For exothermic reaction, $\Delta \mathrm{H}= -ve$ i.e. heat is evolved. The temperature $\mathrm{T}_{2}$ is higher than $\mathrm{T}_{1}$
Thus, $\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)$ is negative.
so, $\log K_{p}^{\prime}-\log K_{p}=-$ ve or $\log K_{p}>\log K_{p}^{\prime}$
$ K_{p}>K_{p}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
${N_2}(g)\, + 3{H_2}(g)\, \rightleftharpoons \,2N{H_3}(g)$
The equilibrium constant of the above reaction is $K_3$. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that $P_{NH_3}<\,< P_{total}$ at equilibrium)
