Page 165 - Physical Chemistry

lev38627_ch05.qxd 3/3/08 9:33 AM Page 146

146

Chapter 5 T and calculation of U from the known U of combustion of benzoic acid then
298
r
Standard Thermodynamic gives us the heat capacity C . The temperature ranges over which the two combus-
Functions of Reaction K P
tions are carried out are very similar. Also, the main contribution to C K P and C K P
comes from the bomb walls and the water bath. For these reasons, it is an excellent ap-
proximation to take C K P C K P . (In precise work, the difference between the two
is calculated using the known heat capacities of the combustion products.) Knowing
C , we find U from Eq. (5.8).
K P r 298
To find the standard internal energy change U° for the reaction, we must allow
298
for the changes in U and U that occur when the reactants and products are brought
R P
from the states that occur in the calorimeter to their standard states. This correction is
typically about 0.1 percent for combustion reactions.
(An analysis similar to Fig. 5.4b enables one to estimate the temperature of a
flame. See Prob. 5.60.)
For reactions that do not involve gases, one can use an adiabatic constant-pressure
calorimeter. The discussion is similar to that for the adiabatic bomb calorimeter, ex-
cept that P is held fixed instead of V, and H of reaction is measured instead of U.

EXAMPLE 5.2 Calculation of U°from calorimetric data
c
Combustion of 2.016 g of solid glucose (C H O ) at 25°C in an adiabatic bomb
6 12 6
calorimeter with heat capacity 9550 J/K gives a temperature rise of 3.282°C.
Find U° of solid glucose.
c 298
With the heat capacity of the products neglected, Eq. (5.8) gives U
(9550 J/K)(3.282 K) 31.34 kJ for combustion of 2.016 g of glucose. The
experimenter burned (2.016 g)/(180.16 g/mol) 0.01119 mol. Hence U per
mole of glucose burned is ( 31.34 kJ)/(0.01119 mol) 2801 kJ/mol, and this
is U° if the difference between conditions in the calorimeter and standard-
c 298
state conditions is neglected.
Exercise

If 1.247 g of glucose is burned in an adiabatic bomb calorimeter whose heat
capacity is 11.45 kJ/K, what will be the temperature rise? (Answer: 1.693 K.)

Relation between H° and U°
Calorimetric study of a reaction gives either U° or H°. Use of H U PV
allows interconversion between H°and U°. For a process at constant pressure,
H U P V. Since the standard pressure P° [Eq. (5.2)] is the same for all sub-
stances, conversion of pure standard-state reactants to products is a constant-pressure
process, and for a reaction we have
¢H° ¢U° P° ¢V° (5.9)
Similar to H° n H° [Eq. (5.3)], the changes in standard-state volume and
i i m,i
internal energy for a reaction are given by V° n V° and U° n U° . A
i i m,i i i m,i
sum like n U° looks abstract, but when we see n , we can translate this into
i i m,i i i
“products minus reactants,” since the stoichiometric number n is positive for products
i
and negative for reactants.
The molar volumes of gases at 1 bar are much greater than those of liquids or
solids, so it is an excellent approximation to consider only the gaseous reactants and
products in applying (5.9). For example, consider the reaction
aA1s2 bB1g2 S cC1g2 dD1g2 eE1l2