Problem 62 - Chemistry: Principles and Reactions

Question

Consider the reaction $$ \mathrm{NH}_{4}{\underline{\phantom{xx}}}^{+}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NH}_{3}(a q) $$ Use \(\Delta G_{i}^{\circ}\) 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 \(25^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

Question: Calculate the standard Gibbs free energy change and the acid dissociation constant for the following reaction at 25 °C using the given standard Gibbs free energy value: \(NH_3{(aq)} + H^+_{(aq)} \rightleftharpoons NH_4^+_{(aq)}\) \(\Delta G_{NH-3}^\circ = -26.7 \frac{kJ}{mol}\) Answer: The standard Gibbs free energy change, \(\Delta G^\circ\), for the reaction at 25 °C is 52.74 \(\frac{kJ}{mol}\) (part a), and the acid dissociation constant, \(K_a\), at 25 °C is approximately 0.1209 (part b).

Step 1: Identifying the Equation

First, identify the equation to be used to find the standard Gibbs free energy change: $$\Delta G^\circ = \sum{(n_i \Delta G_i^\circ)}$$ Step 2: Calculate \(\Delta G^\circ\)

Step 2: Calculate \(\Delta G^\circ\)

Using the equation \(\Delta G^\circ = \sum{(n_i \Delta G_i^\circ)}\), we can calculate \(\Delta G^\circ\) for the given reaction: $$\Delta G^\circ = \Delta G_{H^+}^\circ + \Delta G_{NH_3}^\circ - \Delta G_{NH_4^+}^\circ$$ Now, using the given value for \(\Delta G_{NH_3}^\circ = -26.7 \frac{kJ}{mol}\), and finding the values for \(\Delta G_{H^+}^\circ = 0\) and \(\Delta G_{NH_4^+}^\circ = -79.44 \frac{kJ}{mol}\) from standard tables, we can calculate the standard Gibbs energy change for the reaction: $$\Delta G^\circ = (0) + (-26.7) - (-79.44)$$ $$\Delta G^\circ = 52.74 \frac{kJ}{mol}$$ Step 3: Calculate \(K_a\) using \(\Delta G^\circ\):

Step 3: Calculating \(K_a\)

Next, we need to find \(K_a\) at \(25^\circ \mathrm{C}\). Using the equation \(\Delta G^\circ = -RT\ln{K_a}\), where R is the gas constant (8.314 \(\mathrm{J\,mol^{-1}\,K^{-1}}\)) and T is the temperature in Kelvin (298K): $$K_a = e^{\frac{-\Delta G^\circ}{RT}}$$ Plugging in the values for \(\Delta G^\circ\), R, and T, we get: $$K_a = e^{\frac{-52740\,\mathrm{J\,mol^{-1}}}{8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}(298\,\mathrm{K})}}$$ $$K_a = e^{-2.1157}$$ $$K_a = 0.1209$$ Therefore, the Gibbs free energy change, \(\Delta G^\circ\), at \(25^\circ \mathrm{C}\) is 52.74 \(\frac{kJ}{mol}\) (part a) and the acid dissociation constant, \(K_a\), at \(25^\circ \mathrm{C}\) is approximately 0.1209 (part b).

Key Concepts

Gibbs Free Energy

In thermodynamics, Gibbs free energy is a crucial concept that helps us predict the spontaneity of chemical reactions. The \(\Delta G^{\circ}\) value is an indicator of whether a reaction can occur without external influence. Specifically, when \(\Delta G^{\circ} < 0\), the reaction is spontaneous, meaning it can proceed on its own. If \(\Delta G^{\circ} > 0\), the reaction is non-spontaneous and requires energy input to occur.
To calculate Gibbs free energy change for a reaction, we use the formula:

• \(\Delta G^{\circ} = \sum (n_i \Delta G_i^{\circ})\)

Here, \(n_i\) refers to the stoichiometric coefficients of the reactants and products, and \(\Delta G_i^{\circ}\) represents the standard Gibbs free energy of formation of each component.
For the given reaction involving \mathrm{NH}_4^+\, \mathrm{H}^+\, and \mathrm{NH}_3\, using the known values, we calculated that \(\Delta G^{\circ} = 52.74\) \(\text{kJ/mol}\). This positive number suggests the reaction does not favor the formation of \mathrm{H}^+\ and \mathrm{NH}_3\ under standard conditions.

Acid Dissociation Constant

The acid dissociation constant, denoted as \(K_a\), provides insight into the strength of an acid in solution. It measures the tendency of an acid to donate a proton to the solvent, in this case, water. For weak acids, \(K_a\) values are smaller, indicating less dissociation into ions.
The relationship between \(\Delta G^{\circ}\) and \(K_a\) is given by the equation:

• \(\Delta G^{\circ} = -RT\ln{K_a}\)

Where \(R\) is the gas constant \(8.314 \text{ J/mol}\text{K}\) and \(T\) is the temperature in Kelvin, typically \(298\text{ K}\) for standard conditions. Solving for \(K_a\) with the known \(\Delta G^{\circ}\) (52.74 \(\text{kJ/mol}\)) gives us:
\[ K_a = e^{\frac{-\Delta G^{\circ}}{RT}} \approx 0.1209 \]
This small \(K_a\) value reflects the weak nature of the given acidic solution, with minimal dissociation of \(\mathrm{NH}_4^+\) ions.

Chemical Equilibrium

Chemical equilibrium refers to a state in a reversible reaction where the rates of the forward and backward reactions are equal, resulting in no net change in the concentration of reactants and products. At this point, the reaction has reached a balance and the composition remains constant over time.
For the given reaction:
\(\mathrm{NH}_4^+(aq) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{NH}_3(aq)\)
achieving chemical equilibrium means that the formation of \(\mathrm{H}^+\) and \(\mathrm{NH}_3\) is counterbalanced by their reconversion to \(\mathrm{NH}_4^+\).
In the context of \(K_a\) and \(\Delta G^{\circ}\), when \(\Delta G^{\circ}\) is positive, it indicates the equilibrium position favors the reverse reaction, leading to more reactants than products at equilibrium. The calculated \(K_a = 0.1209\) further supports this, suggesting only a small fraction of \(\mathrm{NH}_4^+\) ionizes to form \(\mathrm{H}^+\) and \(\mathrm{NH}_3\), maintaining the equilibrium predominantly on the side of the reactants.