Question

For a given exothermic reaction, ${\mathrm{K}}_{\mathrm{p}}$​ and ${{\mathrm{K}}_{\mathrm{p}}}^{\text{'}}$​ are the equilibrium constants at temperatures ${\mathrm{T}}_1$and ${\mathrm{T}}_2$respectively. Assuming that heat of reaction is constant in temperatures range between ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$, it is readily observed that

Answer 1

Step 1: Given data

Equilibrium constant at temperature ${\mathrm{T}}_1$and ${\mathrm{T}}_2$$=$ ${\mathrm{K}}_{\mathrm{p}}$​ and ${{\mathrm{K}}_{\mathrm{p}}}^{\text{'}}$​

The heat of the reaction$=$constant

Step 2: Observing ${\mathrm{K}}_{\mathrm{p}}$​ and ${{\mathrm{K}}_{\mathrm{p}}}^{\text{'}}$​

Using formula $\log \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{∆\mathrm{H}°}{2.303\mathrm{R}}(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2})$

${\mathrm{T}}_2>{\mathrm{T}}_1$ So, ${\mathrm{K}}_{\mathrm{p}}>{{\mathrm{K}}_{\mathrm{p}}}^{\text{'}}$(exothermic reaction)

As a result, for an exothermic reaction, the equilibrium will move in the opposite direction as the temperature rises, and the equilibrium constant's value will decrease according to the Le-Chatelier principle.

Therefore, ${\mathrm{K}}_{\mathrm{p}}>{{\mathrm{K}}_{\mathrm{p}}}^{\text{'}}$for a given exothermic reaction, where ${\mathrm{K}}_{\mathrm{p}}$​ and ${{\mathrm{K}}_{\mathrm{p}}}^{\text{'}}$​ are the equilibrium constants at temperatures ${\mathrm{T}}_1$and ${\mathrm{T}}_2$respectively and heat of reaction is constant.