Question

For the reaction equilibrium $2NOBr(g)\rightleftharpoons 2NO(g)+B{{r}_{2}}(g)$, if ${{P}_{B{{r}_{2}}}}=\dfrac{P}{9}$ at equilibrium and P is the initial total pressure, then the ratio $\dfrac{{{K}_{p}}}{P}$ is equal to:
(a) $\dfrac{1}{9}$
(b) $\dfrac{4}{441}$
(c) $\dfrac{1}{27}$
(d) $\dfrac{1}{3}$

Answer 1

Hint: Before the equilibrium reaction the pressure of $NOBr$ will be P and pressure of $NO$ and $B{{r}_{2}}$ will be zero. After equilibrium the pressure of $NOBr$will be $P-2P'$, pressure of $NO$will be $2P'$ and pressure of $B{{r}_{2}}$will be $P'$. The equilibrium constant at pressure can be calculated by the ratio of the product of partial pressure of products to the partial pressure of the reactant.

Complete step by step answer:
We are given a reaction in which two moles of nitrosyl bromide in equilibrium forms two moles of nitrogen oxide and one mole of bromine gas. The reaction is:
$2NOBr(g)\rightleftharpoons 2NO(g)+B{{r}_{2}}(g)$
So, before the equilibrium reaction, the pressure of $NOBr$ will be P and the pressure of $NO$, and $B{{r}_{2}}$ will be zero. After equilibrium the pressure of $NOBr$will be $P-2P'$, pressure of $NO$will be $2P'$ and pressure of $B{{r}_{2}}$will be $P'$.
And the partial pressure of bromine ${{P}_{B{{r}_{2}}}}$ is given as: ${{P}_{B{{r}_{2}}}}=\dfrac{P}{9}$and this is equal to $P'$.
So, by equating this value we can get the partial pressure of $NO$ and $NOBr$.
The partial pressure of $NOBr$= $P-2P'$
$NOBr=P-2\dfrac{P}{9}$
$NOBr=\dfrac{7P}{9}$
The partial pressure of $NO$ = $2P'$
$NO=2\text{ x }\dfrac{P}{9}$
$NO=\dfrac{2P}{9}$
So, the equilibrium constant at pressure can be calculated by the ratio of the product of partial pressure of products to the partial pressure of the reactant.
${{K}_{p}}=\dfrac{{{(NO)}^{2}}(B{{r}_{2}})}{{{(NOBr)}^{2}}}$
So, putting the value of partial pressure in the formula, we get
${{K}_{p}}=\dfrac{{{\left( \dfrac{2P}{9} \right)}^{2}}\left( \dfrac{P}{9} \right)}{{{\left( \dfrac{7P}{9} \right)}^{2}}}$
${{K}_{p}}=\dfrac{4P}{441}$
We have to find the ratio $\dfrac{{{K}_{p}}}{P}$:
$\dfrac{{{K}_{p}}}{P}=\dfrac{4P}{441P}=\dfrac{4}{441}$
Therefore the correct answer is an option (b) $\dfrac{4}{441}$.

Note: When you are writing the formula of the chemical equilibrium constant, the number of moles of the compound must be the power of the compound in the formula, otherwise the value of constant will be wrong.