Reaction between N_2 and O_ 2– takes place as follows: 2N_2(g) + … - Vidyadip

Question

Reaction between $N_2$ and $O_{2–}$ takes place as follows:
$\text{2N}_2\text{(g) + O}_2\text{(g)}\rightleftharpoons2\text{N}_2\text{O(g)}$
If a mixture of 0.482mol $N_2$ and 0.933mol of $O_2$ is placed in a 10L reaction vessel and allowed to form $N_2O$ at a temperature for which $\text{K}_{\text{c}}=2.0\times10^{-37},$ determine the composition of equilibrium mixture.

✓

Answer

Let the concentration of $N_2O$ at equilibrium be x.
The given reaction is:
$\begin{matrix}&2\text{N}_{2(\text{g})}&+&\text{O}_{2\text{(g)}}&\leftrightarrow&2\text{N}_2\text{O}_{\text{(g)}}\\\text{Initial Conc.}&0.482\text{ mol}&&0.933\text{ mol}&&0\\\text{At equiluibrium}&(0.482-\text{x})\text{mol}&&(0.933-\text{x})\text{mol}&&\text{x mol} \end{matrix}$
Therefore, at equilibrium, in the 10L vessel:
$[\text{N}_2]=\frac{0.482-\text{x}}{10},[\text{O}_2]=\frac{0.933-\text{x/2}}{10},[\text{N}_2\text{O]}=\frac{\text{x}}{10}$
The value of equilibrium constant i.e., $\text{K}_{\text{c}}=2.0\times10^{-37}$ is very small. Therefore, the amount of $N_2$ and $O_2$ reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of $N_2$ and $O_2$.
Then,
$[\text{N}_2]=\frac{0.482}{10}=0.0482\text{mol L}^{-1}\text{and}[\text{O}_2]=\frac{0.933}{10}=0.0933\text{mol L}^{-1}$
Now,
$\text{K}_{\text{c}}=\frac{\big[\text{N}_2\text{O}_{\text{(g)}}\big]^2}{\big[\text{N}_{2\text{(g)}}\big]^2\big[\text{O}_{2\text{(g)}}\big]}$
$\Rightarrow2.0\times10^{-37}=\frac{\big(\frac{\text{x}}{10}\big)^2}{(0.0482)^2(0.0933)}$
$\Rightarrow\frac{\text{x}^2}{100}=2.0\times10^{-37}\times(0.0482)^2\times(0.0933)$
$\Rightarrow\text{x}^2=43.35\times10^{-40}$
$\Rightarrow\text{x}=6.6\times10^{-20}$
$[\text{N}_2\text{O}]=\frac{\text{x}}{10}=\frac{6.6\times10^{-20}}{10}$
$=6.6\times10^{-21}$

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