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The value of equilibrium constant of a reaction changes with change of temperature and the change is given by van’t Hoff equation, $\frac{d l n K_{p}}{d T} = \frac{Δ H^{°}}{R T^{2}},$ where enthalpy change, $Δ H °$ is taken as constant in the small temperature range. If for reaction, $A_{(g)} + 3 B_{(g)} ⇌ 2 C_{(g)},$ a plot of $\ln K_{e q}$ versus 1/T is shown, then which of the following condition will be favorable for formation of product C?

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Low temperature and low pressure

Low temperature and high pressure

High temperature and low pressure

High temperature and high pressure

answer is B.

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Detailed Solution

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

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