Question

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

• A. $\left(\frac{101}{18}\right)$
• B. $\left(\frac{101}{9}\right)$
• C. $\left(\frac{202}{9}\right)$
• D. none of these.

Answer 1

Answer: (A)

$\left(\frac{101}{18}\right)$

$N_2+O_2\rightleftharpoons 2NO$ $V=100lit$

At initial $1$ $2$ $3$

Eq'b

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

Assume $x$ moles of $O_2$ added

$N_2+O_2\rightleftharpoons 2NO$

At $1$ $2+x$ $3$

disturbed

state

At new $1-\alpha$ $2+n-\alpha$ $3+2\alpha$

Eq'b

Given new concentration $=0.04mol/lit$

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

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

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

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

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

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