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 \(\mathrm{Cd}^{2+}(a q):\) $$ \begin{array}{c} \mathrm{Cd}^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \rightleftharpoons\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) \\ \Delta H^{\circ}=-57.3 \mathrm{~kJ} ; \quad \Delta S^{\circ}=-67.3 \mathrm{~J} / \mathrm{K} ; \quad \Delta G^{\circ}=-37.2 \mathrm{~kJ} \\\ \mathrm{Cd}^{2+}(a q)+2 \mathrm{en}(a q) \rightleftharpoons\left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q) \\ \Delta H^{\circ}=-56.5 \mathrm{~kJ} ; \quad \Delta S^{\circ}=+14.1 \mathrm{~J} / \mathrm{K} ; \quad \Delta G^{\circ}=-60.7 \mathrm{~kJ} \end{array} $$ (a) Calculate \(\Delta G^{\circ}\) and the equilibrium constant \(K\) for the following ligand exchange reaction: \(\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q)+2 \operatorname{en}(a q) \rightleftharpoons\) $$ \left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) $$ 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 \(\Delta H^{\circ}\) for the following hypothetical reaction: $$ \begin{aligned} \left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) &+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \\ \left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \end{aligned} $$

Step by step solution

01

Write the balanced equation for ligand exchange reaction

We have: Reaction 1: \(\mathrm{Cd}^{2+}(a q)+4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \rightleftharpoons\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) \quad \Delta G_{1}^{\circ}=-37.2 \mathrm{~kJ}\) Reaction 2: \(\mathrm{Cd}^{2+}(a q)+2 \mathrm{en}(a q) \rightleftharpoons\left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q) \quad \Delta G_{2}^{\circ}=-60.7 \mathrm{~kJ}\) We want to find the ΔG° for the following reaction: \(\left[\mathrm{Cd}\left(\mathrm{CH}_{3}\mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q)+2 \operatorname{en}(a q) \rightleftharpoons \left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q)+4 \mathrm{CH}_{3}\mathrm{NH}_{2}(a q)\) This reaction can be obtained by reversing Reaction 1 and adding it to Reaction 2: $$ \begin{aligned} -1 \times \text{Reaction 1} &: -\mathrm{Cd}^{2+}(a q)-4 \mathrm{CH}_{3} \mathrm{NH}_{2}(a q) \rightleftharpoons-\left[\mathrm{Cd}\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) \quad \Delta G_{-1}^{\circ}=37.2 \mathrm{~kJ} \\ \text{Reaction 2} &: \phantom{-}1 \times \mathrm{Cd}^{2+}(a q)+2 \mathrm{en}(a q) \rightleftharpoons \phantom{-}1 \times\left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q) \quad \Delta G_{2}^{\circ}=-60.7 \mathrm{~kJ} \end{aligned} $$ Adding these reactions: \(\left[\mathrm{Cd}\left(\mathrm{CH}_{3}\mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q)+2 \operatorname{en}(a q) \rightleftharpoons \left[\mathrm{Cd}(\mathrm{en})_{2}\right]^{2+}(a q)+4 \mathrm{CH}_{3}\mathrm{NH}_{2}(a q)\) Now, we can find ΔG° by adding the ΔG° of the two reactions: \(\Delta G_{-1}^{\circ} + \Delta G_{2}^{\circ} = 37.2 - 60.7 = -23.5 \mathrm{~kJ}\) Step 2: Calculate K The equilibrium constant K can be calculated using the formula: \(K = e^{-\frac{\Delta G^\circ}{RT}}\) Where R is the gas constant (\(8.314 \mathrm{~J~mol^{-1}K^{-1}}\)) and T is the temperature in K (298 K).

02

Calculate the equilibrium constant

Substitute the values and calculate K: \(K = e^{-\frac{-23.5 \times 10^3}{(8.314)(298)}} \approx 2.97 \times 10^3\) Since \(K > 1\), the reaction is product-favored. Step 3: Analyze the reaction Based on K, the reaction is product-favored, which means that chelates (the product) are more stable compared to other complexes (the reactants). This demonstrates the concept of the Chelate Effect. Step 4: Determine ΔH° and -TΔS° for the ligand exchange reaction We have the thermochemical data for Reaction 1 and 2, so we can find the enthalpy and entropy for the ligand exchange reaction the same way we found the ΔG° in Step 1.

03

Calculate ΔH° and -TΔS° for the ligand_exchange_reaction

For enthalpy: \(\Delta H_{-1}^{\circ} = -(-57.3) = 57.3 \mathrm{~kJ}\) \(\Delta H^{\circ} = \Delta H_{-1}^{\circ} + \Delta H_{2}^{\circ} = 57.3 - 56.5 = 0.8 \mathrm{~kJ}\) For entropy: \(\Delta S_{-1}^{\circ} = -(-67.3) = 67.3 \mathrm{J~K^{-1}}\) \(\Delta S^{\circ} = \Delta S_{-1}^{\circ} + \Delta S_{2}^{\circ} = 67.3 + 14.1 = 81.4 \mathrm{J~K^{-1}}\) Then, \(-T \Delta S^{\circ} = -(298)(81.4 \times 10^{-3}) = -24.3 \mathrm{~kJ}\) Now, we can notice that both \(\Delta H^{\circ}\) and \(-T \Delta S^{\circ}\) have almost equal magnitudes but opposite signs. This indicates that the enthalpic and entropic effects are canceling each other out, but the entropic effect is slightly more important for driving the reaction. Step 5: Predict the sign of ΔH° for the hypothetical reaction The reaction is given by: \(\left[\mathrm{Cd}\left(\mathrm{CH}_{3}\mathrm{NH}_{2}\right)_{4}\right]^{2+}(a q) + 4 \mathrm{NH}_{3}(a q) \rightleftharpoons \left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(a q)+4 \mathrm{CH}_{3}\mathrm{NH}_{2}(a q)\) Based on the chelate effect, the formation of stronger complexes (chelates) is usually accompanied by a negative ΔH°. In this hypothetical reaction, the reactant side has a chelate, while the product side has a weaker complex. Thus, breaking the chelate would result in a positive ΔH° instead of a negative value.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!