Chemical and Ionic Equilibrium - Result Question 2

####2. In which one of the following equilibria, $K _p \neq K _c$ ?

(a) $2 C(s)+O _2(g) \rightleftharpoons 2 CO(g)$

(2019 Main, 12 April II)

(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. (b)

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) $2 C(s)+O _2(g) \rightleftharpoons 2 CO(g)$

$\Delta n _g=n _{\text {product }}-n _{\text {reactant }}=2-(1)=1$

$\Delta n _g \neq 0 \Rightarrow So, K _p \neq K _c$

(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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