Question

An equilibrium mixture, $N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$ in a vessel of capacity $100$ litre contain $1mol{N}_2,2mol{O}_2$ and $3molNO$. Calculate the number of moles of $O_2$ to be added so that at new equilibrium the conc. of $NO$ is found to be $0.04mol/lit$.

• 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 $\left(101/18\right)$

For the equilibrium reaction:

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

At equilibrium (concentration in mol/L)

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

Equilibrium constant for the reaction is

$K_C=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}$

$K_C=\frac{[3/100{]}^2}{\left[1/100\right]\left[2/100\right]}$

$K_C=\frac{9}{2}$

Upon addition of $O_2$, new equilibrium concentration of $NO$ is $0.04mol/L$

From the reaction, $2MNO$ is produced upon consumption of $1M{O}_2$ and $1M{N}_2$

Therefore, the increased concentration of $NO$ =$0.04-0.03=0.01M$ will be obtained by consuming $(\frac{1}{2}\times 0.01)M{N}_2$ and $(\frac{1}{2}\times 0.01)M{O}_2$

Since no additional $N_2$ is added, therefore its concentration at new equilibrium is: $0.01-0.005=0.005M$ and let amount of $O_2$ added is $xM$, then its new equilibrium concentration is $0.02+x-0.005=(x+0.015)M$

Since, $K_C=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}$

$\frac{9}{2}=\frac{[0.04{]}^2}{0.005\times (x+0.015)}$

$x+0.015=\frac{2\times 16}{9\times 50}$

$x=\frac{1.01}{18}$M

Number of moles $=M\times V=xM\times 100L=\frac{101}{18}mol$

Therefore option A is correct.