For N2(g) + 3H2(g) 2NH3(g), K_c = 60 L^2 mol^-2 at 500 K. A gas mixture at this temperature has partial pressures p_N2 = 4.0, p_H2 = 6.0 and p_NH3 = 2.0 atm. Taking R = 0.0821 L atm mol^-1 K^-1, calculate Q_p/K_p (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3ce129

For $N X_{2} (g) + 3 H X_{2} (g) 2 NH X_{3} (g)$ , $K_{c} = 60 L^{2} mo l^{- 2}$ at $500 K$ . A gas mixture at this temperature has partial pressures $p_{N X_{2}} = 4.0$ , $p_{H X_{2}} = 6.0$ and $p_{NH X_{3}} = 2.0$ atm. Taking $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ , calculate $Q_{p} / K_{p}$ (two decimals).

Numerical answer type, tolerance 0.01.

K_{p} = K_{c} (RT)^{Δ n}

with

Δ n = - 2

K_{p} = 60/ (0.0821 \times 500)^{2} = 0.03561 at m^{- 2}

Q_{p} = (2.0)^{2} / (4.0 \times 6. 0^{3}) = 0.00463 at m^{- 2}

. Therefore

Q_{p} / K_{p} = 0.1300

, which rounds to 0.13; as

Q_{p} / K_{p} > 1

the reaction would go backward.