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Chapter 17: Problem 45

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{~S}_{\text {rhombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{SO}_{3}(g) \longrightarrow 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\) c. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

Short Answer

Expert verified
For the given reactions: a) The sign of ΔS° is negative and the magnitude is 257.8 J/mol K. b) The sign of ΔS° is positive and the magnitude is 187.0 J/mol K. c) The sign of ΔS° is positive and the magnitude is 36.2 J/mol K.

Step by step solution

01

Observing reaction components' phases#for a Look at the given reaction: \[2H_2S(g) + SO_2(g) \longrightarrow 3S_{\text{rhombic}}(s) + 2H_2O(g)\] Observe the phases of the reactants and the products. We can see that two gaseous substances are reactants, and a solid and a gaseous substance are the products.

Identifying changes in phases#for a There's a decrease in the number of gaseous molecules on the right-hand side of the equation, although the overall number of gaseous species remains the same.
02

Predicting ΔS° sign#for a Since the gaseous phase exhibits more disorder than the solid phase, the entropy change for this reaction will be negative because the products have more order than the reactants. \[\Delta S^{\circ} < 0\] #a. Calculating ΔS°#

Recall the formula#for a To calculate ΔS°, we need the standard entropies of all substances involved in the reaction. We will use the formula: \[\Delta S^{\circ} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants}\]
03

Using standard entropies#for a From the standard entropy tables, we can find the following values: \(S^{\circ}(H_2S) = 206.1 \frac{J}{mol \cdot K}, S^{\circ}(SO_2) = 248.3 \frac{J}{mol \cdot K}, S^{\circ}(S_{\text{rhombic}}) = 31.8 \frac{J}{mol \cdot K}, S^{\circ}(H_2O) = 188.8 \frac{J}{mol \cdot K}\). Plug these values into the formula: \[\Delta S^{\circ} = [3(31.8) + 2(188.8)] - [2(206.1) + 1(248.3)]\]

Calculating ΔS° value#for a Calculate the entropy change: \[\Delta S^{\circ} = (-257.8) \frac{J}{mol \cdot K}\] For reaction a, the ΔS° is negative, as predicted, and the magnitude is 257.8 J/mol K. #b. Predicting sign of ΔS°#
04

Observing reaction components' phases#for b Look at the given reaction: \[2SO_3(g) \longrightarrow 2SO_2(g) + O_2(g)\] Observe the phases of the reactants and the products. We can see that all substances involved are in the gaseous phase.

Identifying changes in phases#for b There's an increase in the number of gaseous molecules, the gaseous species number changes from 2 to 3, on the right-hand side of the equation.
05

Predicting ΔS° sign#for b Since there are more gaseous molecules when the reaction proceeds, the entropy change for this reaction will be positive because the products have more disorder than the reactants. \[\Delta S^{\circ} > 0\] #b. Calculating ΔS°#

Using standard entropies for b From the standard entropy tables, we can find the following values: \(S^{\circ}(SO_3) = 256.9 \frac{J}{mol \cdot K}, S^{\circ}(SO_2) = 248.3 \frac{J}{mol \cdot K}, S^{\circ}(O_2) = 205.2 \frac{J}{mol \cdot K}\). Plug these values into the formula: \[\Delta S^{\circ} = [2(248.3) + 1(205.2)] - [2(256.9)]\]
06

Calculating ΔS° value#for b Calculate the entropy change: \[\Delta S^{\circ} = (187.0) \frac{J}{mol \cdot K}\] For reaction b, the ΔS° is positive, as predicted, and the magnitude is 187.0 J/mol K. #c. Predicting sign of ΔS°#

Observing reaction components' phases#for c Look at the given reaction: \[Fe_2O_3(s) + 3H_2(g) \longrightarrow 2Fe(s) + 3H_2O(g)\] Observe the phases of the reactants and the products. We can see that solid and gaseous substances are involved in both the reactants and the products.
07

Identifying changes in phases#for c There's an increase in the number of gaseous molecules on the right-hand side of the equation, although they have the same number of gaseous species.

Predicting ΔS° sign#for c Since the gaseous phase exhibits more disorder than the solid phase, the entropy change for this reaction will be positive because the products have more disorder than the reactants. \[\Delta S^{\circ} > 0\] #c. Calculating ΔS°#
08

Using standard entropies#for c From the standard entropy tables, we can find the following values: \(S^{\circ}(Fe_2O_3) = 87.4 \frac{J}{mol \cdot K}, S^{\circ}(H_2) = 130.7 \frac{J}{mol \cdot K}, S^{\circ}(Fe) = 27.3 \frac{J}{mol \cdot K}, S^{\circ}(H_2O) = 188.8 \frac{J}{mol \cdot K}\). Plug these values into the formula: \[\Delta S^{\circ} = [2(27.3) + 3(188.8)] - [1(87.4) + 3(130.7)]\]

Calculating ΔS° value#for c Calculate the entropy change: \[\Delta S^{\circ} = (36.2) \frac{J}{mol \cdot K}\] For reaction c, the ΔS° is positive, as predicted, and the magnitude is 36.2 J/mol K.

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Most popular questions from this chapter

The third law of thermodynamics states that the entropy of a perfect crystal at \(0 \mathrm{~K}\) is zero. In Appendix \(4, \mathrm{~F}^{-}(a q), \mathrm{OH}^{-}(a q)\), and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\). For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text {surr }}\) ? \(\Delta S\) ? \(\Delta S_{\text {univ }} ?\) Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

From data in Appendix 4, calculate \(\Delta H^{\circ}, \Delta S^{\circ}\), and \(\Delta G^{\circ}\) for each of the following reactions at \(25^{\circ} \mathrm{C}\). a. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)\) c. \(\mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(s)\) d. \(\mathrm{HCl}(\mathrm{g})+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\)

Many biochemical reactions that occur in cells require relatively high concentrations of potassium ion \(\left(\mathrm{K}^{+}\right)\). The concentration of \(\mathrm{K}^{+}\) in muscle cells is about \(0.15 \mathrm{M}\). The concentration of \(\mathrm{K}^{+}\) in blood plasma is about \(0.0050 \mathrm{M}\). The high internal concentration in cells is maintained by pumping \(\mathrm{K}^{+}\) from the plasma. How much work must be done to transport \(1.0\) mole of \(\mathrm{K}^{+}\) from the blood to the inside of a muscle cell at \(37^{\circ} \mathrm{C}\), normal body temperature? When \(1.0\) mole of \(\mathrm{K}^{+}\) is transferred from blood to the cells, do any other ions have to be transported? Why or why not?

Consider the following reaction: $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \quad K_{298}=0.090 $$ For \(\mathrm{Cl}_{2} \mathrm{O}(g)\), $$ \begin{aligned} \Delta G_{\mathrm{f}}^{\circ} &=97.9 \mathrm{~kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ} &=80.3 \mathrm{~kJ} / \mathrm{mol} \\ S^{\circ} &=266.1 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} \end{aligned} $$ a. Calculate \(\Delta G^{\circ}\) for the reaction using the equation \(\Delta G^{\circ}=-R T \ln (K)\) b. Use bond energy values (Table 8.4) to estimate \(\Delta H^{\circ}\) for the reaction. c. Use the results from parts a and \(\mathrm{b}\) 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 \(P_{\mathrm{HOCl}}=0.10\) torr.
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