Question

For the reaction $PC{l}_3\left(g\right)+C{l}_2\left(g\right)\to PC{l}_5\left(g\right),$
The mole of each component $PC{l}_5,PC{l}_{3}andC{l}_2$ at equilibrium were found to be 2. If the total pressure is 3 atm. The value of $K_P$ will be-

• A. 1 atm
• B. 2 atm
• C. 3 atm
• D. 1.5 atm

Answer 1

The correct option is A 1 atm

Total moles at equilibrium = 2 + 2 + 2 = 6 mol
$\text{Partial pressure}=\text{mole fraction}\times \text{total pressure}$
$\text{Mole fraction}=\frac{\text{No. of moles}}{\text{Total no. of moles}}$
$P_{PC{l}_3}=\frac{2}{6}\times 3,P_{PC{l}_5}=\frac{2}{6}\times 3,P_{C{l}_2}=\frac{2}{6}\times 3$
$K_P=\frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PC{l}_5}}=\frac{1\times 1}{1}=1atm$