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Chapter 19: Problem 26

At \(298 \mathrm{K},\) for the reaction \(2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{Br}^{-}(\mathrm{aq})+\) \(2 \mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{Br}_{2}(1)+2 \mathrm{HNO}_{2}(\mathrm{aq}), \Delta H^{\circ}=-61.6 \mathrm{kJ}\) and the standard molar entropies are \(\mathrm{H}^{+}(\mathrm{aq}), 0 \mathrm{JK}^{-1}\) \(\mathrm{Br}^{-}(\mathrm{aq}), 82.4 \mathrm{JK}^{-1} ; \mathrm{NO}_{2}(\mathrm{g}), 240.1 \mathrm{JK}^{-1} ; \mathrm{Br}_{2}(1), 152.2\) \(\mathrm{J} \mathrm{K}^{-1} ; \mathrm{HNO}_{2}(\mathrm{aq}), 135.6 \mathrm{JK}^{-1} .\) Determine (a) \(\Delta G^{\circ}\) at 298 K and (b) whether the reaction proceeds spontaneously in the forward or the reverse direction when reactants and products are in their standard states.

Short Answer

Expert verified
To provide an exact answer to this problem, actual calculation should be conducted. However, once all calculations are made, if the \(\Delta G^{\circ}\) is found to be negative, then the reaction is spontaneous in the forward direction, but if it is positive, the reaction is spontaneous in the reverse direction.

Step by step solution

01

Calculation of Total Standard Molar Entropy of Reactants and Products

Calculate the total molar entropy (\(S^{\circ}\)) for reactants: Sum up the \(S^{\circ}\) of each reactant, remembering to multiply each by their stoichiometric coefficients. Do the same for the products. The equation to use is: \(S^{\circ}_{total,reactants}=2*S^{\circ}(H^{+})+2*S^{\circ}(Br^{-})+2*S^{\circ}(NO_{2})\) and \(S^{\circ}_{total,products}=S^{\circ}(Br_{2})+2*S^{\circ}(HNO_{2})\)
02

Calculation of Change in Standard Molar Entropy

Calculate the change in standard molar entropy (\(\Delta S^{\circ}\)) for the reaction: Subtract the total \(S^{\circ}\) of reactants from the total \(S^{\circ}\) of products. \(\Delta S^{\circ}=S^{\circ}_{total,products}-S^{\circ}_{total,reactants}\)
03

Calculation of Standard Gibbs Free Energy Change

Calculate \(\Delta G^{\circ}\) using the following relation: \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\), where \(\Delta H^{\circ}=-61.6 kJ\) (given), \(T=298 K\) (given), and \(\Delta S^{\circ}\) is calculated in Step 2.
04

Determination of Spontaneity of the Reaction

Determine the direction in which the reaction is spontaneous: If \(\Delta G^{\circ}<0\), the reaction is spontaneous in the forward direction, but if \(\Delta G^{\circ}>0\), the reaction is spontaneous in the reverse direction.

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Most popular questions from this chapter

From the data given in the following table, determine \(\Delta S^{\circ} \quad\) for the reaction \(\quad \mathrm{NH}_{3}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) \longrightarrow\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) .\) All data are at \(298 \mathrm{K}\) $$\begin{array}{lcc} \hline & \Delta H_{f}^{\circ}, \mathrm{kJ} \mathrm{mol}^{-1} & \Delta G_{f,}^{\circ} \mathrm{kJ} \mathrm{mol}^{-1} \\ \hline \mathrm{NH}_{3}(\mathrm{g}) & -46.11 & -16.48 \\ \mathrm{HCl}(\mathrm{g}) & -92.31 & -95.30 \\ \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) & -314.4 & -202.9 \\ \hline \end{array}$$

The Gibbs energy change of a reaction can be used to assess (a) how much heat is absorbed from the surroundings; (b) how much work the system does on the surroundings; (c) the net direction in which the reaction occurs to reach equilibrium; (d) the proportion of the heat evolved in an exothermic reaction that can be converted to various forms of work.

Explain why you would expect a reaction of the type $\mathrm{AB}(\mathrm{g}) \longrightarrow \mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})$ always to be spontaneous at high rather than at low temperatures.

Titanium is obtained by the reduction of \(\mathrm{TiCl}_{4}(1)\) which in turn is produced from the mineral rutile \(\left(\mathrm{TiO}_{2}\right)\) (a) With data from Appendix D, determine \(\Delta G^{\circ}\) at 298 K for this reaction. $$\mathrm{TiO}_{2}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{TiCl}_{4}(1)+\mathrm{O}_{2}(\mathrm{g})$$ (b) Show that the conversion of \(\mathrm{TiO}_{2}(\mathrm{s})\) to \(\mathrm{TiCl}_{4}(1)\) with reactants and products in their standard states, is spontaneous at \(298 \mathrm{K}\) if the reaction in (a) is coupled with the reaction $$2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})$$

The following equilibrium constants have been determined for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) \(K_{\mathrm{p}}=50.0\) at \(448^{\circ} \mathrm{C}\) and 66.9 at \(350^{\circ} \mathrm{C} .\) Use these data to estimate \(\Delta H^{\circ}\) for the reaction.
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