Question

$0.1$ mole of $N_2{O}_4\left(g\right)$ was sealed in a tube under one atmosphere conditions at ${25}^{o}C$. Calculate the number of moles of $N{O}_2\left(g\right)$ present, if the equilibrium $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)(K_p=0.14)$ is reached after some time :-

• A. $1.8\times {10}^2$
• B. $2.8\times {10}^2$
• C. $0.037$
• D. $2.8\times {10}^{-2}$

Answer 1

Answer: (C)

$0.037$

$N_2{O}_4\rightleftharpoons 2N{O}_2$

At $t=0$, $0.1$ $0$

At $t_{eq}$ $0.1(1-\alpha )$ $0.1\left(2\alpha \right)$

Total number of moles of gas= $0.1(1-\alpha )+0.1\left(2\alpha \right)=0.1(1+\alpha )$

At $0.1$ mol, pressure was $1atm$, At $0.1(1+\alpha )mol$, pressure will become $1+\alpha atm$

Partial pressure of $N_2{O}_4=\frac{0.1(1-\alpha )}{0.1(1+\alpha )}\times (1+\alpha )$

Partial pressure of $N{O}_2$=$\frac{0.1\times 2\alpha }{0.1(1+\alpha )}\times (1+\alpha )$

$K_P=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}$

After substituting & simplifying, $\alpha \simeq 0.19$

$\therefore$ Number of moles of $N{O}_2$ at equilibrium,

$\Rightarrow 0.1\left(2\alpha \right)=0.1(2\times 0.19)$

$=0.037mol$