For the isomerisation K T at 298 K, Delta G deg = -5.50 kJ mol^-1 (R = 8.314 J mol^-1 K^-1). The mole percent of T in the equilibrium mixture is closest to

JEE Main levelSingle correct optionDifficulty 2 of 3g3185

For the isomerisation $K T$ at 298 K, $Δ G^{\circ} = - 5.50 kJ mo l^{- 1}$ ( $R = 8.314 J mo l^{- 1} K^{- 1}$ ). The mole percent of T in the equilibrium mixture is closest to

K = e^{- Δ G^{\circ} / RT} = exp (5500/ (8.314 \times 298)) = 9.21

. Mole fraction of T

= K / (1 + K) = 90.2%

, which is about 90%.