Question

At some temperature and under a pressure of 4 atm, $PC{l}_5$ is 10% dissociated. Calculated the pressure at which $PC{l}_5$ will be 20% dissociated temperature remaining same.

• A. 0.96 atm
• B. 0.68 atm
• C. 0.54 atm
• D. 0.81 atm

Answer 1

The correct option is A 0.96 atm

Let initially 100 moles were present and after 10% dissociation at equilibrium only 90% are remaining.
At 4 atm pressure and 10% dissociation.
$$\begin{array}{cccc}& PC{l}_5\rightleftharpoons & PC{l}_3& +C{l}_2\\ \text{Equibm mole}& 90& 10& 10\\ \text{Equilibrium partial pressures}& \frac{90}{110}\times 4& \frac{10}{110}\times 4& \frac{10}{110}\times 4\end{array}$$
Hence, $K_p=\frac{P_{PC{l}_5}{P}_{PC{l}_5}}{P_{PC{l}_5}}=\frac{4}{99}$
At 20 % dissociation,
$$\begin{array}{cccc}& PC{l}_5\rightleftharpoons & PC{l}_3& +C{l}_2\\ \text{Equilibrium mole}& 80& 20& 20\\ \text{Equilibrium partial pressures}& \frac{80}{120}\times P& \frac{20}{120}\times P& \frac{20}{120}\times P\\ K_p=\frac{P_{PC{l}_3}{P}_{C{l}_2}}{P_{PC{l}_5}}=\frac{1}{24}\times P& & & \end{array}$$
Thus, $\frac{1}{24}\times P=\frac{4}{99}\text{or}P=0.96$ atm