Question

For the reaction, $N_2{O}_3\rightleftharpoons NO+N{O}_2$, the value of equilibrium constant $K_P$, at fixed temperature is 4. What will be the amount of dissociation at same temperature and 5 atmospheric pressure?

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{7}{9}$
• D. $\frac{1}{2}$.

Answer 1

The correct option is B $\frac{2}{3}$

For the given reaction$N_2{O}_3\rightleftharpoons NO+N{O}_2$, equilibrium constant will be given as $K_p=\frac{P_{NO}\times P_{N{O}_2}}{P_{N_2{O}_3}}$

The $K_p$ for the question is already given as $4$

Now let the dissociation constant be $\alpha$

Then according to the reaction, $N_2{O}_3\rightleftharpoons NO+N{O}_2$

$N_2{O}_3\Rightarrow 1-\alpha$

$NO\Rightarrow \alpha$

$N{O}_2\Rightarrow \alpha$

So, the total number of moles after and before the reaction will be

$n=(1-\alpha )+\alpha +\alpha$

$n=1+\alpha$

Partial pressure of $N_2{O}_3=\frac{1-\alpha }{1+\alpha }\times P$

Partial pressure of $NO=\frac{\alpha }{1+\alpha }\times P$

Partial pressure of $N{O}_2=\frac{\alpha }{1+\alpha }\times P$

Substituting the above values in the formula for equilibrium constant, we get

$K_p=\frac{P_{NO}\times P_{N{O}_2}}{P_{N_2{O}_3}}$

$K_p=\frac{{\left(\frac{\alpha P}{1+\alpha }\right)}^2}{P\left(\frac{1-\alpha }{1+\alpha }\right)}$

Simplifying above, we get

$K_p=\frac{P{\alpha }^2}{1-{\alpha }^2}$

Substituting the values of $K_p$ and $P$ in the above equation we get

$4=\frac{5{\alpha }^2}{1-{\alpha }^2}$

Solving the above to get the value of $\alpha$

$4\times (1-{\alpha }^2)=5{\alpha }^2$

$4-4{\alpha }^2=5{\alpha }^2$

$9{\alpha }^2=4$

${\alpha }^2=\frac{4}{9}$

$\alpha =\frac{2}{3}$

Therefore, the amount of dissociation at same temperature and $5$ atmospheric pressure will be $\alpha =\frac{2}{3}$