5 Steps to a 5 AP Chemistry

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In using this equation be sure to consider the number of moles of each, because ΔHf°

for the individual compounds refer to the formation of 1 mol.

For example, let’s use standard enthalpies of formation to calculate ΔHrxnfor:

6 H 2 O(g) +4 NO(g)→5 O 2 (g) + 4 NH 3 (g)

Answer:

Using tabulated standard enthalpies of formation gives:

ΔHrxn=[5(0.00 kJ) +4(−46.19 kJ)] −[6(−241.85 kJ) +4(90.37)]

=904.68 kJ

People commonly forget to subtract allthe reactants from the products.

The values of ΔHf°will be given to you on the AP exam, or you will be asked to stop

before putting the numbers into the problem.

An alternative means of estimating the heat of reaction is to take the sum of the aver-

age bond energies of the reactant molecules and subtract the sum of the average bond ener-

gies of the product molecules.

Entropies

In much the same way as ΔH°was determined, the standard molar entropies(S°) of ele-

ments and compounds can be tabulated. The standard molar entropy is the entropy asso-

ciated with 1 mol of a substance in its standard state. Entropies are also tabulated, but

unlike enthalpies, the entropies of elements are not zero. For a reaction, the standard

entropy change is calculated in the same way as the enthalpies of reaction:

ΔS°=ΣS°products −ΣS°reactants

Calculate ΔS°for the following. If you do not have a table of S°values, just set up the

problems.

Note: These are thermochemical equations, so fractions are allowed.

a. H 2 (g) +^1 ⁄ 2 O 2 (g) →H 2 O(g)

b. H 2 (g) +^1 ⁄ 2 O 2 (g) →H 2 O(l)

c. CaCO 3 (s) +H 2 SO 4 (l) →CaSO 4 (s) +H 2 O(g) +CO 2 (g)

Answers:

a. H 2 OH 2 O 2

188.7 J/mol K −[131.0 +1/2(205.0)]J/mol K

=−44.8 J/mol K

b. H 2 OH 2 O 2

69.9 J/mol K −[131.0 +1/2(205.0)]J/mol K

=−163.6 J/mol K

ΔΔ Δ

Δ

HH

H

f

f

rxn 2 3

2

Og NHg

HO(g

=°+°

−°

{[ ()] [ ()]}

{[

54

6

Hf

))] 4[+°ΔHf NO(g)]}

128  Step 4. Review the Knowledge You Need to Score High

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