Question

An equilibrium mixture in a vessel of capacity $100$ litre contain $1$ mol $N_2$, $2$ mol $O_2$ amd $3$ mol $NO$. No. of moles of $O_2$ to be added so that at new equilirium the conc. of $NO$ is found to be $0.04$ mol/lit is:

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

Answer 1

The correct option is A $\left(101/18\right)$

Given,

Volume of vessel=$100L$

$N_2=1$ mol, $O_2=2$ mol, $NO=3$ mol

The reaction involved is

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

Here, the equilibrium constant, $K_c$ is

$K_c=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}⟶\left(1\right)$

Now, concentrations of $N_2,O_2$ & $NO$ are

$\left[N_2\right]=\frac{1mol}{100L}=0.01M$ $[\because conc.=\frac{moles}{L}]$

$\left[O_2\right]=\frac{2mol}{100L}=0.02M$

$\left[NO\right]=\frac{3mol}{100L}=0.03M$

From (1) $K_c=\frac{0.03\times 0.03}{0.01\times 0.02}$

$\Rightarrow K_c=\frac{9}{2}=4.5$

Now, it is given, that new equilibrium concentration of $NO=0.04M$

$N_2$ + $O_2$ $\rightleftharpoons 2NO$

c c O

c-x c-x 2x

$\Rightarrow$ This implies $1$ mole of $NO$ requires $\frac{1}{2}$ mole of $O_2$ And $\frac{1}{2}$ mole of $N_2⟶\left(2\right)$

At new equilibrium

$[NO{]}_{new}=0.04M$

$[NO{]}_{initial}=0.03M$

$\therefore$ The increase in concentration at new equilibrium=$0.04-0.03=0.01M$

From (2) To increase $\left[NO\right]$ by $0.01M$, the concentration of $N_2$ consumed=$0.01\times \frac{1}{2}M=0.005M$

$\therefore$ The new concentration of $N_2=0.005M$ at equilibrium. $[\because$ No extra $N_2$ is added$]$

Now, for new $\left[O_2\right]$ calculation,

From (1), $[O_2{]}_{new}=\frac{[NO{]}_{new}^2}{[N_2{]}_{new}{K}_{c_{new}}}$

$=\frac{0.04\times 0.04}{0.005\times 4.5}$

$=\frac{16}{225}M$

Also, the reaction consumed $0.01M\frac{1}{2}$ of oxygen [$\because$ From (2)]

$\therefore$ The $[O_2{]}_{new}=\frac{16}{225}+\frac{0.01}{2}=\frac{137}{1800}M$

Now, the concentration of $O_2$ increased= $[O_2{]}_{new}-[O_2]$

$=\frac{137}{1800}-0.02=\frac{101}{1800}M$

Now, to find number of moles of $O_2$ increased,

concentration=$\frac{mol}{L}$

$\Rightarrow \frac{101}{1800}=\frac{mol}{100}$

$\Rightarrow mol=\frac{101}{18}$.