Question

Under standard condition and 298 K, the free energy difference, ∆Gº between the two chair conformations of a substituted cyclohexane molecule is 5.95 kJ/mol. What percent of the sample at equilibrium represents the most stable conformer?

Answer 1

Delta Go = -RTlnKeq

delta Go = 5.95 kJ/mole = 5.95 X 1000 = 5950 J/mole ( 1 kj = 1000 J )

putting the values and finding Keq

5950 = -8.314 X 298 X ln Keq

ln Keq = -5950 / 2477.572 = -2.4015

Keq = e^-2.402 = 0.0905

suppose the equilibrium reaction is :-

chair 1 <--------> chair 2

now as Keq is less than 1 ....so chair 1 will be more stable

Keq = [chair2]/[chair 1 ] = 0.0905

this means that [chair 2] ~ 0.0905 and [chair 1] ~ 1

[total] = [chair 2] + [chair 1] ~ 1 + 0.0905= 1.0905

percentage of chair 1 = [chair 1] / [total] = 1 / 1.0905 X 100 = 91.70 %