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)

$Q _{ p }=\frac{ p _{ NOBrig }^{2}}{ p _{ NOig }^{2} \cdot p _{ Br _{2}( g )}}$

$=\frac{(0.04 atm )^{2}}{(0 . \operatorname{latm})^{2} \times(0.1 atm )}$

$=\frac{16 \times 10^{-4} atm ^{2}}{1 \times 10^{-4} atm \times 1 \times 10^{-1} atm }$

$=160 atm ^{-1}$

Since $Q_{p}=K_{p},$ therefore there is equilibrium in the flask.