Two products K (kinetic) and T (thermodynamic) interconvert, K T. For K to T, Delta H deg = -12.0 kJ mol^-1 and Delta S deg = -15.0 J mol^-1 K^-1. At 300 K and equilibrium (R = 8.314 J mol^-1 K^-1), what is the mole percent of T, to the nearest integer?

JEE Advanced levelNumerical answerDifficulty 3 of 3g3213

Two products K (kinetic) and T (thermodynamic) interconvert, $K T$ . For K to T, $Δ H^{\circ} = - 12.0 kJ mo l^{- 1}$ and $Δ S^{\circ} = - 15.0 J mo l^{- 1} K^{- 1}$ . At 300 K and equilibrium ( $R = 8.314 J mo l^{- 1} K^{- 1}$ ), what is the mole percent of T, to the nearest integer?

Numerical answer type, tolerance 0.5.

Δ G^{\circ} = - 12000 - 300 (- 15.0) = - 7500 J mo l^{- 1}

K = exp (7500/ (8.314 \times 300)) = 20.23

. Percent T

= 100 K / (1 + K) = 95.29%

, so 95.