Assume the axial conformer of a monosubstituted cyclohexane lies 7.0 kJ/mol above the equatorial conformer (the A-value). Using $R = 8.314 J K^{- 1} mo l^{- 1}$ and 298 K, the percentage of molecules in the equatorial conformer at equilibrium is closest to

Answer: A

K = [eq] / [ax] = e^{7000/ (8.314 \times 298)} = e^{2.825} = 16.87

. Fraction equatorial = K/(1 + K) = 94.4%. The value 5.6% is the axial fraction.