Chemical and Ionic Equilibrium 1 Question 51
2. In which one of the following equilibria, $K_{p} \neq K_{c}$ ?
(a) $2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$
(2019 Main, 12 April II)
(b) $2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}{2}(g)+\mathrm{I}{2}(g)$
(c) $\mathrm{NO}{2}(g)+\mathrm{SO}{2}(g) \rightleftharpoons \mathrm{NO}(g)+\mathrm{SO}_{3}(g)$
(d) $2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}{2}(g)+\mathrm{O}{2}(g)$
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Solution:
Key Idea The relationship between $K_{p}$ and $K_{c}$ is
$$ K_{p}=K_{c}(R T)^{\Delta n_{g}} $$
where, $\Delta n_{g}=n_{\text {products }}-n_{\text {reactants }}$
If $\Delta n_{g}=0$ then $K_{p}=K_{c}$
If $\Delta n_{g}=+$ vethen $K_{p}>K_{c}$
If $\Delta n_{g}=-$ vethen $K_{p}<K_{c}$
Consider the following equilibria reactions
(a) $2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$
$\Delta n_{g}=n_{\text {product }}-n_{\text {reactant }}=2-(1)=1$
$\Delta n_{g} \neq 0 \Rightarrow \mathrm{So}, K_{p} \neq K_{c}$
(b) $2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}{2}(g)+\mathrm{I}{2}(g)$
$\Delta n_{g}=n_{\text {product }}-n_{\text {reactant }}=2-2=0$
$\Delta n_{g}=0 \Rightarrow \mathrm{So}, K_{p}=K_{c}$
(c) $\mathrm{NO}{2}(g)+\mathrm{SO}{2}(g) \rightleftharpoons \mathrm{NO}(g)+\mathrm{SO}_{3}(g)$
$\Delta n_{g}=n_{\text {product }}-n_{\text {reactant }}=2-2=0$
$\Delta n_{g}=0 \Rightarrow$ So, $K_{p}=K_{c}$
(d) $2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}{2}(g)+\mathrm{O}{2}(g)$
$\Delta n_{g}=n_{\text {product }}-n_{\text {reactant }}=2-2=0$
$\Delta n_{g}=0 \Rightarrow$ So, $K_{p}=K_{c}$
