Question

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

• A. 0.97 atm
• B. 0.34 atm
• C. 1.43 atm
• D. 1.67 atm

Answer 1

The correct option is A 0.97 atm

$PC{l}_5\to PC{l}_3+C{l}_2$
$K_p=\left(0.1/1.1\right)\times 4\times \left(0.1/1.1\right)\times 4/0.9\times 4/1.1=0.2\times 0.2\times P/1.2\times 0.8$
So new pressure is $P=0.97$ atm.