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Section 6.3
N (g) 3H (g) ∆ 2NH (g) Temperature Dependence of the
3
2
2
Equilibrium Constant
K P
Figure 6.6
Thermodynamic quantities for
N (g) 3H (g) ∆ 2NH (g). In
2
2
3
the T → 0 limit, S° → 0.

At high temperatures, RT ln K° G° T S°, so R ln K° S° in the
P
P
high-T limit—note the approach of the R ln K° curve to the S° curve at high T. At
P
low T, RT ln K° G° H°, so ln K° H°/RT. Therefore ln K° and K° go
P
P
P
P
to infinity as T → 0. The products 2NH (g) have a lower enthalpy and lower internal
3
energy (since U° H° in the T 0 limit) than the reactants N (g) 3H (g), and
2
2
in the T 0 limit, the equilibrium position corresponds to complete conversion to the
low-energy species, the products. The low-T equilibrium position is determined by the
internal-energy change U°. The high-T equilibrium position is determined by the
entropy change S°.
Figure 6.6b plots ln K° versus 1/T for N (g) 3H (g) ∆ 2NH (g) for the range
2
P
3
2
200 to 1000 K. The plot shows a very slight curvature, as a result of the small tem-
perature variation of H°.
EXAMPLE 6.3 H°from K° versus T data
P
Use Fig. 6.6b to estimate H° for N (g) 3H (g) ∆ 2NH (g) for temperatures
2
2
3
in the range 300 to 500 K.
Since only an estimate is required, we shall ignore the slight curvature of the
plot and treat it as a straight line. The line goes through the two points
1
1
T 1 0.0040 K , ln K° 20.0 and T 1 0.0022 K , ln K° 0
P
P
1
Hence the slope (Sec. 1.6) is (20.0 0)/(0.0040 K 1 0.0022 K ) 1.11
4
10 K. Note that the slope has units. From Eq. (6.40), the slope of a ln K°-versus-
P
1/T plot equals H°/R, so
4
1
1
¢H° R slope 11.987 cal mol K 211.11 10 K2
22 kcal>mol
in agreement with Fig. 6.6a.