Question

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

• A. $\frac{101}{9}$
• B. None of these
• C. $\frac{101}{18}$
• D. $\frac{202}{9}$

Answer 1

The correct option is C $\frac{101}{18}$

Find the equilibrium constant

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

No. of moles at equilibrium

$1mol2mol3mol$

Concentration at equilibrium

$\frac{1}{100}\frac{2}{100}\frac{3}{100}$

Since volume $=100L$

Therefore, $K_C=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}=\frac{(3{)}^2}{1\times 2}=\left(\frac{9}{2}\right)$

Find the number of moles of $O_2$ to be added for required $\left[NO\right]$

Let ‘$a$’ mole of $O_2$ be added, then equilibrium will get disturbed. And let $x$ be the number of moles of reactant consumed. But the value of the new equilibrium constant will remain the same.

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

𝐴𝑡 $t=0$

$1mol2mol3mol$

At new equilibrium point

$(1-x)(2+a)-x(3+2x)$

$\left[NO\right]=\left[\frac{(3+2x)}{100}\right]=0.04;(3+2x)=4$

$2x=1,x=0.5$

$K_C=\frac{\frac{(3+2x{)}^2}{{100}^2}}{\frac{(1-x)(2+a-x)}{{100}^2}}=\frac{9}{2}$

$K_C=\frac{4^2}{0.5(1.5+a)}=\frac{9}{2}$

$=\frac{16}{0.5(1.5+a)}=\frac{9}{2}$

$7.11=1.5+a$

$a=5.61mol$ or $\frac{101}{18}mol$

Hence, option (A) is the correct option.