Chemical and Ionic Equilibrium 1 Question 52
53. One mole of $\mathrm{N}{2}$ and 3 moles of $\mathrm{PCl}{5}$ are placed in a $100 \mathrm{~L}$ vessel heated to $227^{\circ} \mathrm{C}$. The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate the degree of dissociation for $\mathrm{PCl}{5}$ and $K{p}$ for the reaction,
$$ \mathrm{PCl}{5}(g) \rightleftharpoons \mathrm{PCl}{3}(g)+\mathrm{Cl}_{2}(g) $$
$(1984,6 M)$
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Solution:
- Total moles of gases at equilibrium $=\frac{p V}{R T}=\frac{2.05 \times 100}{0.082 \times 500}=5.0$
Out of this 5 moles, 1.0 mole is for $\mathrm{N}{2}(g)$ and remaining 4 moles for $\mathrm{PCl}{5}$ and its dissociation products.
$$ \begin{gathered} \mathrm{PCl}{5} \rightleftharpoons \underset{x}{\mathrm{PCl}{3}}+\underset{x}{\mathrm{Cl}_{2}} \ 3+x=4 \Rightarrow x=1 \end{gathered} $$
Degree of dissociation $=\frac{1}{3}=0.33$
54
$$ \begin{array}{lllc} & \mathrm{N}{2}+ & 3 \mathrm{H}{2} & \rightleftharpoons 2 \mathrm{NH}{3} \ \text { Initial : } & 1.0 & 3.0 & 0 \ \text { Equilibrium } & 1-0.25 & 3-0.75 & 0.05 \ & =0.75 & =2.25 \ {\left[\mathrm{~N}{2}\right]=} & \frac{0.75}{4},\left[\mathrm{H}{2}\right]= & \frac{2.25}{4},\left[\mathrm{NH}{3}\right]=\frac{0.50}{4} \end{array} $$
$$ \begin{gathered} K_{c}=\frac{\left[\mathrm{NH}{3}\right]^{2}}{\left[\mathrm{~N}{2}\right]\left[\mathrm{H}{2}\right]^{3}}=\frac{(0.50)^{2}}{(0.75)(2.25)^{3}} \times 16 \ =0.468 \mathrm{~L}^{2} \mathrm{~mol}^{-2} \ \text { Also for : } \frac{1}{2} \mathrm{~N}{2}+\frac{3}{2} \mathrm{H}{2} \rightleftharpoons \mathrm{NH}{3} \ K_{c}^{\prime}=\sqrt{K_{c}}=0.68 \end{gathered} $$
