Question

In the reaction, $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$, the amount of each $PC{l}_5$, $PC{l}_3$ and $C{l}_2$ is $2$ mole at equilibrium and total pressure is $3$ atmosphere. The value of $K_p$ will be:

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

The correct option is D 1

Total pressure is 3 atm.

Total number of moles $=2+2+2=6$

Number of moles of $PC{l}_5=2$

Mole fraction of $PC{l}_5=\frac{2}{6}=0.333$

Partial pressure of $PC{l}_5=0.333\times 3atm=1atm$.

Number of moles of $PC{l}_3=2$

Mole fraction of $PC{l}_3=\frac{2}{6}=0.333$

Partial pressure of $PC{l}_3=0.333\times 3atm=1atm$.

Number of moles of $C{l}_2=2$

Mole fraction of $C{l}_2=\frac{2}{6}=0.333$

Partial pressure of $C{l}_2=0.333\times 3atm=1atm$.

The equilibrium constant $K_p=\frac{P_{PCl3}\times P_{Cl2}}{P_{PCl5}}$

$K_p=\frac{1\times 1}{1}=1$