Question
For a given exothermic reaction, $${K_p}$$ and \[K_{p}^{'}\] are the equilibrium constants at temperatures $${T_1}$$ and $${T_2},$$ respectively. Assuming that heat of reaction is constant in temperature range between $${T_1}$$ and $${T_2},$$ it is readily observed that
A.
\[{{K}_{p}}>K_{p}^{'}\]
B.
C.
\[{{K}_{p}}=K_{p}^{'}\]
D.
\[{{K}_{p}}=\frac{1}{K_{p}^{'}}\]
Answer :
\[{{K}_{p}}>K_{p}^{'}\]
Solution :
The equilibrium constant at two different temperatures for a thermodynamic process is given by
$$\log \frac{{{K_2}}}{{{K_1}}} = \frac{{\Delta {H^ \circ }}}{{2.303R}}\left[ {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right]$$
Here, $${K_1}$$ and $${K_2}$$ are replaced by $${K_p}$$ and \[K_{p}^{'}.\]
Therefore, \[\log \frac{{{K}^{'}}_{p}}{{{K}_{p}}}=\frac{\Delta {{H}^{\circ }}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\]
For exothermic reaction,
\[\begin{align}
& {{T}_{2}}>{{T}_{1}}\,\,\text{and}\,\,H=-ve \\
& \therefore \,\,{{K}_{p}}>{{K}^{'}}_{p} \\
\end{align}\]