The Foundations of Chemistry

Plan

(i)We want one mole of C(graphite) as reactant, so we write down equation (1).

(ii)We want two moles of H 2 (g) as reactants, so we multiply equation (2) by 2 [designated

below as 2(2)].

(iii)We want one mole of CH 4 (g) as product, so we reverse equation (3) to give (3).

(iv)We do the same operations on each 
H^0 value.

(v)Then we add these equations term by term. The result is the desired thermochemical

equation, with all unwanted substances canceling. The sum of the 
H^0 values is the


H^0 for the desired reaction.

Solution

H^0

C(graphite)O 2 (g)88nCO 2 (g) 393.5 kJ/mol rxn) (1)

2H 2 (g)O 2 (g)88n2H 2 O(
)2(285.8 kJ/mol rxn) 2 (2)

CO 2 (g)2H 2 O(
)88nCH 4 (g)2O 2 (g) 890.3 kJ/mol rxn) (3)

C(graphite)2H 2 (g)88nCH 4 (g) 
H^0 rxn74.8 kJ/mol rxn)

CH 4 (g) cannot be formed directly from C(graphite) and H 2 (g), so its
H^0 fvalue cannot be

measured directly. The result of this example tells us that this value is 74.8 kJ/mol.

EXAMPLE 15-8 Combining Thermochemical Equations: Hess’s Law

Given the following thermochemical equations, calculate the heat of reaction at 298 K for the

reaction of ethylene with water to form ethanol.

C 2 H 4 (g)H 2 O(
)88nC 2 H 5 OH(
)

H^0

C 2 H 5 OH(
)3O 2 (g)88n2CO 2 (g)3H 2 O(
) 1367 kJ/mol rxn (1)

C 2 H 4 (g)3O 2 (g)88n2CO 2 (g)2H 2 O(
) 1411 kJ/mol rxn (2)

Plan

We reverse equation (1) to give (1); when the equation is reversed, the sign of 
H^0 is

changed because the reverse of an exothermic reaction is endothermic. Then we add it to

equation (2).

Solution

H^0

2CO 2 (g)3H 2 O(
)88nC 2 H 5 OH(
)3O 2 (g) 1367 kJ/mol rxn (1)

C 2 H 4 (g)3O 2 (g)88n2CO 2 (g)2H 2 O(
) 1411 kJ/mol rxn (2)

C 2 H 4 (g)H 2 O(
)88nC 2 H 5 OH(
) 
H^0 rxn44 kJ/mol rxn

You should now work Exercises 30 and 32.

We have used a series of reactions
for which
H^0 values can be easily
measured to calculate
H^0 for a
reaction that cannot be carried out.

606 CHAPTER 15: Chemical Thermodynamics