Question

The molecule methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) can act as a monodentate ligand. The following are equilibrium reactions and the thermochemical data at \(298 \mathrm{~K}\) for reactions of methylamine and en with \(\mathrm{Cd}^{2+}(a q)\) : $$ \begin{aligned} \mathrm{Cd}^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \rightleftharpoons & {\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) } \\ \Delta H^{\circ}=&\left.-57.3 \mathrm{~kJ} ; \Delta S^{\circ}=-67.3 \mathrm{~J} / \mathrm{K} ; \Delta G^{\circ}=-37.2 \mathrm{k}\right] \\ & \mathrm{Cd}^{2+}(a q)+2 \mathrm{en}(a q) \rightleftharpoons\left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q) \\ \Delta H^{\circ}=&\left.-56.5 \mathrm{k} ; ; \Delta S^{\circ}=+14.1 \mathrm{~J} / \mathrm{K} ; \Delta G^{\circ}=-60.7 \mathrm{k}\right] \end{aligned} $$ (a) Calculate \(\Delta G^{\circ}\) and the equilibrium constant \(K\) for the following ligand exchange reaction: $$ \begin{aligned} {\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q)+} & 2 \mathrm{en}(a q) \rightleftharpoons \\ & {\left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+(a q)}+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) } \end{aligned} $$ Based on the value of \(K\) in part (a). what would you conclude about this reaction? What concept is demonstrated? (b) Determine the magnitudes of the enthalpic \(\left(\Delta H^{\circ}\right)\) and the entropic \(\left(-T \Delta S^{\circ}\right)\) contributions to \(\Delta G^{\circ}\) for the ligand exchange reaction. Explain the relative magnitudes. (c) Based on information in this exercise and in the "A Closer Look" box on the chelate effect, predict the sign of \(\Delta H^{2}\) for the following hypothetical reaction: $$ \begin{aligned} {\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q)+} & 4 \mathrm{NH}_{3}(a q) \rightleftharpoons \\ & {\left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) } \end{aligned} $$

Short answer

In the ligand exchange reaction between \(\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}\) and ethylenediamine, the change in Gibbs free energy is \(\Delta G^{\circ}_{exchange}=-84.2 \mathrm{kJ/mol}\), the equilibrium constant is \(K \approx 8.24\times10^{11}\), indicating a favorable reaction towards the products. It is enthalpically driven with \(\Delta H^{\circ}_{exchange} = -170.3 \mathrm{kJ/mol}\) and \(-T\Delta S^{\circ}_{exchange} = 11.6 \mathrm{kJ/mol}\). For the hypothetical reaction involving \(\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}\) and \(\mathrm{NH_{3}}\), it is predicted that the enthalpy change will be negative.

Step-by-step solution

(Calculate the change in Gibbs free energy)

We can use Hess's law to find the change in Gibbs free energy of the ligand exchange reaction. Given the thermodynamic data for both initial reactions, the ligand exchange reaction can be constructed as a linear combination of these reactions. The ligand exchange reaction can be found by reversing the first reaction, so that $$ \begin{aligned} \left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) & \rightleftharpoons \mathrm{Cd}^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \end{aligned} $$ has a \(\Delta G^{\circ} = 37.2 \mathrm{kJ}\). Then, adding twice the second reaction, we arrive at the desired ligand exchange reaction. Thus, the total \(\Delta G^{\circ}\) for the ligand exchange reaction is $$\Delta G^{\circ}_{exchange} = 37.2 \mathrm{kJ} + 2(-60.7 \mathrm{kJ}) = -84.2 \mathrm{kJ}.$$ Step 2: Calculate the equilibrium constant

(Use the Gibbs free energy change to find the equilibrium constant)

To calculate the equilibrium constant \(K\), use the formula $$ K = \exp \left(-\frac{\Delta G^{\circ}}{RT}\right), $$ where \(R = 8.314 \mathrm{J/(mol~K)}\) and \(T = 298 \mathrm{K}\). Hence, $$K = \exp{\left(-\frac{(-84.2\times10^3\,\mathrm{J/mol})}{(8.314 \mathrm{J/(mol~K)}) (298 \mathrm{K})}\right)} \approx 8.24\times10^{11}.$$ Since \(K\) is very large, we can conclude that the ligand exchange reaction is favorable and will proceed mostly toward the products. Step 3: Calculate enthalpic and entropic contributions

(Compute the enthalpy and entropy changes for the ligand exchange reaction)

As in Step 1, we can use Hess's law to find the change in enthalpy and entropy of the ligand exchange reaction. For the enthalpy change, we find $$\Delta H^{\circ}_{exchange} = -57.3 \mathrm{kJ} + 2(-56.5 \mathrm{kJ}) = -170.3 \mathrm{kJ}.$$ For the entropy change, we find $$\Delta S^{\circ}_{exchange} = -67.3 \mathrm{J/K} + 2(14.1 \mathrm{J/K}) = -39.1 \mathrm{J/K}.$$ The entropic contribution to the Gibbs free energy change is given by $$-T\Delta S^{\circ}_{exchange} = -(298 \mathrm{K})(-39.1 \mathrm{J/K}) = 11.6 \mathrm{kJ}.$$ Thus, the enthalpy change is negative and larger in magnitude than the entropy change, making the ligand exchange reaction favorable and enthalpically driven. Step 4: Predict the sign of \(\Delta H^{\circ}\) for the hypothetical reaction

(Predict the enthalpy change for the given hypothetical reaction)

Based on the information given, we know that both \(\mathrm{NH_3}\) and \(\mathrm{CH_3NH_2}\) act as monodentate ligands. Since the ligand exchange in this case also involves monodentate ligands, and similar chelate effects as in the previously examined reaction can be expected, we can predict that the enthalpy change for the hypothetical reaction will also likely be negative, making the reaction favorable.

Key concepts

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G^{\circ} \)) is a key thermodynamic quantity used to predict the direction of chemical reactions. It tells us if a reaction will proceed spontaneously. If \( \Delta G^{\circ} \) is negative, the reaction is favorable and can occur without external energy input.

Calculating Gibbs Free Energy
To find \( \Delta G^{\circ} \) for a ligand exchange reaction, you can use Hess's Law. This involves combining the reactions and their corresponding \( \Delta G^{\circ} \) values.

• Reverse the initial reaction to change sign of \( \Delta G^{\circ} \).
• Add the adjusted reactions and \( \Delta G^{\circ} \)
• Resulting \( \Delta G^{\circ} \) gives the energy change for the exchange.

In our example, this results in \( \Delta G^{\circ}_{exchange} = -84.2 \, \mathrm{kJ} \), indicating the reaction is favorable.

Equilibrium Constant

The Equilibrium Constant (\( K \)) provides insight into the concentration of reactants and products at equilibrium. A large \( K \) means the reaction favors products, while a small \( K \) favors reactants.

Linking to Gibbs Free Energy
The equilibrium constant can be computed from \( \Delta G^{\circ} \) using the formula:\[K = \exp \left(-\frac{\Delta G^{\circ}}{RT}\right)\]where \( R \) is the universal gas constant and \( T \) is temperature in Kelvin.

• A large negative \( \Delta G^{\circ} \) (like -84.2 kJ) results in a large \( K \)
• This indicates that the ligand exchange reaction strongly favors product formation.
• In our example, \( K \approx 8.24\times10^{11} \), showing a highly favorable reaction.

Enthalpy and Entropy

Enthalpy (\( \Delta H^{\circ} \)) and Entropy (\( \Delta S^{\circ} \)) are thermodynamic properties that influence \( \Delta G^{\circ} \). Enthalpy is the heat absorbed or released, while entropy measures disorder.

Enthalpic and Entropic Contributions
For the ligand exchange reaction, calculate:

• Enthalpy change: \( \Delta H^{\circ}_{exchange} = -170.3 \, \mathrm{kJ} \)
• Entropy change: \( \Delta S^{\circ}_{exchange} = -39.1 \, \mathrm{J/K} \)
• Entropic contribution to Gibbs Free Energy: \(-T\Delta S^{\circ}_{exchange} = 11.6 \, \mathrm{kJ} \)

Here, the reaction is driven primarily by the enthalpic term, as \( \Delta H^{\circ} \) has a larger influence, making the reaction exothermic.

Entropy plays a smaller, yet crucial role by resisting disorganization, although the overall enthalpic effect prevails.