Question

The decalinols A and B can be equilibrated using aluminum isopropoxide in 2-propanol (isopropyl alcohol) containing a small amount of acetone. Assuming a value of \(\Delta G^{0}\) (equatorial to axial) for cyclohexanol is \(4.0 \mathrm{~kJ}(0.95 \mathrm{kcal}) / \mathrm{mol}\), calculate the percent of each decalinol in the equilibrium mixture at \(25^{\circ} \mathrm{C}\).

Short answer

Answer: In the equilibrium mixture at 25°C, approximately 28.98% of Decalinol A and 71.02% of Decalinol B are present.

Step-by-step solution

Relationship between \(\Delta G^0\) and equilibrium constant

We will utilize the relationship between the standard Gibbs free energy change (\(\Delta G^0\)) and the equilibrium constant (K) for the reaction: $$ \Delta G^0 = -RT \ln K $$ Where \(R\) is the gas constant (\(8.314 \mathrm{J/(mol \cdot K)}\)), \(T\) is the temperature in Kelvin, and \(K\) is the equilibrium constant. We need to solve for K, given \(\Delta G^0\) and the temperature.

Calculating Temperature in Kelvin

Convert the given temperature (\(25^{\circ} \mathrm{C}\)) to Kelvin: $$ T = 25^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 298.15 \mathrm{K} $$

Calculate the Equilibrium Constant (K)

Use the provided \(\Delta G^0 = 4.0 \mathrm{kJ/mol}\) and the calculated temperature in Kelvin to solve for the equilibrium constant: $$ K = e^{-\frac{\Delta G^0}{RT}} = e^{-\frac{4.0 \times 10^3 \mathrm{J/mol}}{(8.314 \mathrm{J/(mol \cdot K)})(298.15 \mathrm{K})}} \approx 2.45 $$

Determine the Equilibrium Concentrations and Percentages

Let's denote the initial concentrations of Decalinol A as [A] and Decalinol B as [B]. Since the reaction reaches equilibrium, we should have: $$ \frac{[B]}{[A]} = K = 2.45 $$ As the initial concentrations should add up to \(100 \%\), we can write: $$ [A] + [B] = 100 \% $$ We can now solve these two equations simultaneously: $$ [A] = \frac{100\%}{2.45 + 1} \approx 28.98 \% \\ [B] = \frac{100\% - 28.98\%}{1} \approx 71.02 \% $$

Final Answer

In the equilibrium mixture at \(25^{\circ} \mathrm{C}\), approximately \(28.98 \%\) of Decalinol A and \(71.02 \%\) of Decalinol B are present.