Question

The equilibrium constant at $25^\circ C$ for the reaction, $2NO\left(\right.g\left.\right)+Br_{2}\left(\right.g\left.\right)\rightleftharpoons2NOBr\left(\right.g\left.\right)$ is equal to 160 $atm^{- 1}$ . If the partial pressures of NO, $Br_{2}$ and NOBr in a flask at $25^\circ $ C are 0.01, 0.1 and 0.04 atm respectively. It can be said that

• A. there is equilibrium in the flask
• B. there reaction will proceed in the forward direction
• C. the reaction will proceed in the backward direction
• D. the partial pressure of NOBr finally will be 0.05 atm

Answer 1

Answer: (A)

$Q_{p}=\frac{p_{N O B r \left(\right. g \left.\right)}^{2}}{p_{N O \left(\right. g \left.\right)}^{2} . p_{B r_{2} \left(\right. g \left.\right)}}$
$=\frac{\left(\right. 0.04 a t m \left(\left.\right)^{2}}{\left(\right. 0.1 a t m \left(\left.\right)^{2} \times \left(\right. 0.1 a t m \left.\right)}$
$=\frac{16 \times 1 0^{- 4} a t m^{2}}{1 \times 1 0^{- 4} a t m \times 1 \times 1 0^{- 1} a t m}$
$=160atm^{- 1}$
Since $Q_{p}=K_{p}$ , therefore there is equilibrium in the flask.