Question

Consider the reaction $$ \mathrm{NH}_{4}^{+}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NH}_{3}(a q) $$ Use \(\Delta G_{f}^{\circ}\) for \(\mathrm{NH}_{3}(a q)\) at \(25^{\circ} \mathrm{C}=-26.7 \mathrm{~kJ} / \mathrm{mol}\) and the appropriate tables to calculate (a) \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) (b) \(K_{\mathrm{a}}\) at \(25^{\circ} \mathrm{C}\)

Short answer

Question: Calculate the standard Gibbs free energy change and the equilibrium constant for the dissociation of ammonium ions into hydrogen ions and ammonia at 25°C, given that the standard Gibbs free energy of formation for aqueous ammonia at 25°C is -26.7 kJ/mol. Answer: The standard Gibbs free energy change for the reaction is 52.61 kJ/mol, and the equilibrium constant is \(5.69 \times 10^{-10}\) at 25°C.

Step-by-step solution

Calculate the Gibbs free energy change for the reaction.

Using the given value of the Gibbs free energy of formation for aqueous ammonia (\(\Delta G_{f}^{\circ}(\mathrm{NH}_{3}) = -26.7\) kJ/mol) and obtaining the values for the other species involved in the reaction from the tables: \(\Delta G_{f}^{\circ}(\mathrm{NH}_{4}^{+}) = -79.31\) kJ/mol and \(\Delta G_{f}^{\circ}(\mathrm{H}^{+}) = 0\) kJ/mol (since it's a reference state). We can calculate the Gibbs free energy change for the reaction (\(\Delta G^{\circ}\)) by using the following equation: \(\Delta G^{\circ} = \sum [\Delta G_{f}^{\circ} (\text{products})] - \sum [\Delta G_{f}^{\circ} (\text{reactants})]\) That is: \(\Delta G^{\circ} = [\Delta G_{f}^{\circ} (\mathrm{H}^{+}) + \Delta G_{f}^{\circ} (\mathrm{NH}_{3})] - [\Delta G_{f}^{\circ} (\mathrm{NH}_{4}^{+})]\)

Substitute the values and calculate \(\Delta G^{\circ}\)

Plugging in the appropriate values, \(\Delta G^{\circ} = [(0) + (-26.7)] - [-79.31]\) \(\Delta G^{\circ} = 52.61 \; \text{kJ/mol}\) So the standard Gibbs free energy change for the reaction is 52.61 kJ/mol.

Calculate the equilibrium constant \(K_{a}\) using \(\Delta G^{\circ}\)

We can relate the standard Gibbs free energy change with the equilibrium constant (\(K_{a}\)) using the following equation: \(\Delta G^{\circ} = -RT \ln K_{a}\) Here, R is the gas constant (8.314 J/mol K) and T is the temperature in Kelvin (25°C = 298.15 K). To find \(K_{a}\), we can rearrange the equation: \(K_{a} = e^{-\frac{\Delta G^{\circ}}{RT}}\)

Substitute the values and calculate \(K_{a}\)

Substituting the values, \(K_{a} = e^{-\frac{(52.61 \times 10^{3}\;\text{J/mol})}{(8.314\;\text{J/mol}\cdot\text{K})(298.15\;\text{K})}}\) \(K_{a} = 5.69 \times 10^{-10}\) So the equilibrium constant for the reaction is \(5.69 \times 10^{-10}\). To summarize, the standard Gibbs free energy change for the reaction is 52.61 kJ/mol, and the equilibrium constant is \(5.69 \times 10^{-10}\) at 25°C.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time. It's a dynamic balance, not a static one, meaning that the reactions continue to occur, but because they are happening at the same rate, the overall concentrations remain constant.

For the reaction \( NH_4^+(aq) \rightleftharpoons H^+(aq) + NH_3(aq) \), equilibrium is reached when the rate at which ammonium ions (\( NH_4^+ \) ) dissociate into hydrogen ions (\( H^+ \) ) and ammonia (\( NH_3 \) ) equals the rate at which \( H^+ \) and \( NH_3 \) combine to form \( NH_4^+ \). Understanding this balance is crucial to predict the behavior of the reaction under different conditions and to calculate the equilibrium concentrations of each species involved.

Equilibrium Constant (Ka)

The equilibrium constant (\( K_a \)) is a numerical value that represents the ratio of the concentrations of products to reactants at chemical equilibrium, each raised to the power of their stoichiometric coefficients. It is specific to a particular reaction at a specific temperature. For the given reaction, \( K_a \) indicates the strength of the acid (\( NH_4^+ \) ) in its ability to donate a proton to water.

The smaller the value of \( K_a \)—as in this case, where \( K_a = 5.69 \times 10^{-10} \)—the weaker the acid and the more incomplete the dissociation. Conversely, a larger \( K_a \) value would signify a stronger acid and a more complete dissociation into ions at equilibrium. \( K_a \) helps in understanding not just the direction of the reaction but also the extent to which reactants are converted into products.

Standard Gibbs Free Energy Change

The standard Gibbs free energy change (\( \Delta G^\circ \)) for a reaction is a thermodynamic property that indicates the spontaneity of a reaction under standard conditions (1 atm pressure and 298.15 K). It combines the concepts of enthalpy, entropy, and temperature to provide a value that predicts whether a reaction will proceed without outside intervention.

In our example, \( \Delta G^\circ \) for the dissociation of ammonium ion into hydrogen ion and ammonia is 52.61 kJ/mol. A positive value for \( \Delta G^\circ \) indicates that the reaction is non-spontaneous under standard conditions, which means external energy is required to drive the reaction. Conversely, a negative \( \Delta G^\circ \) would symbolize a spontaneous process. It is essential to note that these are standard conditions and that \( \Delta G^\circ \) can vary with different conditions such as temperature and concentration.

Thermodynamics in Chemistry

Thermodynamics in chemistry is the study of energy changes accompanying chemical reactions and physical processes. It is primarily concerned with the concepts of energy, enthalpy, entropy, and Gibbs free energy. These quantities help chemists understand the energetics of reactions—why some reactions occur spontaneously while others do not, and how changes in conditions such as temperature and pressure can influence the direction and extent of a reaction.

The linkage between \( \Delta G^\circ \) and \( K_a \) is a prime example of thermodynamics in action. The equation \( \Delta G^\circ = -RT \ln K_a \) shows how the equilibrium constant is directly related to the free energy change. These concepts are fundamental not only to predict whether a reaction will occur but also to design the conditions necessary to make a reaction proceed in a desired direction.