Question

${PCl}_5$, is 40% dissociated according to the following reaction, when equilibrium pressure is 2 atm, it will be 80% dissociated,when equilibrium pressure is approximately :
${PCl}_5\left(g\right)\rightleftharpoons {PCl}_3\left(g\right)+{Cl}_2\left(g\right)$

• A. 0.2 atm
• B. 0.5 atm
• C. 0.4 atm
• D. 4 atm

Answer 1

The correct option is A 0.2 atm

solution:

PCl5 dissociates into PCl3 and Cl2 as shown below in reaction..

$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

1 0 0 at t = 0,

at eq. (1 - d) d d

here, d is degree of dissociation.

so, $Kp=\frac{p_{\left[PCl3\right]}\times p_{\left[Cl2\right]}}{p_{\left[PCl5\right]}}$

= $d/(1+d)P\times d/(1+d)P/(1-d)/(1+d)P$

$=d²P/(1+d)(1-d)$

$=d²P/(1-d²)$

here, Kp is equilibrium constant.

so, $P\propto 1/d²/(1-d²)$

i.e., $P/2=\left[\right(0.4\left)²/\right(1-0.4²\left)\right]/\left[\right(0.8\left)²/\right(1-0.8²\left)\right]$

$P/2=\left[0.16/0.84\right]/\left[0.64/0.36\right]$

$P/2=(0.16\times 0.36)/(0.84\times 0.64)$

$P/2=\left(0.09\right)/\left(0.84\right)=9/84$

P = 18/84 = 0.21428 atm $\approx$ 0.2 atm

hence the correct option : A