The Foundations of Chemistry

At constant pressure, wP
V,so

Hq(P
V)P
V

Hqp (constant Tand P)

The difference between
Eand
His the amount of expansion work (P
Vwork) that
the system can do. Unless there is a change in the number of moles of gas present, this
difference is extremely small and can usually be neglected. For an ideal gas, PVnRT.
At constant temperature and constant pressure, P
V(
n)RT,a work term. Substituting
gives

H
E(
n)RT or 
E
H(
n)RT (constant Tand P)

As usual, 
nrefers to the number of

moles of gaseous productsminus the

number of moles of gaseous reactants.

15-11 Relationship of Hand E 619

Problem-Solving Tip:Two Equations Relate H and E—Which

One Should Be Used?

The relationship 
H
EP
Vis valid for anyprocess that takes place at constant

temperature and pressure. It is very useful for physical changes that involve volume

changes, such as expansion or compression of a gas. When a chemical reaction occurs

and causes a change in the number of moles of gas, it is more convenient to use the rela-

tionship in the form 
H
E(
n)RT.You should always remember that 
nrefers to

the change in number of moles of gasin the balanced chemical equation.

In Example 15-14 we found that the change in internal energy,
E,for the combus-
tion of ethanol is 1365 kJ/mol ethanol at 298 K. Combustion of one mole of ethanol at
298 K and constant pressure releases 1367 kJ of heat. Therefore (see Section 15-5)

H 1367 

mol e

k

t

J

hanol



The difference between
Hand
Eis due to the work term, P
Vor (
n)RT.In this
balanced equation there are fewer moles of gaseous products than of gaseous reactants:

n 2  3 1.

C 2 H 5 OH(
)3O 2 (g)88n2CO 2 (g)3H 2 O(
)

Thus, the atmosphere does work on the system (compresses it). Let us find the work done
on the system per mole of reaction.

wP
V(
n)RT

(1 mol)


8

m

.3

o

1

l

4

K

J

(298 K)2.48^103 J

w2.48 kJ or (
n)RT2.48 kJ

We can now calculate
Efor the reaction from
Hand (
n)RTvalues.

E
H(
n)RT[ 1367 (2.48)]1365 kJ/mol rxn

This value agrees with the result that we obtained in Example 15-14. The size of the work
term (2.48 kJ) is very small compared with
H(1367 kJ/mol rxn). This is true for

The positive sign is consistent with the

fact that work is done on the system.

The balanced equation involves one

mole of ethanol, so this is the amount

of work done when one mole of

ethanol undergoes combustion.