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Chapter 23: Problem 75

Consider the following two ligand exchange reactions: \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+6 \mathrm{NH}_{3} \rightleftharpoons\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}+6 \mathrm{H}_{2} \mathrm{O}\) $$ \left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+3 \mathrm{en} \rightleftharpoons\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}+6 \mathrm{H}_{2} \mathrm{O} $$ (a) Which of the reactions should have a larger \(\Delta S^{\circ} ?\) (b) Given that the Co-N bond strength is approximately the same in both complexes, which reaction will have a larger equilibrium constant? Explain your choices.

Short Answer

Expert verified
(a) The reaction that replaces 6 water molecules with 6 ammonia molecules will have a larger \(\Delta S^{\circ}\), meaning it has a greater decrease in randomness. (b) The reaction that replaces 6 water molecules with 3 ethylenediamine molecules will have a larger equilibrium constant, because with a smaller decrease in randomness (smaller negative \(\Delta S^{\circ}\)), this reaction will be more spontaneous.

Step by step solution

01

Analyzing Entropy Change

Consider the number of particles in the reactants and products. In both reactions reactants are a single complex ion and a small molecule, and products are a single complex ion and water. However, the first reaction involves 6 small ammonia molecules being taken up to form the product, while the second reaction involves 3 larger ethylenediamine molecules. As such, the first reaction should show a larger decrease in entropy, meaning its \(\Delta S^{\circ}\) value is more negative.
02

Analyzing Equilibrium Constant

According to Le Chatelier’s Principle, an equilibrium will shift to minimize disturbances such as changes in concentration, temperature, volume, or pressure. If we consider that the Co-N bond strength is approximately the same for both complexes, the reaction with the larger equilibrium constant will be the one where the system can best minimize the disturbance of the reaction, which in this case is the formation of the complex ion. Given the first reaction's larger negative \(\Delta S^{\circ}\) (greater decrease in randomness), this reaction will be less spontaneous than the second, leading to a smaller equilibrium constant. As a result, the second reaction should have a larger equilibrium constant.

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Most popular questions from this chapter

For the same type of ligands, explain why the crystal field splitting for an octahedral complex is always greater than that for a tetrahedral complex.

The \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) ion, which has a square- planar geometry, is diamagnetic, whereas the \(\left[\mathrm{NiCl}_{4}\right]^{2-}\) ion, which has a tetrahedral geometry, is paramagnetic. Show the crystal field splitting diagrams for those two complexes.

Write the formulas for each of the following ions and compounds: (a) bis(ethylenediamine)dichlorochromium(III), (b) pentacarbonyliron(0), (c) potassium tetracyanocuprate(II), (d) tetraammin eaquachlorocobalt(III) chloride.

A student in 1895 prepared three coordination compounds containing chromium, with the following properties: $$ \begin{array}{llc} \hline & & \begin{array}{l} \mathrm{Cl}^{-} \text {Ions in } \\ \text { Solution per } \\ \text { Formula Unit } \end{array} \\ \text { Formula } & \text { Color } & 3 \\ \hline \text { (a) } \mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O} & \text { Violet } & 3 \\ \text { (b) } \mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O} & \text { Light green } & 2 \\ \text { (c) } \mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O} & \text { Dark green } & 1 \\ \hline \end{array} $$ Write modern formulas for these compounds and suggest a method for confirming the number of \(\mathrm{Cl}^{-}\) ions present in solution in each case. (Hint: Some of the compounds may exist as hydrates and \(\mathrm{Cr}\) has a coordination number of 6 in all the compounds.

The \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) complex is more labile than the \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) complex. Suggest an experiment that would prove that \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) is a labile complex.
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