The equilibrium constant at constant pressure, \(K _{ p }\) for the reaction, \(2 NO _{( g )}+ Cl _{2( g )} \leftrightarrow 2 NOCl _{( g )}\) is 800 . The partial pressures of the gas mixture at equilibrium are \(0.02 atm\) each for \(NO\) and \(Cl _{2}\). What is the partial pressure of \(NOCl\) in the mixture?
A. \(0.02 atm\) B. \(6.41 \times 10\) \({ }^{4} atm\) C. \(0.04 atm\) D. \(0.08 atm\)
Correct Answer: D
Explanation
\begin{aligned} &2 NO _{( g )}+ Cl _{2( g )} \leftrightarrow 2 NOCl _{( g )}, kp =800 \\ &k p=\frac{P^{2}, NOCl _{( g )}}{P_{ NO ( g )}^{2} \cdot P_{ Cl _{2}( g )}} \\ &800=\frac{P^{2} \cdot NOCl _{( g )}}{(0.02)^{2} \cdot 0.02} \\ &P^{2} \cdot NOCl =800 \times(0.02)^{3}=0.0064 \\ &P_{ NOCl }=\sqrt{0.0064}=0.08 atm \end{aligned}