For a reaction $Δ H^{\circ} = - 92.0 kJ mo l^{- 1}$ and $Δ S^{\circ} = - 199 J mo l^{- 1} K^{- 1}$ , both assumed independent of temperature. Find the temperature in K above which the reaction is non-spontaneous under standard conditions (nearest integer).

Numerical answer type, answer in K, tolerance 1.

Answer: 462 K

Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = 0

T = Δ H^{\circ} /Δ S^{\circ} = (- 92000) / (- 199) = 462.31 K

. Since

Δ S^{\circ} < 0

Δ G^{\circ}

becomes positive above this temperature, about 462 K.