Chemical and Ionic Equilibrium - Result Question 54
####53. One mole of $N _2$ and 3 moles of $PCl _5$ are placed in a $100 L$ vessel heated to $227^{\circ} C$. The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate the degree of dissociation for $PCl _5$ and $K _p$ for the reaction,
$$ PCl _5(g) \rightleftharpoons PCl _3(g)+Cl _2(g) $$
$(1984,6 M)$
Topic 2 Ionic Equilibrium
Objective Questions I (Only one correct option)
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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 $N _2(g)$ and remaining 4 moles for $PCl _5$ and its dissociation products.
$$ \begin{gathered} PCl _5 \rightleftharpoons \underset{x}{PCl _3}+\underset{x}{Cl _2} \ 3+x=4 \Rightarrow x=1 \end{gathered} $$
Degree of dissociation $=\frac{1}{3}=0.33$
54
$$ \begin{array}{lllc} & N _2+ & 3 H _2 & \rightleftharpoons 2 NH _3 \ \text { Initial : } & 1.0 & 3.0 & 0 \ \text { Equilibrium } & 1-0.25 & 3-0.75 & 0.05 \ & =0.75 & =2.25 \ {\left[N _2\right]=} & \frac{0.75}{4},\left[H _2\right]= & \frac{2.25}{4},\left[NH _3\right]=\frac{0.50}{4} \end{array} $$
$$ \begin{gathered} K _c=\frac{\left[NH _3\right]^{2}}{\left[N _2\right]\left[H _2\right]^{3}}=\frac{(0.50)^{2}}{(0.75)(2.25)^{3}} \times 16 \ =0.468 L^{2} mol^{-2} \ \text { Also for : } \frac{1}{2} N _2+\frac{3}{2} H _2 \rightleftharpoons NH _3 \ K _c^{\prime}=\sqrt{K _c}=0.68 \end{gathered} $$
