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A vessel (A) contains 1 mole each of $N_{2}$ and $O_{2}$ and another vessel (B) contains 2 mol each of $N_{2}$ and $O_{2}$ . Both vessels are heated to the same temperature till equilibrium is established in both cases. Then, correct statement is

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$K_{P}$ for $N_{2} + O_{2} ⇋ 2 N O$ in A and B are in the ratio $1 : 2$

$K_{P}$ for $N_{2} + O_{2} ⇋ 2 N O$ in A and B are in the ratio $2 : 1$

$K_{C}$ for $N_{2} + O_{2} ⇋ 2 N O$ in A and B are in the ratio $1 : 2$

$K_{C}$ for $N_{2} + O_{2} ⇋ 2 N O$ in A and B are equal

answer is C.

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Detailed Solution

According to the law of chemical equilibrium,

$K_{C} = \frac{[N O_{2}]^{2}}{[N_{2}] [O_{2}]} K_{C} = \frac{[2]^{2}}{[2] [2]} K_{C} = 1$

Therefore, the correct answer is option (C) $K_{C}$ for $N_{2} + O_{2} ⇋ 2 N O$ in A and B are equal.

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