For PCl5(g) PCl3(g) + Cl2(g), K_c = 0.04 mol L^-1 at a certain temperature. A 5.0 L vessel contains 0.50 mol PCl5, 0.30 mol PCl3 and 0.40 mol Cl2. Choose the correct statement.

Backward reaction occurs; Q_c = 0.048

Forward reaction occurs; Q_c = 0.048

Backward reaction occurs; Q_c = 0.24

Backward reaction occurs; Q_c = 0.048

Forward reaction occurs; Q_c = 20.8

JEE Main levelSingle correct optionDifficulty 2 of 3ce122

For $PCl X_{5} (g) PCl X_{3} (g) + Cl X_{2} (g)$ , $K_{c} = 0.04 mol L^{- 1}$ at a certain temperature. A $5.0 L$ vessel contains $0.50$ mol $PCl X_{5}$ , $0.30$ mol $PCl X_{3}$ and $0.40$ mol $Cl X_{2}$ . Choose the correct statement.

Concentrations:

[PCl X_{5}] = 0.10

[PCl X_{3}] = 0.060

[Cl X_{2}] = 0.080 mol L^{- 1}

Q_{c} = (0.060) (0.080) /0.10 = 0.048 mol L^{- 1}

, which is greater than

K_{c} = 0.04

. The net reaction is therefore backward. Using moles instead of concentrations ignores the volume, since

Δ n \neq = 0