For the reaction N_2 O_4( g) rightleftarrows 2 NO_2( g), K_ p=0.492 atm at 300 K . K_ c for the reaction at same temperature is _________ x 10^-2. (Given : R=0.082 L atm mol^-1 K^-1)

K_ P= K_ C * ( RT)^Delta n_ g \ Delta n_ g=1 \ => K_ c=frac K_ P RT=(0.492 / 0.082 x 300)=2 x 10^-2 \

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For the reaction N2 O4g leftarrows 2 NO2g Kp0492 atm at... | (Chemistry) | DivineJEE

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Chemistry

Exam

Chemistry

For the reaction $\mathrm{N}_2 \mathrm{O}_{4(\mathrm{~g})} \rightleftarrows 2 \mathrm{NO}_{2(\mathrm{~g})}, \mathrm{K}_{\mathrm{p}}=0.492 \mathrm{~atm}$ at $300 \mathrm{~K} . \mathrm{K}_{\mathrm{c}}$ for the reaction at same temperature is _________ $\times 10^{-2}$ .

(Given : $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ )

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Explanation

$\begin{aligned} & \mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}} \cdot(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \\ & \Delta \mathrm{n}_{\mathrm{g}}=1 \\ & \Rightarrow \mathrm{K}_{\mathrm{c}}=\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{RT}}=\frac{0.492}{0.082 \times 300}=2 \times 10^{-2} \end{aligned}$