Physical Chemistry Third Edition

1312 K Answers to Numerical Exercises and Odd-Numbered Numerical Problems

2.29
a. percent difference− 41 .9%
b. T 2  425 .7K
c.q 0
∆Uw−5103 J

2.31
a. T 2  282 .8K
q 0
w∆U−1878 J
b. T 2  235 .1K
q 0
w∆U−1722 J

2.33

298.15 K 500 K 1000 K 2000 K

Ar 0 0 0 0

H − 0. 01 0

He 0 0 0 0

O5. 41 2. 27 0. 62 0. 19

C0. 25 0. 087 0. 024 0. 80

2.35
a. T 2  403 .2K
b. ∆U2599 J
w2599 J
c.T 2  403 .6K

2.37
a. T 2  346 .4K
b. w∆U−4310 J
c.q 0
∆Uw−2770 J
T 2  383 .4K
d.∆U 0
q−w6860 J

2.41
a. w 0
q∆U2806 J
b. ∆U2806 J
q4677 J
w−1871 J

2.43
a. q∆H2079 J
w− 831 .45 J
∆U1248 J

b. q∆U1247 J

w 0

∆H2079 J

2.45

∆H5745 J

2.47

q∆H 24 .80 kJ

2.49

a. ∆H◦ 1236 .79 kJ mol−^1

∆U◦ 1234 .31 kJ mol−^1

b. ∆H◦− 565 .99 kJ mol−^1

∆U◦− 563 .53 kJ mol−^1

c. ∆H◦− 1366 .84 kJ mol−^1

∆U◦− 1364 .37 kJ mol−^1

2.51

a. ∆H◦348K− 1233 .38 kJ mol−^1

b. ∆H◦348K− 566 .66 kJ mol−^1

c.∆H◦348K− 1368 .19 kJ mol−^1

2.53

a. ∆H◦− 3119 .71 kJ mol−^1

b. ∆H◦− 2875 .76 kJ mol−^1

c. ∆H− 2855 .68 kJ mol−^1 z

2.55

a. ∆fH◦(C 12 H 22 O 11 )− 2224 .5kJmol−^1

∆U◦− 5640 .9kJmol−^1

b. ∆fH◦(C 18 H 36 O 2 )− 947 .7kJmol−^1

The enthalpy change per gram is− 39 .65 kJ g−^1

The enthalpy change per gram of sucrose

− 16 .48 kJ g−^1

2.57

a. ∆H◦− 890 .309 kJ mol−^1

b. ∆U◦− 885 .351 kJ mol−^1

c. ∆U◦− 885 .347 kJ mol−^1

2.59

a. ∆U◦≈−2796 kJ mol−^1

∆H◦≈−2794 kJ mol−^1

b. ∆H298K◦ − 2855 .68 kJ mol−^1

∆H373K◦ − 2852 .74 kJ mol−^1

c.∆U◦≈−606 kJ mol−^1

∆H◦≈−608 kJ mol−^1