Chemical and Ionic Equilibrium - Result Question 2
2. In which one of the following equilibria, $K _p \neq K _c$ ?
(2019 Main, 12 April II)
(a) $2 C(s)+O _2(g) \rightleftharpoons 2 CO(g)$
(b) $2 HI(g) \rightleftharpoons H _2(g)+I _2(g)$
(c) $NO _2(g)+SO _2(g) \rightleftharpoons NO(g)+SO _3(g)$
(d) $2 NO(g) \rightleftharpoons N _2(g)+O _2(g)$
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Answer:
Correct Answer: 2. (a)
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 $\Delta n _g=+$ ve then $K _p > K _c$
If $\Delta n _g=-$ ve then $K _p < K _c$
Consider the following equilibria reactions
(a) $2 C(s)+O _2(g) \rightleftharpoons 2 CO(g)$
$\Delta n _g=n _{\text {product }}-n _{\text {reactant }}=2-(1)=1$
$\Delta n _g \neq 0 \Rightarrow So, K _p \neq K _c$
(b) $2 HI(g) \rightleftharpoons H _2(g)+I _2(g)$
$\Delta n _g=n _{\text {product }}-n _{\text {reactant }}=2-2=0$
$\Delta n _g=0 \Rightarrow So, K _p=K _c$
(c) $NO _2(g)+SO _2(g) \rightleftharpoons NO(g)+SO _3(g)$
$\Delta n _g=n _{\text {product }}-n _{\text {reactant }}=2-2=0$
$\Delta n _g=0 \Rightarrow$ So, $K _p=K _c$
(d) $2 NO(g) \rightleftharpoons N _2(g)+O _2(g)$
$\Delta n _g=n _{\text {product }}-n _{\text {reactant }}=2-2=0$
$\Delta n _g=0 \Rightarrow$ So, $K _p=K_c$
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BETA
