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Q.

The $K_{c}$ for the reaction $N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$ is $4.62 \times 10^{- 3} M$ at 298 K. Its $K_{P}$ value would be

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a

$1.86 \times 10^{- 6} kPa$

b

11.45 kPa

c

$0.07 {kPa}^{- 1}$

d

0.038 kPa

answer is A.

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Detailed Solution

$K_{p} = K_{c} (R T)^{Δ ν_{E};}$ where $Δ ν_{g} = + 1$

$= (4.62 \times 10^{- 3} {molL}^{- 1}) (8.314 {kPaLK}^{- 1} {mol}^{- 1}) (298 K) = 11.45 kPa$

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The Kc for the reaction N2O4(g)⇌2NO2(g) is 4.62×10−3M at 298 K. Its KP value would be