Question

Given that the free energy of the twist-boat conformer of cyclohexane is 5.3 kcal/mol greater than that of the chair conformer, calculate the percentage of twist-boat conformers present in a sample of cyclohexane at 25 °C. Does your answer agree with the statement made about the relative number of molecules in these two conformations?

Answer 1

Answer:

0.013%

Yes, it does. The answer agrees with the statement.

Explanation:

Both conformers are in equilibrium, and it can be represented by the equilibrium equation K:

K = [twist-boat]/[chair]

The free energy between them can be calculated by:

ΔG° = -RTlnK

Where R is the gas constant (8.314 J/mol.K), and T is the temperature (25°C + 273 = 298 K).

ΔG° = 5.3 kcal/mol * 4.182 kJ/kcal = 22.165 kJ/mol = 22165 J/mol

22165 = -8.314*298*lnK

-2477.572lnK = 22165

lnK = -8.946

K = $e^{-8.946}$

K = 1.30x10⁻⁴

[twist-boat]/[chair] = 1.30x10⁻⁴

[twist-boat] = 1.30x10⁻⁴[chair]

The percentage of the twist-boat conformer is:

[twist-boat]/([twist-boat] + [chair]) * 100%

1.30x10⁻⁴[chair]/(1.30x10⁻⁴[chair] + [chair]) *100%

0.013%

The statement about the conformers is that the chair conformer is more stable, and because of that is more present. So, the answer agrees with it.