Question

An equilibrium mixture in a vessel of capacity $100$ litre contain $1mol{N}_2,2mol{O}_2$ and $3molNO$. The number of moles of $O_2$ to be added so that at new equilibrium, the conc. of $NO$ is found to be $0.04mol/lit$.

$\left[N_2\right(g)+O_2(g)\rightleftharpoons 2NO(g\left)\right]$

• A. $\left(101/18\right)$
• B. $\left(101/9\right)$
• C. $\left(202/9\right)$
• D. None of these

Answer 1

Answer: (A)

$\left(101/18\right)$

$N_2+O_2\rightleftharpoons 2NO$

At eq. $1$ $2$ $3$

$K_C=\frac{{\left(\frac{3}{100}\right)}^2}{\left(\frac{2}{100}\right)\left(\frac{1}{100}\right)}=\frac{9}{2}$

Assume $x$ moles of $O_2$ added:

$N_2+O_2\rightleftharpoons 2NO$

$1$ $2+x$ $3$

New eq. $(1-\alpha )$ $(2+x-\alpha )$ $(3+2\alpha )$

Given new concentration of [NO]$=0.04mol/lit$

$\frac{3+2\alpha }{100}=0.04$

$K_C=\frac{{\left(\frac{4}{100}\right)}^2}{\frac{\left(\frac{1}{2}\right)}{100}\times \frac{(\frac{3}{2}+x)}{100}}=\frac{9}{2}$

$\Rightarrow \frac{32}{\frac{3}{2}+x}=\frac{9}{2}$

$\Rightarrow 64=\frac{27}{2}+9x$

$x=\frac{101}{18}$

Hence $\left(\frac{101}{18}\right)$ moles of $O_2$ added.