Question

In a $0.25$ litre tube dissociation of $4$ moles of NO is take place. If its degree of dissociation is $10\%$. The value of $K_p$ for reaction $2NO\rightleftharpoons N_2+O_2$ is:

• A. $\frac{1}{(18{)}^2}$
• B. $\frac{1}{(8{)}^2}$
• C. $\frac{1}{16}$
• D. $\frac{1}{32}$

Answer 1

The correct option is A $\frac{1}{(18{)}^2}$

Given, $V=0.25l$, initial molar mass of $NO=4mol,\alpha =0.1$

$2NO\leftrightharpoons N_2+O_2$

$4$

$4(1-0.1)$ $2\times 0.1$ $2\times 0.1$

$△n_g=0$ (change in gaseous molar)

$K_c=K_n{\left(\frac{1}{V}\right)}^{△n_g}$

$K_c=\frac{2\times 0.1\times 2\times 0.1}{16\times 0.92}\times {\left(\frac{1}{0.25}\right)}^0=\frac{1}{{18}^2}{\left(\frac{1}{0.25}\right)}^o$

$K_p=K_c(RT{)}^{△n_g}$

$K_p=K_c=\frac{1}{{18}^2}$