Physical Chemistry Third Edition

98 2 Work, Heat, and Energy: The First Law of Thermodynamics

For 1 mol of reaction

∆U−

(

4. 956 × 104 J

)

(

1 .234 g

)

(

128 .19 g

1 mol

)

− 5. 148 × 106 J mol−^1

∆H− 5. 148 × 106 J mol−^1 +(−2)

(

8 .3145 J K−^1 mol−^1

)

(298.15 K)

− 5. 153 × 106 J mol−^1 −5153 kJ mol−^1

This value compares with the published value,∆H◦− 5156 .8kJmol−^1 , which applies to

a constant-temperature reaction at 298.15 K.

The calculation in the previous example ignores the fact that the final temperature

is not equal to the initial temperature and that the pressure is not equal to exactly

1 bar. It also neglects the contribution to qfrom the combustion of part of the

wire fuse.

EXAMPLE2.35

a.Estimate the error due to the inequality of the temperatures of the reactant and products

in the previous example.

b.Assume that the mass of iron wire burned in the combustion experiment of the pre-

vious example is 8.0 mg. The energy of combustion of iron is− 6 .68 kJ g−^1. Find

the percent error in Example 2.34 caused by neglect of the energy of the wire’s

combustion.

Solution

a.The initial amount of naphthalene was 0.00963 mol, giving 0.0963 mol of CO 2 and

0.0385 mol of H 2 O. From the heat capacities in Table A.8, the heat capacity of the products

is 6.55 J K−^1. The heat required to bring the products back to 298.15 K is

q 2 

(

6 .55JK−^1

)

(3.484 K) 22 .8J 0 .0228 kJ

This means thatqcal 4. 956 × 104 J+ 22 .9J 4. 958 × 104 J for the reaction with the

products at the same temperature as the reactants, for a difference of about .04%.

b.From the example

q− 4. 956 × 104 J

The energy of combustion of the iron wire is

∆Uwire(0.0080 g)

(

− 6 .68 kJ g−^1

)

 0 .0534 kJ53 J

The percent error is

% error

53 J

4. 956 × 104 J

×100% 0 .11%