At 500 K, K_c = 0.60 L^2 mol^-2 for N2(g) + 3H2(g) 2NH3(g). For the reaction NH3(g) 1/2 N2(g) + 3/2 H2(g) at the same temperature, calculate K_p in atm using R = 0.0821 L atm mol^-1 K^-1, correct to one decimal place.

JEE Advanced levelNumerical answerDifficulty 3 of 3ce094

At 500 K, $K_{c} = 0.60 L^{2} mo l^{- 2}$ for $N X_{2} (g) + 3 H X_{2} (g) 2 NH X_{3} (g)$ . For the reaction $NH X_{3} (g) \frac{1}{2} N X_{2} (g) + \frac{3}{2} H X_{2} (g)$ at the same temperature, calculate $K_{p}$ in atm using $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ , correct to one decimal place.

Numerical answer type, answer in atm, tolerance 0.1.

Reversing and halving the equation gives

K_{c}^{'} = (K_{c})^{- 1/2} = 0.6 0^{- 1/2} = 1.2910 mo l^{1} L^{- 1}

. For the new reaction

Δ n = 2 - 1 = + 1

, so

K_{p}^{'} = K_{c}^{'} RT = 1.2910 \times 41.05 = 52.995

atm.