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Chapter 17: Problem 74

Theoretically, one can obtain zinc from an ore containing zinc sulfide, \(\mathrm{ZnS}\), by the reaction $$ \mathrm{ZnS}(s) \longrightarrow \mathrm{Zn}(s)+\mathrm{S}(s) $$ (a) Show by calculation that this reaction is not feasible at \(25^{\circ} \mathrm{C}\). (b) Show that by coupling the above reaction with the reaction $$ \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) $$ the overall reaction, in which \(\mathrm{Zn}\) is obtained by roasting in oxygen, is feasible at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Based on the step-by-step solution above, we can determine that Reaction 1 (ZnS -> Zn + S) is not feasible at 25°C since ΔG° > 0. However, when coupled with Reaction 2 (S + O2 -> SO2), the overall reaction of obtaining Zn by roasting in oxygen becomes feasible at 25°C as the ΔG° for the overall reaction is negative (-92.1 kJ/mol).

Step by step solution

01

Gather the necessary data

Obtain the \(\Delta H^{\circ}_f\) and \(S^{\circ}_m\) values for the relevant substances involved in both reactions (from an appendix or chemistry database): Substance | \(\Delta H^{\circ}_f\) (kJ/mol) | \(S^{\circ}_m\) (J/mol·K) ------------------|-----------------------------|------------------------ \(\mathrm{ZnS}(s)\) | \(-206.0\) | \(80.5\) \(\mathrm{Zn}(s)\) | \(0\) | \(41.6\) \(\mathrm{S}(s)\) | \(0\) | \(31.8\) \(\mathrm{O}_{2}(g)\) | \(0\) | \(205.2\) \(\mathrm{SO}_{2}(g)\) | \(-296.9\) | \(248.2\)
02

Compute the \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for Reaction 1

Use the \(\Delta H^{\circ}_f\) and \(S^{\circ}_m\) values to determine the standard enthalpy change \((\Delta H^{\circ})\) and standard entropy change \((\Delta S^{\circ})\) for the reaction \(\mathrm{ZnS}(s) \longrightarrow \mathrm{Zn}(s)+\mathrm{S}(s)\). The equation is: \(\Delta H^{\circ} = \Sigma \Delta H^{\circ}_{\text{products}} - \Sigma \Delta H^{\circ}_{\text{reactants}}\) \(\Delta S^{\circ} = \Sigma S^{\circ}_{\text{m, products}} - \Sigma S^{\circ}_{\text{m, reactants}}\) For Reaction 1: \(\Delta H^{\circ} = [(0) + (0)] - [(-206.0)] = 206.0\) kJ/mol \(\Delta S^{\circ} = [(41.6) + (31.8)] - [(80.5)] = -7.1\) J/mol·K
03

Compute the \(\Delta G^{\circ}\) for Reaction 1

Now, we will compute the standard Gibbs free energy change \((\Delta G^{\circ})\) using the equation: \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\) For Reaction 1 at \(25^{\circ} \mathrm{C}\) (298 K): \(\Delta G^{\circ} = (206.0\,\text{kJ/mol}) - (298\,\text{K})(-7.1\,\text{J/mol·K})(\frac{1\,\text{kJ}}{1000\,\text{J}}) = 206.0\,\text{kJ/mol} - 2.1\text{ kJ/mol} = 208.1\,\text{kJ/mol}\) Since \(\Delta G^{\circ} > 0\), Reaction 1 is not feasible at \(25^{\circ} \mathrm{C}\).
04

Compute the \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for Reaction 2

Next, compute the standard enthalpy change \((\Delta H^{\circ})\) and standard entropy change \((\Delta S^{\circ})\) for the reaction \(\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)\): For Reaction 2: \(\Delta H^{\circ} = [(-296.9)] - [(0) + (0)] = -296.9\) kJ/mol \(\Delta S^{\circ} = [(248.2)] - [(31.8) + (205.2)] = 11.2\) J/mol·K
05

Compute the \(\Delta G^{\circ}\) for Reaction 2

Compute the standard Gibbs free energy change \((\Delta G^{\circ})\) for Reaction 2 at \(25^{\circ} \mathrm{C}\) (298 K): \(\Delta G^{\circ} = (-296.9\,\text{kJ/mol}) - (298\,\text{K})(11.2\,\text{J/mol·K})(\frac{1\,\text{kJ}}{1000\,\text{J}}) = -296.9\,\text{kJ/mol} - 3.3\,\text{kJ/mol} = -300.2\,\text{kJ/mol}\)
06

Compute the overall reaction by coupling Reaction 1 and Reaction 2

Couple Reaction 1 and Reaction 2 to obtain the following overall reaction: \(\mathrm{ZnS}(s) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{Zn}(s) + \mathrm{SO}_{2}(g)\) To determine the \(\Delta G^{\circ}\) for the overall reaction, sum the \(\Delta G^{\circ}\) values from Reaction 1 and Reaction 2: \(\Delta G^{\circ}_{\text{overall}} = \Delta G^{\circ}_{1} + \Delta G^{\circ}_{2} = 208.1\,\text{kJ/mol} - 300.2\,\text{kJ/mol} = -92.1\,\text{kJ/mol}\) Since \(\Delta G^{\circ}_{\text{overall}} < 0\), the overall reaction of obtaining \(\mathrm{Zn}\) by roasting in oxygen is feasible at \(25^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

Consider the reaction $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ At \(500^{\circ} \mathrm{C}\), a flask initially has all three gases, each at a partial pressure of \(0.200\) atm. When equilibrium is established, the partial pressure of HI is determined to be \(0.48\) atm. What is \(\Delta G^{\circ}\) for the reaction at \(500^{\circ} \mathrm{C}\) ?

On the basis of your experience, predict which of the following reactions are spontaneous. (a) \(\mathrm{Zn}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) (c) \(\mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)\)

Predict the sign of \(\Delta S\) for the following. (a) a lake freezing (b) precipitating lead chloride (c) a candle burning (d) weeding a garden

How many moles of ATP must be converted to ADP by the reaction \(\mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ADP}(a q)+\mathrm{HPO}_{4}^{2-}(a q)+2 \mathrm{H}^{+}(a q)\) $$ \Delta G^{\circ}=-31 \mathrm{~kJ} $$ to bring about a nonspontaneous biochemical reaction in which \(\Delta G^{\circ}=+372 \mathrm{~kJ}\) ?

\(K_{a}\) for acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is \(1.754 \times 10^{-5} .\) At \(50^{\circ} \mathrm{C}, K_{\mathrm{a}}\) is \(1.633 \times 10^{-5}\). Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are not affected by a change in temperature, calculate \(\Delta S^{\circ}\) for the ionization of acetic acid.
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