PCl5(g) is 40% dissociated at a total pressure of 2.0 atm at equilibrium: PCl5(g) PCl3(g) + Cl2(g). The value of K_p (in atm) is

JEE Main levelSingle correct optionDifficulty 2 of 3ce016

$PCl X_{5} (g)$ is 40% dissociated at a total pressure of 2.0 atm at equilibrium: $PCl X_{5} (g) PCl X_{3} (g) + Cl X_{2} (g)$ . The value of $K_{p}$ (in atm) is

For 1 mol of

PCl X_{5}

n_{PCl X_{5}} = 1 - α

n_{PCl X_{3}} = n_{Cl X_{2}} = α

, total

= 1 + α

. So

K_{p} = α^{2} P / (1 - α^{2}) = (0.16) (2.0) /0.84 = 0.381 atm