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Chapter 17: Problem 101

Consider the reactions $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$ $$\mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q)$$ where $$\mathrm{en}=\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$$ The \(\Delta H\) values for the two reactions are quite similar, yet \(\mathrm{K}_{\text { reaction } 2}>K_{\text { reaction }}\) . Explain.

Short Answer

Expert verified
In summary, the difference in stability between the two coordination complexes is due to the stronger interaction between the bidentate ethylenediamine ligand and the nickel ion compared to the monodentate ammonia ligand. This results in a higher equilibrium constant (\(K\)) for reaction 2, as the more stable complex \(Ni(en)_3^{2+}\) is favored compared to the less stable complex \(Ni(NH_3)_6^{2+}\).

Step by step solution

01

Identify the differences between the ligands in the two reactions

The first reaction involves forming a coordination complex with ammonia \(NH_3\) ligands whereas the second reaction involves forming a coordination complex with ethylenediamine \(en\) ligands. The main difference between the two ligands is their structure, with ammonia being a monodentate ligand and ethylenediamine being a bidentate ligand.
02

Explain the significance of bidentate ligands

Bidentate ligands like ethylenediamine can form two coordinate bonds with the central metal ion, in this case nickel. This leads to a more stable complex, as the bidentate ligand can effectively hold the metal ion in place through two bonds, resulting in a stronger interaction between the metal ion and the ligand.
03

Discuss the difference in stability between the coordination complexes

Since the complex formed with ethylenediamine is more stable due to the stronger interaction between the metal ion and the ligand, the equilibrium for reaction 2 will favor the formation of the \(Ni(en)_3^{2+}\) complex. In contrast, the \(Ni(NH_3)_6^{2+}\) complex formed with ammonia ligands is less stable, due to the weaker interaction between the metal ion and the ligand.
04

Relate the difference in stability to equilibrium constants

The greater the stability of the coordination complex, the more the equilibrium will favor its formation. In other words, the equilibrium constant for the formation of the complex will be higher when the complex is more stable. In this case, since the \(Ni(en)_3^{2+}\) complex is more stable than the \(Ni(NH_3)_6^{2+}\) complex, the equilibrium constant (\(K\)) for reaction 2 will be greater than the equilibrium constant for reaction 1.
05

Conclude

In summary, the equilibrium constant \(K\) for reaction 2 is greater than the equilibrium constant for reaction 1 because the ethylenediamine ligand forms a more stable coordination complex with the nickel ion compared to the ammonia ligand. This leads to a stronger interaction between the metal ion and the ligand, resulting in a higher equilibrium constant for the reaction involving ethylenediamine.

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Most popular questions from this chapter

The melting point for carbon diselenide \(\left(\mathrm{CSe}_{2}\right)\) is \(-46^{\circ} \mathrm{C}\) . At a temperature of \(-75^{\circ} \mathrm{C},\) predict the signs for \(\Delta S_{\mathrm{surr}}\) and $\Delta S_{\mathrm{univ}}\( for the following process: \)\operatorname{CSe}_{2}(l) \rightarrow \operatorname{CSe}_{2}(s)$

Consider the following reaction: $\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \qquad K_{298}=0.090\( For \)\mathrm{Cl}_{2} \mathrm{O}(g)$ $$\Delta G_{\mathrm{f}}^{\circ}=97.9 \mathrm{kJ} / \mathrm{mol}$$ $$\Delta H_{\mathrm{f}}^{\circ}=80.3 \mathrm{kJ} / \mathrm{mol}$$ $$S^{\circ}=266.1 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}$$ a. Calculate \(\Delta G^{\circ}\) for the reaction using the equation $\Delta G^{\circ}=-R T \ln (K)$ b. Use bond energy values (Table 8.5\()\) to estimate \(\Delta H^{\circ}\) for the reaction. c. Use the results from parts a and b to estimate \(\Delta S^{\circ}\) for the reaction. d. Estimate \(\Delta H_{\mathrm{f}}^{\circ}\) and \(S^{\circ}\) for \(\mathrm{HOCl}(g)\) e. Estimate the value of \(K\) at \(500 . \mathrm{K}\) . f. Calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\) when $P_{\mathrm{H}_{2} \mathrm{O}}=18\( torr, \)P_{\mathrm{Cl}_{2} \mathrm{O}}=$ 2.0 torr, and \(P_{\mathrm{HOC}}=0.10\) torr.

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. $\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)$ b. $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$ c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

You have a 1.00 -L sample of hot water \(\left(90.0^{\circ} \mathrm{C}\right)\) sitting open in a \(25.0^{\circ} \mathrm{C}\) room. Eventually the water cools to \(25.0^{\circ} \mathrm{C}\) while the temperature of the room remains unchanged. Calculate \(\Delta S_{\mathrm{sur}}\) for this process. Assume the density of water is 1.00 $\mathrm{g} / \mathrm{cm}^{3}$ over this temperature range, and the heat capacity of water is constant over this temperature range and equal to 75.4 $\mathrm{J} / \mathrm{K} \cdot$ mol.

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ} ?\) a. \(\mathrm{C}_{\text { graphite }}(s)\) or \(\mathrm{C}_{\text { diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)$ c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)
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