For the reaction $2 NO (g) + {Cl}_{2} (g) ⇌ 2 NOCl (g)$ , a reaction mixture containing $NO (g)$ and ${Cl}_{2} (g)$ at partial pressures of0.373 atm and 0.310 atm, respectively, is taken and at equilibrium the total pressure of the system is found to be 0.544 atm The value of $K_{p}$ for the reaction is

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$50.08 {atm}^{- 1}$

$(1 / 0.544) {atm}^{- 1}$

$25.04 {atm}^{- 1}$

$0.544 {atm}^{- 1}$

answer is D.

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Detailed Solution

$2 NO (g) + {Cl}_{2} (g) ⇌ 2 NOCl (g)$

$0.373 atm-2p 0.310 atm - p 2 p$

$\begin{array}{l} p_{total} = (0.373 atm - 2 p) + (0.310 atm - p) + 2 p = 0.683 atm - p \\ p = 0.683 atm - p_{total} = (0.683 - 0.544) atm = 0.139 atm \end{array}$

$K_{p} = \frac{{(p_{NOCI})}^{2}}{{(p_{NO})}^{2} (p_{{Cl}_{2}})} = \frac{(2 \times 0.139 atm)^{2}}{(0.373 - 2 \times 0.139)^{2} (0.310 - 0.139) {atm}^{3}} = 50.08 {atm}^{- 1}$

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