For the reaction atm at 300 K. for the reaction at same temperature is

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Solution

Q.
For the reaction $N_{2} O_{4 (g)} ⇄ 2 {NO}_{2 (g)},$ $K_{p} = 0.492$ atm at 300 K. $K_{c}$ for the reaction at same temperature is _______ $\times 10^{- 2}$ . (Given: R = 0.082 L atm ${mol}^{- 1}$ $K^{- 1}$ ) [2024]

Ans.
(2)

$N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$

$Δ n_{g} =$ gaseous moles of products - gaseous moles of reactants

$= 2 - 1 = 1$

$K_{p} = K_{c} {(R T)}^{Δ n_{g}}$

$0.492 atm = K_{c} \times {(0.082 atm L {mol}^{- 1} K^{- 1} \times 300 K)}^{1}$

$K_{c} = \frac{0.492}{0.082 \times 300} mol L^{- 1}$

$= 0.02 mol L^{- 1} = 2 \times 10^{- 2} mol L^{- 1}$

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