Page 202 - Physical Chemistry

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Since G° H° T S°, we end up with Section 6.3
Temperature Dependence of the
d ln K° ¢H° Equilibrium Constant
P (6.36)*
dT RT 2
This is the van’t Hoff equation. [Since ln K° G°/RT, Eq. (6.36) follows from
P
the Gibbs–Helmholtz equation (Prob. 4.13) (
(G/T)/
T) H/T .] In (6.36), H°
2
P
H° is the standard enthalpy change for the ideal-gas reaction at temperature T
T
[Eq. (5.3)]. The greater the value of H° , the faster the equilibrium constant K°
P
changes with temperature.
The degree superscript in (6.36) is actually unnecessary, since H of an ideal gas is
independent of pressure and the presence of other ideal gases. Therefore, H per mole
of reaction in the ideal gas mixture is the same as H°. However, S of an ideal gas de-
pends strongly on pressure, so S and G per mole of reaction in the mixture differ
quite substantially from S° and G°.
Multiplication of (6.36) by dT and integration from T to T gives
1 2
¢H°
d ln K° dT
P
RT 2
K°1T 2 T 2 ¢H°1T2
P
2
ln 2 dT (6.37)
K°1T 2 T 1 RT
P
1
To evaluate the integral in (6.37), we need H° as a function of T. H°(T) can be
found by integration of C° (Sec. 5.5). Evaluation of the integral in Eq. (5.19) leads
P
to an equation with the typical form (see Example 5.6 in Sec. 5.5)
3
2
¢H° A BT CT DT ET 4 (6.38)
T
where A, B, C, D, and E are constants. Substitution of (6.38) into (6.37) allows K° at
P
any temperature T to be found from its known value at T .
2 1
H° for gas-phase reactions usually varies slowly with T, so if T T is reason-
2 1
ably small, it is generally a good approximation to neglect the temperature dependence
of H°. Moving H° outside the integral sign in (6.37) and integrating, we get
K°1T 2 ¢H° 1 1
2
P
ln a b (6.39)
K°1T 2 R T 1 T 2
P
1
EXAMPLE 6.2 Change of K° with T
P
Find K° at 600 K for N O (g) ∆ 2NO (g) (a) using the approximation that H°
P 2 4 2
is independent of T; (b) using the approximation that C° is independent of T;
P
(c) using the NIST-JANAF tables (Sec. 5.9).
(a) If H° is assumed independent of T, then integration of the van’t
Hoff equation gives (6.39). Appendix data for NO (g) and N O (g) give
2 2 4
H° 57.20 kJ/mol and G° 4730 J/mol. From G° RT ln K°, we
298 298 P
find K° 0.148. Substitution in (6.39) gives
P,298
K° 57200 J>mol 1 1
P,600
ln a b 11.609
0.148 8.314 J>mol-K 298.15 K 600 K
K° 1.63 10 4
P,600