Question

Calculate pressure required for 50% dissociation of $PC{l}_5$. If the $K_p$ at that temprature is 1.8 atm.

• A. $3$ atm
  
• B. $5.4$atm
  
• C. $9$atm
  
• D. none
  

Answer 1

Answer: (B)

$5.4$atm

${PCl}_5\rightleftharpoons {PCl}_3+{Cl}_2$

given, $K_p=1.8$

Moles of different gases at equilibrium are ${PCl}_5=0.5$ moles

${PCl}_3=0.5$ moles

${Cl}_2=0.5$ moles

therefore, total number of moles $=1.5$ moles

Partial pressure are $P_{{PCl}_3}=P_{{PCl}_5}=P_{{Cl}_2}=\frac{0.5}{1.5}P$

where $P$ is total pressure in atmosphere.

$K_p=\frac{\left(P_{{PCl}_3}\right)\left(P_{{Cl}_2}\right)}{\left(P_{{PCl}_5}\right)}=\frac{{\left(\frac{0.5}{1.5}\right)}^2{P}^2}{\left(\frac{0.5}{1.5}\right)P}=\frac{1}{3}P$

$\frac{1}{3}P=1.8$

$P=5.4$ atm

$\Rightarrow$ Pressure is $5.4$ atmosphere.