Under standard condition and 298 K, the free energy difference, ∆Gº between the two chair conformations of a substituted cyclohexane molecule is 4.93 kJ/mol. What percent of the sample at equilibrium represents the most stable conformer?

Question

aaronq · Answer

From the data given, you can find the equilibrium constant.

$\sfΔG^o=−RTln(K)→\large K=e^{−\frac{ΔG^o}{RT}}$

Use the eq. constant to determine the percent of the sample that is the stable conformer.

aaronq · Answer

Since you're not likely to comeback i'll just do it anyway,

$\large \sf K=e^{\frac{-4930~ J/mol}{8.314~J/K*mol*298~K}}=0.136715\approx 0.137$

$\sf stable ~conformer \rightleftharpoons~less~stable~conformer$

$\sf K=\dfrac{[less~stable]}{[stable]}=\dfrac{0.137}{1}$

$\sf \%~of~stable~conformer=\dfrac{1}{1.137}*100\%=87.9727\% \approx88\%$