Page 114 - Essentials of physical chemistry

76 Essentials of Physical Chemistry

TABLE 4.3
Polynomial Fits to Selected Heat Capacities, Temperature Range 298.158K–15008K
Compound a b (E–2) c (E–5) d (E–9) e (E–12) R 2
26.78 1.224 2.299 19.40 5.261 0.9992
H 2
O 2 28.28 2.347 2.647 24.15 6.679 0.9997
CO 31.30 1.699 4.053 28.34 6.686 1.0000
CO 2 19.29 7.756 6.818 31.75 6.123 1.0000
53.24 3.795 1.887 7.421 5.209 0.9896
C 2gas
C graphite 6.227 6.520 6.046 2.908 5.881 1.0000
12.16 8.158 7.889 40.09 8.537 0.9999
C diamond
HCCH 14.29 14.09 16.54 102.2 24.20 0.9999
4.279 15.04 7.623 14.87 0 1.0000
H 2 CCH 2
4.084 18.34 7.623 11.31 0 1.0000
H 3 CCH 3
26.86 2.817 11.50 94.51 23.74 1.0000
CH 4
31.41 1.653 3.643 24.00 5.383 1.0000
N 2
NH 3 27.38 2.210 2.487 24.26 6.197 1.0000
HCl 30.68 0.9687 1.756 8.024 1.087 0.9999
Cl 2 26.58 3.722 5.035 31.09 7.137 0.9994
H 2 O 32.20 0.4494 2.286 12.43 2.252 1.0000
H 2 CO 27.11 1.052 7.635 70.77 18.84 0.9999

This is surely a formidable expression but we can simplify it by doing the integrals term by term.

2 2 3 3
T (298:15) T (298:15)
DH rxn (T) ¼ DH 0 þ (Da)(T 298:15) þ (Db) þ (Dc)
f,298:15
2 3
4 4 5 5
T (298:15) T (298:15)
þ (Dd) þ (De) :
4 5
n
n
n
A common error students make is to use (T 298.15) instead of the correct form as [T (298.15) ].
This sort of problem often appeared on final examinations in the days when students had only slide
rules so it should be easier with a calculator. The easy (?) way to set up this problem is to first calculate
the D terms in the heat capacity polynomial.

Keep in mind the coefficients in the balanced equation 1=2H 2 þ 1=2Cl 2 ! HCl.
T
ð
4
2
3
The key formula is DH rxn (T) ¼ DH 0 (Da þ DbT þ DcT þ DdT þ DeT )dt.
298:15 þ
298:15
Use data from Table 4.3
Compound a b (E 2) c (E 5) d (E 9) e (E 12) R 2
HCl 30.68 0.9687 1.756 8.024 1.087 0.9999
H 2 26.78 1.224 2.299 19.40 5.261 0.9992
26.58 3.722 5.035 31.09 7.137 0.9994
Cl 2