The Foundations of Chemistry

CH 4 (g)2O 2 (g)88nCO 2 (g)2H 2 O(
)887 kJ

indicates release of energy

We can write the change in energythat accompanies this reaction as

CH 4 (g)2O 2 (g)88nCO 2 (g)2H 2 O(
) 
E887 kJ/mol rxn

As discussed in Section 15-2, the negative sign indicates a decreasein energy of the system,

or a releaseof energy by the system.

The reverse of this reaction absorbsenergy. It can be written as

CO 2 (g)2H 2 O(
)887 kJ88nCH 4 (g)2O 2 (g)

indicates absorption of energy

or

CO 2 (g)2H 2 O(
)88nCH 4 (g)2O 2 (g) 
E887 kJ/mol rxn

If the latter reaction could be forced to occur, the system would have to absorb 887 kJ of

energy per mole of reaction from its surroundings.

When a gas is produced against constant external pressure, such as in an open vessel

at atmospheric pressure, the gas does work as it expands against the pressure of the atmos-

phere. If no heat is absorbed during the expansion, the result is a decrease in the internal

energy of the system. On the other hand, when a gas is consumed in a process, the atmos-

phere does work on the reacting system.

Let us illustrate the latter case. Consider the complete reaction of a 2
1 mole ratio of

H 2 and O 2 to produce steam at some constant temperature above 100°C and at one atmos-

phere pressure (Figure 15-7).

2H 2 (g)O 2 (g)88n2H 2 O(g)heat

Assume that the constant-temperature bath surrounding the reaction vessel completely

absorbs all the evolved heat so that the temperature of the gases does not change. The

volume of the system decreases by one third (3 mol gaseous reactants n2 mol gaseous

At 25°C the change in internal energy
for the combustion of methane is
887 kJ/mol CH 4. The change in heat
content is 890 kJ/mol CH 4 (see
Section 15-1). The small difference is
due to work done on the system as it is
compressed by the atmosphere.

614 CHAPTER 15: Chemical Thermodynamics

Figure 15-7 An illustration of the
one-third decrease in volume that
accompanies the reaction of H 2 with
O 2 at constant temperature. The
temperature is above 100°C. 2H 2 (g) + O 2 (g)

P = 1 atm P = 1 atm

2H 2 O(g)

2 mol H 2

+

1 mol O (^2) 2 mol
H 2 O(g)
Constant pressure,
T of gases > 100°C
Constant
temperature bath
(heated mineral oil)