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Chapter 17: Problem 41

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{SO}_{2}(\mathrm{~g}) \longrightarrow 3 \mathrm{~S}_{\text {thombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\) 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 reaction a, the sign of ΔS° is negative with a value of -439.6 J/molK. For reaction b, the sign of ΔS° is positive with a value of 187.6 J/molK. For reaction c, the sign of ΔS° is slightly positive with a value of 11.3 J/molK.

Step by step solution

01

Predict the sign of ΔS° for reaction a

For reaction a, we have: \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{SO}_{2}(\mathrm{~g}) \longrightarrow 3 \mathrm{S_{thombic}}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\) We can see that there are 3 moles of gaseous reactants and 2 moles of gaseous products. As a result, the reaction leads to more order because we have fewer moles of gas products than the reactants. So, the sign of ΔS° will be negative.
02

Calculate ΔS° for reaction a

To calculate ΔS° for reaction a, we need the standard entropy values for each substance involved in the reaction: \(S^{\circ}_{H2S(g)} = 206.8 \, J/molK\) \(S^{\circ}_{SO2(g)} = 248.2 \, J/molK\) \(S^{\circ}_{S_{thombic}(s)} = 31.9 \, J/molK\) \(S^{\circ}_{H2O(g)} = 188.7 \, J/molK\) Now, we can use the equation to calculate ΔS°: \(\Delta S^{\circ}=\sum S^{\circ}_{\text {products }}-\sum S^{\circ}_{\text {reactants }}\) \(\Delta S^{\circ} = [3(31.9) + 2(188.7)] - [2(206.8) + 1(248.2)]\) \(\Delta S^{\circ} = -439.6 \, J/molK\) So, ΔS° for reaction a is -439.6 J/molK
03

Predict the sign of ΔS° for reaction b

For reaction b, we have: \(2 \mathrm{SO}_{3}(g) \longrightarrow 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\) We can notice that there are 3 moles of gaseous products and 2 moles of gaseous reactants. So, the reaction is leading to more disorder with more moles of gas products than the reactants. Thus, the sign of ΔS° will be positive.
04

Calculate ΔS° for reaction b

To calculate ΔS° for reaction b, we need the standard entropy values for each substance involved in the reaction: \(S^{\circ}_{SO3(g)} = 256.8 \, J/molK\) \(S^{\circ}_{SO2(g)} = 248.2 \, J/molK\) \(S^{\circ}_{O2(g)} = 205.0 \, J/molK\) Now, we can use the equation to calculate ΔS°: \(\Delta S^{\circ}=\sum S^{\circ}_{\text {products }}-\sum S^{\circ}_{\text {reactants }}\) \(\Delta S^{\circ} = [2(248.2) + 1(205)] - 2(256.8)\) \(\Delta S^{\circ} = 187.6 \, J/molK\) So, ΔS° for reaction b is 187.6 J/molK
05

Predict the sign of ΔS° for reaction c

For reaction c, we have: \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\) Comparing the moles of gas in reactants and products, we have 3 moles of gaseous reactants and 3 moles of gaseous products. So, there is no significant change in the number of moles of gas. As a result, the sign of ΔS° will be close to zero or slightly positive due to the formation of water molecules.
06

Calculate ΔS° for reaction c

To calculate ΔS° for reaction c, we need the standard entropy values for each substance involved in the reaction: \(S^{\circ}_{Fe2O3(s)} = 87.4 \, J/molK\) \(S^{\circ}_{H2(g)} = 130.6 \, J/molK\) \(S^{\circ}_{Fe(s)} = 27.2 \, J/molK\) \(S^{\circ}_{H2O(g)} = 188.7 \, J/molK\) Now, we can use the equation to calculate ΔS°: \(\Delta S^{\circ}=\sum S^{\circ}_{\text {products }}-\sum S^{\circ}_{\text {reactants }}\) \(\Delta S^{\circ} = [2(27.2) + 3(188.7)] - [1(87.4) + 3(130.6)]\) \(\Delta S^{\circ} = 11.3 \, J/molK\) So, ΔS° for reaction c is 11.3 J/molK

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Most popular questions from this chapter

A mixture of hydrogen gas and chlorine gas remains unreacted until it is exposed to ultraviolet light from a burning magnesium strip. Then the following reaction occurs very rapidly: $$\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g)$$ Explain.

Hydrogen cyanide is produced industrially by the following exothermic reaction: $$2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \stackrel{\mathrm{low}^{\circ} \mathrm{c}}{\mathrm{Pt} \cdot \mathrm{Rh}} 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ Is the high temperature needed for thermodynamic or kinetic reasons?

The Ostwald process for the commercial production of nitric acid involves three steps: a. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for each of the three steps in the Ostwald process (see Appendix 4 ). b. Calculate the equilibrium constant for the first step at \(825^{\circ} \mathrm{C}\), assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. c. Is there a thermodynamic reason for the high temperature in the first step, assuming standard conditions?

Given the values of \(\Delta H\) and \(\Delta S\), which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=300 . \mathrm{K}\) b. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) c. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=298 \mathrm{~K}\) d. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)

For the reaction at \(298 \mathrm{~K}\), $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ}\) and \(-176.6 \mathrm{~J} / \mathrm{K}, \mathrm{re}-\) spectively. What is the value of \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) ? Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G\) negative above or below this temperature?
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