulate Hrxn values for every single
reaction known to science – some reactions are too
dangerous or too difficult to attempt.....
Also: Hrxn depends on temperature and pressure –
tabulating heats of reaction for every single reaction at
different values of T and P would be rather nightmarish
How to get around these difficulties?
Heats of Formation
Define Hf : heat of formation
Hf is the enthalpy change associated with the
formation of a compound from its constituent
elements
REMEMBER: H depends on states of products and
reactants
We define Ho as a standard enthalpy change - takes
place with all elements in their standard states
Standard state: most stable form of a substance at 1
atm pressure
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So.... we can have
Hovap = standard heat of vaporization
Hofus = standard heat of fusion
The “o” means ‘standard’, which means 1 atm
pressure
Standard enthalpies are generally tabulated at 25oC
(298 K)
E.g., What does Horxn mean?
So... Hof is a standard heat of formation
Enthalpy change for the formation of 1 mol of
substance from its elements, all substances in
their standard states
KNOW the common standard states (e.g., O2(g),
C(graphite), H2(g))
By definition, Hof for the most stable form of an
element = 0 kJ
(think about this – write the formation
reaction for C(graphite))
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Hof values are tabulated in Appendix C, e.g.,
AlCl3(s)
-705.6 kJ/mol
CO2(g)
-393.5 kJ/mol
NH3(g)
-46.19 kJ/mol
E.g. write the formation reactions for AlCl3(s), CO2(g), and
NH3(g)
Now: can we use Hess's Law and tabulated Hof values to
calculate enthalpy changes for reactions?
Horxn : standard enthalpy change for a reaction is given by

Horxn =  nHof(products) -  mHof(reactants)
Where n and m are the stoichiometric coefficients from the
reaction......
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E.g., given the following Hof ,
NH4NO3(s)
-356.6 kJ/mol
N2O(g)
H2O(g)
81.6 "
"
-241.8 " "
Calculate Horxn for
NH4NO3(s)  N2O(g) + 2H2O(g)
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Problems du Jour
From the following enthalpies of reaction:
H2(g) + F2(g)  2HF(g)
-537 kJ
C(s) + 2F2(g)  CF4(g)
-680 kJ
2C(s) + 2H2(g)  C2H4(g)
52.3 kJ
Find Hrxn for the reaction
C2H4(g) + 6F2(g)  2CF4(g) + 4HF(g)
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Problem du Jour
Complete combustion of 1 mol of acetone, C3H6O, results
in the liberation of 1790 kJ of heat:
C3H6O(g) + 4O2(g)  3CO2(g) + 3H2O(l)
Horxn = -1790 kJ
Using this information, along with the following
thermochemical data:
Substance
Hof, kJ
CO2(g)
-393.5
H2O(l)
-285.83
Calculate Hof for acetone.
Problems du Jour
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Using data from Appendix C, calculate Horxn for the
following reactions:
2KOH(s) + CO2(g)  K2CO3(s) + H2O(g)
Fe2O3(s) + 6HCl(g)  2FeCl3(s) + 3H2O(g)
Problem du Jour
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When a 9.55 g sample of solid sodium hydroxide dissolves
in 100.0 g of water in a coffee-cup calorimeter, the
temperature rises from 23.6 oC to 47.4 oC. Calculate H (in
kJ/mol NaOH) for the solution process
NaOH(s)  Na+(aq) + Cl-(aq)
Assume that the specific heat of the solution is the same
as that of pure water.
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