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Chapter 17: Problem 94

Consider two reactions for the production of ethanol: $$\begin{array}{l}\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \\\ \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+\mathrm{H}_{2}(g) \end{array}$$ Which would be the more thermodynamically feasible at standard conditions? Why?

Short Answer

Expert verified
The more thermodynamically feasible reaction at standard conditions is Reaction 1 (C2H4 + H2O → CH3CH2OH) because it has a negative Gibbs free energy change (ΔG° = -14.5 kJ/mol), indicating that it is thermodynamically favorable.

Step by step solution

01

Identify the components of the reactions

The first reaction involves the reactants: ethene (C2H4) and water (H2O) and the product: ethanol (CH3CH2OH). The second reaction involves the reactants: ethane (C2H6) and water (H2O) and the products: ethanol (CH3CH2OH) and hydrogen (H2).
02

Gather standard enthalpy change and standard entropy change values

Consult a standard thermodynamic table to obtain the following standard enthalpy change (ΔH°) and standard entropy change (ΔS°) values for each substance: ΔHf° (C2H4) = +52.3 kJ/mol ΔS° (C2H4) = 219.24 J/mol⋅K ΔHf° (H2O) = -241.8 kJ/mol (as gas) ΔS° (H2O) = 188.7 J/mol⋅K ΔHf° (CH3CH2OH) = -277.7 kJ/mol ΔS° (CH3CH2OH) = 160.7 J/mol⋅K ΔHf° (C2H6) = -84 kJ/mol ΔS° (C2H6) = 229.32 J/mol⋅K ΔHf° (H2) = 0 kJ/mol ΔS° (H2) = 130.68 J/mol⋅K
03

Calculate the enthalpy and entropy changes for each reaction

For Reaction 1: ΔH° (reaction 1) = ΔHf° (CH3CH2OH) - ΔHf° (C2H4) - ΔHf° (H2O) = (-277.7) - 52.3 - (-241.8) = -88.2 kJ/mol ΔS° (reaction 1) = ΔS° (CH3CH2OH) - ΔS° (C2H4) - ΔS° (H2O) = 160.7 - 219.24 - 188.7 = -247.24 J/mol⋅K For Reaction 2: ΔH° (reaction 2) = ΔHf° (CH3CH2OH) + ΔHf° (H2) - ΔHf° (C2H6) - ΔHf° (H2O) = (-277.7) + 0 - (-84) - (-241.8) = 48.1 kJ/mol ΔS° (reaction 2) = ΔS° (CH3CH2OH) + ΔS° (H2) - ΔS° (C2H6) - ΔS° (H2O) = 160.7 + 130.68 - 229.32 - 188.7 = -126.64 J/mol⋅K
04

Compute Gibbs free energy change for each reaction at standard conditions

Standard conditions imply a temperature of 298 K. For Reaction 1: ΔG° (reaction 1) = ΔH° (reaction 1) - TΔS° (reaction 1) = (-88.2 kJ/mol) - (298 K)(-247.24 J/mol⋅K)/1000 = -88.2 + 73.7 = -14.5 kJ/mol For Reaction 2: ΔG° (reaction 2) = ΔH° (reaction 2) - TΔS° (reaction 2) = (48.1 kJ/mol) - (298 K)(-126.64 J/mol⋅K)/1000 = 48.1 + 37.7 = 85.8 kJ/mol
05

Compare the Gibbs free energy change values for both reactions

ΔG° (reaction 1) = -14.5 kJ/mol (negative, indicating a thermodynamically feasible reaction) ΔG° (reaction 2) = 85.8 kJ/mol (positive, indicating a thermodynamically unfavorable reaction) The more thermodynamically feasible reaction at standard conditions is Reaction 1 because it has a negative Gibbs free energy change (ΔG° = -14.5 kJ/mol).

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Most popular questions from this chapter

Consider the reactions $$\begin{aligned}\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) & \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) \\ \mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) & \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q)\end{aligned}$$ where $$\text { en }=\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$$ The \(\Delta H\) values for the two reactions are quite similar, yet \(K_{\text {reaction } 2}>K_{\text {reaction } 1 .}\) Explain.

Using data from Appendix 4, calculate \(\Delta G\) for the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{~S}_{\text {mombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ for the following conditions at \(25^{\circ} \mathrm{C}\) : $$\begin{array}{l}P_{\mathrm{H}_{2} \mathrm{~S}}=1.0 \times 10^{-4} \mathrm{~atm} \\\P_{\mathrm{SO}_{2}}=1.0 \times 10^{-2} \mathrm{~atm} \\ P_{\mathrm{H}_{2} \mathrm{O}}=3.0 \times 10^{-2} \mathrm{~atm}\end{array}$$

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\) using the following data: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}\) at \(600 . \mathrm{K}\) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}\) at 600. \(\mathrm{K}\)

Which of the following processes are spontaneous? a. A house is built. b. A satellite is launched into orbit. c. A satellite falls back to earth. d. The kitchen gets cluttered.

The enthalpy of vaporization of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) is \(31.4\) \(\mathrm{kJ} / \mathrm{mol}\) at its boiling point \(\left(61.7^{\circ} \mathrm{C}\right) .\) Determine \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{sur}}\), and \(\Delta S_{\text {univ }}\) when \(1.00 \mathrm{~mol}\) chloroform is vaporized at \(61.7^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\)
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