Question 23.11: One step in the conversion of coal to liquid fuels involves ......
One step in the conversion of coal to liquid fuels involves the reaction between H_{2}(g) and CO(g) to produce methanol, CH_{3}OH(g).
CO(g) + 2\ H_{2}(g) ⇋ CH_{3}OH(g) \qquad ΔH^{\circ}_{rxn} = –90.5\ kJ·mol^{–1} (23.29)
The equilibrium constant at 25°C for this chemical equation is K = 2.5 × 10^4. Calculate the value of K at 325°C.
We can apply the van’t Hoff equation (Equation 23.28) to the reaction described by Equation 23.29, using the value of ΔH^{\circ}_{rxn} given. At T_{1} = 298\ K, K_{1} = 2.5 × 10^4. Thus, at T_{2} = 573\ K, we have for K_{2}
ln\left(\frac{K_2}{K_{1}} \right) =\frac{ΔH^{\circ}_{rxn}}{R} \left(\frac{T_{2} – T_{1}}{T_{1}T_{2}} \right) (23.28)
ln\left(\frac{K_2}{2.5 × 10^4} \right)=\left(\frac{–90.5 × 10^3\ J·mol^{–1}}{8.3145\ J^{–1}·K·mol^{–1}} \right) \left(\frac{598\ K – 298\ K}{(598\ K)(298\ K)} \right)and
K_{2}= (2.5 × 10^4)(e^{ –18.3}) = 3 × 10^{–4}The large decrease in K with increasing T for this chemical reaction is a consequence of the large negative value of ΔH^{\circ}_{rxn} for the equation.