5 Steps to a 5 AP Chemistry 2019

Thermodynamics ❮ 127

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

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

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

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

Answer:

∆=∆° +∆°

−∆°+∆°

{5[O(g)] 4[ NH(g)]}

{[6HO(g)]4[NO(g)]}

rxn2 3

2

HH H

HH

ff

ff

Using tabulated standard enthalpies of formation gives:

DHrxn = [5(0.00 kJ) + 4(-46.19 kJ)] - [6(-241.85 kJ) + 4(90.37)]

= 904.68 kJ

People commonly forget to subtract all the reactants from the products.

The values of DHf° 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 average

bond energies of the reactant molecules and subtract the sum of the average bond energies

of the product molecules.

Entropies

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

elements and compounds can be tabulated. The standard molar entropy is the entropy

associated 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:

DS° = S S° products - S S° reactants

Calculate DS° 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 O H 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 O H 2 O 2

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

= - 163.6 J/mol K

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