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 \(25^{\circ} \mathrm{C}\), there are approximately 27.7% decalinol A (equatorial) and 72.3% decalinol B (axial).

Step-by-step solution

Calculate the equilibrium constant

Using the given \(\Delta G^{0}\) and the relationship between standard Gibbs energy change and the equilibrium constant, we can calculate the equilibrium constant, K, for the process: \(\Delta G^{0} = -RT \ln{K}\) where \(\Delta G^{0}\) is the standard Gibbs energy change, R is the gas constant (8.314 J/(mol*K)), T is the temperature in Kelvin (here, \(25^{\circ} \mathrm{C}\) = 298.15 K), and K is the equilibrium constant. Rearrange the equation to solve for K: \(K = e^{\frac{-\Delta G^{0}}{RT}}\) Plug in the given values: \(K = e^{\frac{-(4.0 \times 10^3 \mathrm{J/mol})}{(8.314 \mathrm{J/(mol*K)})(298.15 \mathrm{K})}}\) \(K \approx 2.62\)

Calculate the percent of each decalinol in the equilibrium mixture

Now that we have the equilibrium constant, let's denote the equilibrium mole fractions of decalinol A (equatorial) and B (axial) as \(x_A\) and \(x_B\), respectively. Since K represents the ratio of the concentration of the products (decalinol B) to that of reactants (decalinol A), the equilibrium constant can be written as: \(K = \frac{x_B}{x_A}\) As the sum of mole fractions should be 1, we have: \(x_A + x_B = 1\) We now have two equations and two unknowns that we need to solve for \(x_A\) and \(x_B\): 1. \(K = \frac{x_B}{x_A}\) 2. \(x_A + x_B = 1\) Rearrange the first equation to isolate \(x_B\): \(x_B = Kx_A\) Now, substitute this expression for \(x_B\) into the second equation: \(x_A + Kx_A = 1\) Solve for \(x_A\): \((1 + K)x_A =1\) \(x_A = \frac{1}{1+K} = \frac{1}{1+2.62} \approx 0.277\) Now, calculate \(x_B\): \(x_B = 1 - x_A = 1 - 0.277 \approx 0.723\)

Calculate the percent of each decalinol in the equilibrium mixture

Now that we have the mole fractions, we can calculate the percent of each decalinol in the equilibrium mixture: Percent of decalinol A (equatorial): \(x_A \times 100 = 0.277 \times 100 \approx 27.7\%\) Percent of decalinol B (axial): \(x_B \times 100 = 0.723 \times 100 \approx 72.3\%\) Thus, at equilibrium, the mixture consists of approximately 27.7% decalinol A and 72.3% decalinol B.

Key concepts

Gibbs Free Energy

Understanding Gibbs free energy is crucial when analyzing chemical reactions and their spontaneity. It's denoted by the symbol \( G \) and represents the maximum amount of work a thermodynamic system can perform at constant temperature and pressure. When we discuss chemical reactions, the term \( \Delta G^{0} \) is commonly used, representing the change in Gibbs free energy under standard conditions.

Gibbs free energy is intrinsically related to the feasibility of reactions. When \( \Delta G < 0 \) for a process, it indicates that the reaction is spontaneous in the given direction; conversely, when \( \Delta G > 0 \) the process is non-spontaneous. A zero change in Gibbs free energy \( (\Delta G = 0) \) denotes a system in equilibrium.

In the context of the exercise, the given \( \Delta G^{0} \) provides a starting point to calculate the equilibrium constant, since a direct mathematical relationship exists between \( \Delta G^{0} \) and the equilibrium constant \( K \) for the reaction in question. This relationship is described by the equation \( \Delta G^{0} = -RT \ln{K} \) where \( R \) is the gas constant and \( T \) is the temperature in Kelvin. The knowledge of \( \Delta G^{0} \) allows us to predict the behavior and composition of the equilibrium mixture.

Equilibrium Constant

The equilibrium constant, represented by \( K \), is a number that expresses the ratio of concentrations of products to reactants for a reversible reaction at equilibrium at a given temperature. This constant is crucial as it helps predict the extent of a reaction and determine the composition of the final equilibrium mixture.

The equation \( K = e^{\frac{-\Delta G^{0}}{RT}} \) directly relates the Gibbs free energy to \( K \) and provides a way to calculate it, as seen in the exercise. If \( K > 1 \) the products are favored at equilibrium, indicating a large amount of the products will be present compared to the reactants. Conversely, if \( K < 1 \) the reactants are favored.

Once \( K \) is known, it can be applied to determine the mole fractions of each chemical species at equilibrium. As illustrated, this mathematical relationship is the key to solving for the equilibrium position of a given chemical system and ultimately enables us to calculate the specific composition percentages of the involved substances.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture. It is the ratio of the number of moles of a particular substance to the total number of moles of all substances in the mixture. Denoted by \( x \), it is a dimensionless quantity that represents the fraction of a substance within the entirety of the mixture.

For a binary mixture like the one in the exercise, consisting of decalinols A and B, the mole fraction of decalinol A (\( x_A \) ) and decalinol B (\( x_B \) ) would add up to 1, since they are the only components. The relationship \( x_A + x_B = 1 \) holds true for all mixtures irrespective of the number of components.

Once the mole fractions are known, these can be easily converted into percentages, providing a clear picture of the composition of the mixture. This is particularly useful when predicting the outcome of a reaction at equilibrium and analyzing its composition, just as the exercise requires. Understanding mole fractions is essential not only for solving chemistry problems but also for practical applications in fields like pharmacology, where dosage and concentration can be critical.