Nitric oxide reacts with $\mathrm{Br}_2$ and gives nitrosyl bromide as per reaction given below: $2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NOBr}(\mathrm{g})$ When $0.087 \mathrm{~mol}$ of $\mathrm{NO}$ and $0.0437 \mathrm{~mol}$ of $\mathrm{Br}_2$ are mixed in a closed container at constant temperature, $0.0518 \mathrm{~mol}$ of $\mathrm{NOBr}$ is obtained at equilibrium. Calculate equilibrium amount of $\mathrm{NO}$ and $\mathrm{Br}_2$.

Question

Nitric oxide reacts with $\mathrm{Br}_2$ and gives nitrosyl bromide as per reaction given below: $2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NOBr}(\mathrm{g})$ When $0.087 \mathrm{~mol}$ of $\mathrm{NO}$ and $0.0437 \mathrm{~mol}$ of $\mathrm{Br}_2$ are mixed in a closed container at constant temperature, $0.0518 \mathrm{~mol}$ of $\mathrm{NOBr}$ is obtained at equilibrium. Calculate equilibrium amount of $\mathrm{NO}$ and $\mathrm{Br}_2$.,

MARKS App · Accepted Answer

$0.0518 \mathrm{~mol}$ of $\mathrm{NOBr}$ is formed from $0.087 \mathrm{~mol}$ of $\mathrm{NO}$ and $0.0437 \mathrm{~mol}$ of $\mathrm{Br}_2$. At equilibrium, amount of $\mathrm{NO}=0.087-0.0518$ $=0.0352 \mathrm{~mol}$ Amount of $\mathrm{Br}_2=0.0437-0.0259=0.0178 \mathrm{~mol}$.

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