Physical Chemistry , 1st ed.

If the chemical process occurs isothermally, then by assuming the gases in-

volved are acting ideally,

d(pV) d(nRT) dnRT

where dnrefers to the change in the number of moles of gas that accompanies

the chemical reaction. Since both Rand Tare constant, the chain rule of cal-

culus does not provide additional terms. Therefore, for isothermal chemical

processes, equations 2.58 and 2.59 can be written as

HU+ RTn (2.60)

For equation 2.60, pressure and volume are not constrained to be constant.

Example 2.18

One mole of ethane, C 2 H 6 , is burned in excess oxygen at constant pressure

and 600°C. What is the Uof the process? The amount of heat given off by

the combustion of 1 mole of ethane is 1560 kJ (that is, it is an exothermic

reaction).

Solution

For this constant-pressure process,Hq, so H1560 kJ. It is nega-

tive because heat is given off. In order to determine RTn, we need the bal-

anced chemical reaction. For the combustion of ethane in oxygen, it is

C 2 H 6 (g) + ^72 O 2 (g) →2CO 2 (g) + 3H 2 O (g)

The fractional coefficient is necessary for oxygen in order to balance the re-

action. The water product is listed as a gas because the temperature of the

process is well above its boiling point! The change in the number of moles of

gas,n, is nproductsnreactants(2 + 3) (1 + ^72 )  5 4.5 0.5 mole.

Therefore,

1560 kJ U+ (0.5 mol)8.314 

mo

J

lK

(873 K)

1

1

00

k

0

J

J



Solving:

1560 kJU+ 1.24 kJ

U1561 kJ

In this example,rxnUand rxnHare only slightly different. This shows that

someof the change in the total energy went into work, and the rest went

into heat.

2.11 Changing Temperatures

For a process that occurs under constant pressure (which includes most

processes of interest to the chemist), the Hof the process is easy to measure.

It is equal to the heat,q, of the process. But the temperature of the process can

change, and we expect that Uand, more importantly,Hwill vary with the

temperature. How do we figure Hfor a different temperature?

Since enthalpy is a state function, we can select any convenient path to de-

termine Hfor the reaction at the desired temperature. We can use an idea

similar to Hess’s law to determine the change in the state function Hfor a

58 CHAPTER 2 The First Law of Thermodynamics