Chemistry - Chemical Equilibrium Question with Solution | TestHub

Pressure at $t=0$ is $0.02$.

$$\begin{array}{rl}& \text{ Pressure at eq. }\frac{0.02\times 99}{100}P\frac{0.02\times 1}{100}\\ & \because K_P=\frac{{\left[P_{\mathrm{NOCl}}\right]}^2}{{\left[P_{\mathrm{NO}}\right]}^2\left[P_{{\mathrm{Cl}}_2}\right]}=\frac{(0.02\times 0.01{)}^2}{(0.99\times 0.02{)}^2\times P}=1\times 10^{-3}\\ & \therefore P=0.102\mathrm{atm}\end{array}$$

For Cl₂ in the vessel, using $PV=nRT$:

$0.102\times V=n\times R\times T$

For a gas in the vessel, $1\times V=9.2\times R\times T\dots \dots (ii)$

By Eqs. (i) and (ii), the mole of Cl₂(n) at equilibrium is $0.0204$.

$2atm$ of NO reacts with $1atm$ of Cl₂ (if $V,T$ are constant).

$\frac{0.02\times 1\text{ atm }}{100}$ NO reacts with $\frac{0.02\times 1}{2\times 100}=0.0001atm$ Cl₂.

Thus, $0.0001\times V=n_{C{l}_2\text{ used }}\times RT$.

By Eqs. (ii) and (iii), the $n_{C{l}_2}$ which has reacted is $2\times 10^{-5}\mathrm{mol}$.

Therefore, the total Cl₂ added is $0.0204+2\times 10^{-5}=0.02042\mathrm{mol}$.