For N2(g) + 3H2(g) 2NH3(g), Delta H deg = -92.0 kJ/mol (assume constant) and K = 6.0 x 10^5 at 298 K. Taking R = 8.314 J mol^-1 K^-1, find the temperature in K (nearest integer) at which K = 1.0.

JEE Advanced levelNumerical answerDifficulty 3 of 3ce217

For $N X_{2} (g) + 3 H X_{2} (g) 2 NH X_{3} (g)$ , $Δ H^{\circ} = - 92.0 kJ/mol$ (assume constant) and $K = 6.0 \times 1 0^{5}$ at $298 K$ . Taking $R = 8.314 J mo l^{- 1} K^{- 1}$ , find the temperature in K (nearest integer) at which $K = 1.0$ .

Numerical answer type, answer in K, tolerance 1.

ln (K_{2} / K_{1}) = - (Δ H / R) (1/ T_{2} - 1/ T_{1})

with

ln (1/6.0 \times 1 0^{5}) = - 13.3047

. So

1/ T_{2} = 1/298 - R ln (K_{2} / K_{1}) /Δ H = 0.0021534 K^{- 1}

and

T_{2} = 464.39

K, i.e. 464 K.