Question

The value of $K_P$ is $1\times {10}^{-3}{\text{atm}}^{-1}$ at $25{}^{\circ }\text{C}$ for the reaction $2NO\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons 2NOCl\left(g\right).$ A flask contains $NO$ at $0.02\text{atm}$ and at $25{}^{\circ }\text{C}$. The mole of $C{l}_2$ that must be added if $1\%$ of the $NO$ is to be converted to $NOCl$ at equilibrium is $x\times {10}^{-5}$. The volume of the flask is such that $0.2$ mole of gas produces $1\text{atm}$ pressure at $25{}^{\circ }\text{C}$ (Ignore probable association of $NO$ to $N_2{O}_2$). Find the value of $x$.
(Given: $\frac{1}{(99{)}^2}\approx 1.02\times {10}^{-4}$)

Answer 1

$$\begin{array}{ccccc}& 2NO\left(g\right)& +& C{l}_2\left(g\right)& \rightleftharpoons 2NOCl\\ t=0& 0.02& & 0& 0\\ t_{eq.}& \frac{0.02\times 99}{100}& & p& \frac{0.02\times 1}{100}\end{array}$$
$\therefore K_p=\frac{[P_{NOCl}{]}^2}{[P_{NO}{]}^2{P}_{C{l}_2}}=\frac{(0.02\times 0.01{)}^2}{(0.99\times 0.02{)}^2\times P}={10}^{-3}\therefore P=0.102\text{atm}PV=nRT\Rightarrow 0.102\times V=n\times R\times T.........\left(i\right)\text{From the question we have}1\times V=0.2\times R\times T...........\left(ii\right)$
From eqs. (i) and (ii), we get
$n=0.0204$ (no. of moles of $C{l}_2$ at equilibrium)
Pressue of $C{l}_2$ involved in reaction
$=\frac{1}{2}\times \text{pressure of}NOCl$
$=\frac{1}{2}\times \frac{0.02}{100}=0.0001\text{atm}$
$PV=nRT\Rightarrow 0.0001\times V=n\times RT........\left(iii\right)$
From eqs. (ii) and (iii), $n=2\times {10}^{-5}$ (moles of $C{l}_2$ involved in reaction)
Initial moles of $C{l}_2$ taken $=0.0204+2\times {10}^{-5}=0.02042=2042\times {10}^{-5}$