Question

The value of equilibrium. constant of a reaction changes with change of temperature and the change is given by van’t Hoff equation, $\frac{d\ln K_p}{dT}=\frac{\Delta H^{\circ }}{dT}=\frac{\Delta H^0}{R{T}^2}$ where enthalpy change, $\Delta H^{\circ }$ is taken as constant in the small temperature range
If for reaction, $A_{(g)}+3B_{(g)}\rightleftharpoons 2C_{(g)}$, a plot of In $K_{eq}$ versus for $1/T$ a reaction is shown, then which of the following condition will be favourable for formation of product $C$ ?

• A. Low temperature and high pressure
• B. High temperature and high pressure
• C. High temperature and low pressure
• D. Low temperature and low pressure

Answer 1

Answer: (B)

Since slope is negative hence reaction is endothermic. So high temperature favours forward reaction similarly high pressure favours forward reaction