Question

One mole of ${\mathrm{N}}_2{\mathrm{O}}_4$ in a $1\mathrm{L}$ flask decomposes to attain the equilibrium ${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right)$. At the equilibrium the mole fraction of ${\mathrm{N}}_2{\mathrm{O}}_4$ is $1/2$ . then ${\mathrm{K}}_{\mathrm{C}}$ will be:

Answer 1

Step 1: Reaction for the decomposition:

$\underset{}{{\mathrm{N}}_2{\mathrm{O}}_4}\rightleftharpoons \underset{}{2{\mathrm{NO}}_2}$

$$\begin{gathered} \mathrm{t}=01\mathrm{mol}0 \\ \mathrm{t}={\mathrm{t}}_{\mathrm{eq}}1-\mathrm{x}2\mathrm{x} \end{gathered}$$

Step 2: ${\mathrm{K}}_{\mathrm{C}}$ for the reaction:

As the mole fraction of ${\mathrm{NO}}_2$ would be= $\frac{2\mathrm{x}}{(1-\mathrm{x}+2\mathrm{x})}$

So, $=\frac{2\mathrm{x}}{(\mathrm{x}+1)}$$=\frac{1}{2}$.

Therefore, $4\mathrm{x}=\mathrm{x}+1$

And so $3\mathrm{x}=1$

$\mathrm{x}=\frac{1}{3}$

So, it's $\frac{\left(\frac{4}{9}\right)}{\left(\frac{2}{3}\right)}$ = $\frac{4\times 3}{9\times 2}$= $\frac{2}{3}$.

Hence, ${\mathrm{K}}_{\mathrm{C}}$ for the reaction would be: $\frac{2}{3}$