Question

In the reaction ${PCl}_5\leftrightharpoons {PCl}_3+{Cl}_2$, the amount of each ${PCl}_5,{PCl}_{3}and{Cl}_2$ is $2mole$ at equilibrium and total pressure is $3$ atmosphere. The value of $K_p$ will be?

• A. $1.0atm$
• B. $3.0atm$
• C. $2.9atm$
• D. $6.0atm$

Answer 1

Answer: (A)

$1.0atm$

$Pc{l}_5=pc{l}_3+c{l}_2$

at eqm: 2 2 2

$n_T=2+2+2=6$

$K_P=\frac{P_{pc{l}_3}.P_{pc{l}_2}}{P_{pc{l}_5}}$

$=\frac{\frac{2}{3}\times 3.\frac{2}{3}\times 3}{\frac{2}{6}\times 3}=1$