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When the decomposition of one mole of $N_{2} O_{4} (g)$ occurs at 1 atm and 280 K then $k_{p} = 1 a t m$ . If the same reaction starting with one mole of $N_{2} O_{4} (g)$ is carried out at 6 atm and at same temperature then the value of $k_{p}$ would be

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$\frac{1}{3} a t m$

$6 a t m$

$\frac{1}{6} a t m$

$1 a t m$

answer is B.

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Detailed Solution

$N_{2} O_{4} ⇌ 2 N O_{2}$

$k_{p}$ or $k_{c}$ do not changes with pressure, $k_{p}$ value would be 1 atm.

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