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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…
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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…
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Feb 16, 2019
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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?
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Adam Jenkin
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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 %
Jonothon
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Feb 21, 2019
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