For PCl5(g) PCl3(g) + Cl2(g) an equilibrium mixture in a 10.0 L vessel contains 0.40 mol PCl5, 0.20 mol PCl3 and 0.20 mol Cl2. At constant temperature the volume is then compressed to 5.0 L and, at once, 0.13 mol of Cl2 is withdrawn. Calculate Q_c/K_c for the new composition (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3ce152

For $PCl X_{5} (g) PCl X_{3} (g) + Cl X_{2} (g)$ an equilibrium mixture in a $10.0 L$ vessel contains $0.40$ mol $PCl X_{5}$ , $0.20$ mol $PCl X_{3}$ and $0.20$ mol $Cl X_{2}$ . At constant temperature the volume is then compressed to $5.0 L$ and, at once, $0.13$ mol of $Cl X_{2}$ is withdrawn. Calculate $Q_{c} / K_{c}$ for the new composition (two decimals).

Numerical answer type, tolerance 0.01.

K_{c} = (0.020) (0.020) /0.040 = 0.0100 mol L^{- 1}

. New composition in

5.0 L

[PCl X_{5}] = 0.080

[PCl X_{3}] = 0.040

[Cl X_{2}] = 0.07/5.0 = 0.014 mol L^{- 1}

Q_{c} = (0.040) (0.014) /0.080 = 0.0070

, so

Q_{c} / K_{c} = 0.700

, i.e. 0.70 (less than 1, so the reaction goes forward).