Question

\(K_{a}\) for acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is \(1.754 \times 10^{-5} .\) At \(50^{\circ} \mathrm{C}, K_{\mathrm{a}}\) is \(1.633 \times 10^{-5}\). Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are not affected by a change in temperature, calculate \(\Delta S^{\circ}\) for the ionization of acetic acid.

Short answer

The standard entropy change (ΔS°) for the ionization of acetic acid at both temperatures is approximately -109.94 J·K⁻¹·mol⁻¹.

Step-by-step solution

Calculate \(\Delta G^{\circ}\) at both temperatures

To get the \(\Delta G^{\circ}\) values for the given two temperatures we will use \(\Delta G^{\circ}=-RT \ln \left(K_{a}\right)\) for each one, where K is the equilibrium constant, R is the gas constant, and T is the temperature in Kelvin. For \(25^{\circ} \mathrm{C}\) (or \(298\,\mathrm{K}\)): \(\Delta G_1^{\circ}=-R(298\,\mathrm{K})\ln\left(1.754 \times 10^{-5}\right)\) For \(50^{\circ} \mathrm{C}\) (or \(323\,\mathrm{K}\)): \(\Delta G_2^{\circ}=-R(323\,\mathrm{K})\ln\left(1.633 \times 10^{-5}\right)\)

Set up the \(\Delta G^{\circ}\) equations with \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\)

Now we are going to set up equations for each of them using \(\Delta G^{\circ}= \Delta H^{\circ} - T\Delta S^{\circ}\). For \(25^{\circ} \mathrm{C}\) (\(298\,\mathrm{K}\)): \(\Delta G_1^{\circ}=\Delta H^{\circ}-298\,\mathrm{K}\Delta S^{\circ}\) For \(50^{\circ} \mathrm{C}\) (\(323\,\mathrm{K}\)): \(\Delta G_2^{\circ}=\Delta H^{\circ}-323\,\mathrm{K}\Delta S^{\circ}\)

Solve the system of equations for \(\Delta S^{\circ}\)

Now we have a system of two linear equations with two variables, \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\). \(\Delta G_1^{\circ}-\Delta G_2^{\circ}=-25\,\mathrm{K}\Delta S^{\circ}\) Next, we can substitute the \(\Delta G^{\circ}\) values calculated in Step 1 and solve the equation for \(\Delta S^{\circ}\). \((-R(298\,\mathrm{K})\ln\left(1.754 \times 10^{-5}\right))-(-R(323\,\mathrm{K})\ln\left(1.633 \times 10^{-5}\right))=-25\,\mathrm{K}\Delta S^{\circ}\) Now, we can isolate \(\Delta S^{\circ}\) and get its value: \(\Delta S^{\circ}=\frac{-R(298\,\mathrm{K})\ln\left(1.754 \times 10^{-5}\right)+R(323\,\mathrm{K})\ln\left(1.633 \times 10^{-5}\right)}{-25\,\mathrm{K}}\) Finally, plugging the values of R(\(8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}\)) to get the numerical value for \(\Delta S^{\circ}\): \(\Delta S^{\circ}=\frac{-8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}(298\,\mathrm{K})\ln\left(1.754 \times 10^{-5}\right)+8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}(323\,\mathrm{K})\ln\left(1.633 \times 10^{-5}\right)}{-25\,\mathrm{K}}\) After calculating, we get: \(\Delta S^{\circ}\approx -109.94\, \mathrm{J\,K^{-1}\,mol^{-1}}\)

Key concepts

Gibbs Free Energy

Gibbs Free Energy is a crucial energy function in thermodynamics that allows us to predict the direction of chemical reactions and determine their spontaneity. In chemistry, the Gibbs Free Energy change, denoted as \( \Delta G^{\circ} \), provides insight into the feasibility of a reaction. It can be calculated using the formula: \( \Delta G^{\circ} = -RT \ln K \), where

• R is the universal gas constant \( (8.314 \, \mathrm{J\,K^{-1}\,mol^{-1}}) \)
• T is the temperature in Kelvin
• K is the equilibrium constant

For spontaneous reactions, \( \Delta G^{\circ} \) should be negative, meaning the reaction can proceed without additional energy input. For the ionization of acetic acid at two different temperatures, the change in \( \Delta G^{\circ} \) is calculated at both 25°C and 50°C using their respective equilibrium constants. This calculation is integral because it sets the foundation for determining the entropy change \( \Delta S^{\circ} \), as shown in our exercise.

Equilibrium Constant

The equilibrium constant, symbolized as \( K \), is a fundamental measure in chemistry that quantifies the position of equilibrium in a chemical reaction at a certain temperature. It essentially reflects the ratio of the concentrations of the products to the reactants at equilibrium. It's critical to note that the equilibrium constant is temperature-dependent, which explains the difference in \( K_a \) values for acetic acid at 25°C and 50°C.

In the context of calculating \( \Delta G^{\circ} \), the equilibrium constant embodies how the system's free energy landscape molds upon changing the temperature. A higher equilibrium constant often indicates that product formation is favored. This relationship between \( \Delta G^{\circ} \) and \( K \) can be incredibly insightful for predicting how temperature changes impact reaction spontaneity. Understanding equilibrium constants helps chemists to control reaction conditions to achieve desired results.

Entropy Change

Entropy change, denoted as \( \Delta S^{\circ} \), is a thermodynamic function that describes the degree of disorder or randomness in a system. In chemical reactions, changes in entropy can indicate the system's transition from order to disorder (or vice versa).

• Entropy increases when molecules gain more freedom of movement, such as gases expanding or solutes dissolving.
• Conversely, entropy decreases in processes where order is created, like crystallization.

In the case of acetic acid ionization examined in the exercise, determining \( \Delta S^{\circ} \) allows us to understand how the disorder in the system changes with ionization at different temperatures. Once \( \Delta G^{\circ} \) values are known, \( \Delta S^{\circ} \) is computed by using the linear equations derived from the relation \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \). This process provides insights into the entropic factors that contribute to the acid's ionization and aids in predicting the directionality of chemical reactions under varying thermal conditions.