Question

0.1 mole of $N_2{O}_4\left(g\right)$ was sealed in a tube under one atmospheric conditions at ${25}^{0}C$. 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. 0.034
• D. $2.8\times {10}^{-2}$

Answer 1

Answer: (C)

According to the equation

$\begin{array}{cccc}& N_2{O}_4& \rightleftharpoons & 2N{O}_2\\ \text{Initial}& 1atm& & 0\\ \text{change}& -x& & +2x\\ \text{Equilibrium}& 1-x& & 2x\end{array}$

$K_P=\frac{(P_{N{O}_2}{)}^2}{P_{N_2{O}_4}}$ $[\therefore K_P$ is similar to $K_C$ in aspect of setting up rate quotient]

$0.14=\frac{(2x{)}^2}{1-x}$

$0.14(1-x)=4x^2$

$x=0.17$

From ideal gas equation

$PV=nRT$

$V=\frac{nRT}{P}$

$V=\frac{0.1\times 0.082\times (273+25)}{1}$

$V=2.45lit$

Moiles of $N{O}_2=\frac{P_{N{O}_2}\times V}{RT}$

Moles of $N{O}_2=\frac{0.34\times 2.45}{0.0821\times (273+25)}$

Moles of $N{O}_2=0.034$

$\therefore$ option C is correct