Question

At an equilibrium pressure of $4atm$, $N_2{O}_4$ dissociates $15\%$ into $N{O}_2$. At same temperature what will be equilibrium pressure required for $25\%$ dissociation.
(Given, $N_2{O}_{4\left(g\right)}\rightleftharpoons 2N{O}_{2\left(g\right)}$)

• A. $3.9atm$
• B. $5.3atm$
• C. $0.7atm$
• D. $1.38atm$

Answer 1

The correct option is D $1.38atm$

The given equilibrium is,
$\begin{array}{cccc}& N_2{O}_4& \rightleftharpoons & 2N{O}_2\\ t=0& n& & 0\\ t=t_{eq}& n(1-\alpha )& & 2n\alpha \end{array}$
Therefore, total number of moles $=n-n\alpha +2n\alpha =n(1+\alpha )$
$\text{Again the equilibrium constant}$,
$K_p=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}$
We know,
$P_{\text{gas}}={\chi }_{\text{gas}}{P}_T$
$\text{Where},P_T=\text{total pressure}$
$P_{\text{gas}}=\text{partial pressure of the gas}$
${\chi }_{\text{gas}}=\text{mole fraction of the gas}$

Now,
$\text{Mole fraction},{\chi }_{N_2{O}_4}=\frac{n(1-\alpha )}{n(1+\alpha )}=\left(\frac{1-\alpha }{1+\alpha }\right)$
$\text{Mole fraction},{\chi }_{N{O}_2}=\frac{2n\alpha }{n(1+\alpha )}=\left(\frac{2\alpha }{1+\alpha }\right)$
$\therefore K_p=\frac{{\left(\frac{2\alpha \times P_T}{1+\alpha }\right)}^2}{\frac{(1-\alpha )\times P_T}{(1+\alpha )}}$
$\Rightarrow K_p=\frac{4{\alpha }^2}{(1-{\alpha }^2)}{P}_T$
$=\frac{4\times (0.15{)}^2\times 4}{1-(0.15{)}^2}$
According to the question,

$\frac{4\times (0.15{)}^2\times 4}{1-(0.15{)}^2}=\frac{4\times (0.25{)}^2\times P}{1-(0.25{)}^2}$
$\therefore P=1.38\text{atm}$