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?

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 =

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.

"the periodic table is arranged in order of atomic number."

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