Question

Among the many complex ions of cobalt are the following: $$ \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}(a q)+3 \mathrm{en}(a q) \rightleftharpoons \mathrm{Co}(\mathrm{en})_{3}^{3+}(a q)+6 \mathrm{NH}_{3}(a q) $$ where "en" stands for ethylenediamine, \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2} .\) Six \(\mathrm{Co}-\mathrm{N}\) bonds are broken and \(\operatorname{six} \mathrm{Co}-\mathrm{N}\) bonds are formed in this reaction, so \(\Delta H_{\mathrm{rxn}}^{\circ} \approx 0,\) yet \(K>1 .\) What are the signs of \(\Delta S^{\circ}\) and \(\Delta G^{\circ} ?\) What drives the reaction?

Short answer

For the reaction, \( \Delta S^{\circ} > 0 \) and \( \Delta G^{\circ} < 0 \). The increase in entropy drives the reaction.

Step-by-step solution

Understanding the Reaction

The reaction given is $$ \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}(aq) + 3 \mathrm{en}(aq) \rightleftharpoons \mathrm{Co}(\mathrm{en})_{3}^{3+}(aq) + 6 \mathrm{NH}_{3}(aq).$$ In this substitution reaction, 'en' (ethylenediamine) replaces 'NH\textsubscript{3}' ligands around the cobalt ion.

Assess Enthalpy Change (ΔH)

The problem states that 6 Co-N bonds are broken and 6 Co-N bonds are formed. Since the number and type of bonds broken and formed are the same, \( \Delta H_{\mathrm{rxn}}^{\circ} \approx 0 \).

Determine Entropy Change (ΔS)

Examine the change in the number of particles: On the left: 1 complex ion + 3 'en' molecules = 4 particles.On the right: 1 complex ion + 6 NH\textsubscript{3} molecules = 7 particles. Since the number of particles increases, the disorder or randomness of the system increases. Hence, \( \Delta S^{\circ} > 0 \).

Determine Gibbs Free Energy Change (ΔG)

Using the relationship \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \), substitute the values: \( \Delta H^{\circ} \) is approximately 0 and \( \Delta S^{\circ} > 0 \). Therefore, \( \Delta G^{\circ} = 0 - T \Delta S^{\circ} \). Since \( \Delta S^{\circ} > 0 \), \( \Delta G^{\circ} < 0 \). Because \( \Delta G^{\circ} \) is negative, the reaction is spontaneous.

Key concepts

enthalpy change

Enthalpy change, denoted as \( \Delta H \), represents the heat absorbed or released during a chemical reaction at constant pressure. It provides vital information about the energy changes in a chemical process.

In the given reaction: \[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}(aq) + 3 \(en\)(aq) \rightleftharpoons \mathrm{Co}\left(en\right)_{3}^{3+}(aq) + 6 \mathrm{NH}_{3}(aq).\], six Co-N bonds are broken and six new Co-N bonds are formed.

Since the number and type of bonds broken and formed are the same, the enthalpy change \( \Delta H_{\mathrm{rxn}}^{\circ} \) is approximately zero. This means there is no significant net energy change in the system due to bond breaking and formation.

entropy change

Entropy change, denoted as \( \Delta S \), measures the degree of disorder or randomness in a system. It helps in understanding how the energy distribution within a system changes during a reaction.

To determine the entropy change in this reaction, we compare the number of particles in the reactants and products. Initially, we have 1 complex ion and 3 ethylenediamine molecules, totaling 4 particles. On the product side, there is 1 complex ion and 6 ammonia molecules, totaling 7 particles.

The increase in the number of particles indicates an increase in disorder (entropy). Thus, \( \Delta S^{\circ} > 0 \).

gibbs free energy

Gibbs free energy change, denoted as \( \Delta G \), is used to predict the spontaneity of a reaction. It combines enthalpy, entropy, and temperature into one quantity, showing the total amount of useful work obtainable from a reaction. The formula for Gibbs free energy is:
\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]

In our reaction: because \( \Delta H^{\circ} \) is approximately zero and \( \Delta S^{\circ} \) is positive, the term \( -T \Delta S^{\circ} \) becomes negative. Hence, \( \Delta G^{\circ} = 0 - T \Delta S^{\circ} < 0 \). A negative \( \Delta G^{\circ} \) implies that the reaction is spontaneous under standard conditions.

spontaneous reaction

A spontaneous reaction is one that occurs naturally without needing external energy input. It is driven by the changes in Gibbs free energy. If \( \Delta G^{\circ} \) is negative, the reaction is spontaneous.

For the cobalt complex ion reaction, the relationship between enthalpy and entropy changes yields a \( \Delta G^{\circ} \) that is negative. Despite the enthalpy change being negligible (\( \Delta H^{\circ} \) ≈ 0), the considerable increase in entropy (due to more product particles being formed) ensures \( \Delta S^{\circ} \) positive, leading to a negative \( \Delta G^{\circ} \). Thus, the reaction is spontaneous and will proceed without external energy.

coordination chemistry

Coordination chemistry focuses on the properties and reactions of complex ions, which consist of a central metal atom or ion bonded to surrounding molecules or ions called ligands. In this context, 'en' (ethylenediamine) and 'NH\textsubscript{3}' (ammonia) are examples of ligands that can donate electron pairs to the central cobalt ion.

The given reaction highlights a substitution process in coordination chemistry:

\[ \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{3+}(aq) + 3 \(en\)(aq) \rightleftharpoons \mathrm{Co}\left(en\right)_{3}^{3+}(aq) + 6 \mathrm{NH}_{3}(aq). \]

Here, ethylenediamine (\