Question

For the equilibrium reaction
$2 NOBr ( g ) \rightleftharpoons 2 NO ( g )+ Br _2( g )$
If $P_{B r_2}=\frac{P}{9}$, where $P$ is total pressure at equilibrium. The ratio $\frac{K_p}{P}$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{1}{81}$
• C. $\frac{1}{9}$
• D. $\frac{1}{27}$

Answer 1

Answer: (B)

$\underset{x}{2 NOBr ( g ) } \rightleftharpoons \underset{\frac{2P}{9}}{2 NO ( g )} + \underset{\frac{P}{9}}{Br _2( g )}$
Since total pressure is $P$.
So, $x+\frac{2 P}{9}+\frac{P}{9}=P \Rightarrow x=\frac{6 P}{9}$
$\therefore K_p=\frac{p_{ NO }^2 \times p_{ Br _2}}{p_{ NOBr }^2}=\frac{\left(\frac{2 P}{9}\right)^2 \times \frac{P}{9}}{\left(\frac{6 P}{9}\right)^2}$
$\therefore \frac{K_p}{P}=\frac{1}{81}$