1. Tardigrade
  2. Question
  3. Chemistry
  4. The equilibrium constant at 25° C for the reaction, 2NO(.g.)+Br2(.g.)leftharpoons2NOBr(.g.) is equal to 160 atm- 1 . If the partial pressures of NO, Br2 and NOBr in a flask at 25° C are 0.01, 0.1 and 0.04 atm respectively. It can be said that

Q. 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

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Solution:

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

Questions from NTA Abhyas 2022

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