Question

At ${227}^{o}C$ one mole of $C{l}_2$ and 3 moles of $PC{l}_5$ are placed in a 100 litre container and allowed to reach equilibrium. It is found that the equilibrium pressure is 2.05 atmospheres. Assuming ideal behaviour, the degree of dissociation 'x' for $PC{l}_5$ and $K_p$ for the reaction $PC{l}_5$ (g) $\rightleftharpoons$ $PC{l}_3$ (g) + $C{l}_2$ (g) is:

• A. x = 0.25, $K_p$ = 0.12
• B. x = 0.64, $K_p$ = 2.42
• C. x = 0.33, $K_p$ = 0.41
• D. None of the above

Answer 1

The correct option is C

x = 0.33, $K_p$ = 0.41

Let x be the degree of dissociation (we usually use α, here it is given as x)

$\begin{array}{cccccc}Atstart& 3& & 0& & 1\\ & PC{l}_5\left(g\right)& \leftrightharpoons & PC{l}_3\left(g\right)& +& C{l}_2\left(g\right)\\ Atequilibrium& 3(1-x)& & 3x& & 1+3x\end{array}$

$\therefore \text{Total number of moles at equilibrium}$

$=3(1-x)+3x+1+3x=3(1+x)+1$
From ideal gas equation,
PV = nRT;
We use P = 2.05 atm, T = 500K,
V = 100 litres and
R = 0.0821 Latm / (Kmol)
We get n = 5 = number of moles at equilibrium = 3(1+x) + 1
Solving we get x = $\frac{1}{3}$ = 0.3333 $\approx$ 0.33
We can't rule out option d here since there is no mention of $K_p$
Now let us try and derive the expression of $K_p$ in terms of x and equilibrium pressure P.

$K_p=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$

$K_p=\frac{\left[\frac{3xP}{3(1+x)+1}\right]\left[\frac{(1+3x)P}{3(1+x)+1}\right]}{\left[\frac{3(1-x)}{3(1+x)+1}P\right]}$

= $\frac{3x(3x+1)}{4+3x}\times \frac{1P}{3(1-x)}$

= $\frac{(3x^2+x)\times P}{(4+3x)(1-x)}=\frac{x(3x+1)\times P}{(4+3x)(1-x)}$

Substituting the values of x and P, we obtain $K_p$ = 0.42atm