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Chapter 19: Problem 99

The oxidation of glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\( in body tissue produces \)\mathrm{CO}_{2}$ and \(\mathrm{H}_{2} \mathrm{O} .\) In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\( and \)\mathrm{CO}_{2}$. (a) Using data given in Appendix \(\mathrm{C}\), compare the equilibrium constants for the following reactions: $$ \begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)+2 \mathrm{CO}_{2}(g) \end{array} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

Short Answer

Expert verified
The equilibrium constants for the two reactions are: \(K_1 = e^{-\dfrac{ΔG°_1}{RT}}\) \(K_2 = e^{-\dfrac{ΔG°_2}{RT}}\) Compare the calculated equilibrium constants (K_1 and K_2). The larger equilibrium constant corresponds to the more spontaneous reaction. The maximum work that can be obtained from each process under standard conditions are: \(W_{max1}=-ΔG°_1\) \(W_{max2}=-ΔG°_2\) Compare the maximum works (W_max1 and W_max2) calculated for both reactions. The reaction with higher maximum work represents the process that can produce more work under standard conditions.

Step by step solution

01

Write down the equilibrium constant formula for both reactions

The equilibrium constant (K) for a reaction can be expressed in terms of standard Gibbs free energy change (ΔG°) by the following formula: $$ K = e^{-\dfrac{ΔG°}{RT}} $$ Where: \(K\) = equilibrium constant \(ΔG°\) = standard Gibbs free energy change in J/mol \(R\) = universal gas constant, 8.314 J/(K·mol) \(T\) = temperature, 298K (since, standard conditions) Now, we can write the equilibrium constants for the given reactions: Reaction 1: $$ C_6H_{12}O_6(s) + 6O_2(g) \rightleftharpoons 6CO_2(g) + 6H_2O(l) $$ Reaction 2: $$ C_6H_{12}O_6(s) \rightleftharpoons 2C_2H_5OH(l) + 2CO_2(g) $$
02

Calculate the ΔG° for both reactions

To calculate the ΔG° for both reactions, we'll use the following equation: $$ ΔG° = ΔG°_{(products)} - ΔG°_{(reactants)} $$ For reaction 1: $$ ΔG°_1 = [6ΔG°(CO_2(g)) + 6ΔG°(H_2O(l))] - [ΔG°(C_6H_{12}O_6(s)) + 6ΔG°(O_2(g))] $$ For reaction 2: $$ ΔG°_2 = [2ΔG°(C_2H_5OH(l)) + 2ΔG°(CO_2(g))] - ΔG°(C_6H_{12}O_6(s)) $$ Using the values of ΔG° provided in Appendix C, calculate ΔG° for both reactions.
03

Calculate the equilibrium constants

Now, we can use the calculated ΔG° for both reactions and the previously mentioned equation to find the equilibrium constants for both reactions: For reaction 1: $$ K_1 = e^{-\dfrac{ΔG°_1}{RT}} $$ For reaction 2: $$ K_2 = e^{-\dfrac{ΔG°_2}{RT}} $$
04

Compare the equilibrium constants

Compare the calculated equilibrium constants (K_1 and K_2). The larger equilibrium constant corresponds to the more spontaneous reaction.
05

Calculate the maximum work

The maximum work (W_max) that can be obtained from a process under standard conditions can be calculated using the following equation: $$ W_{max}=-ΔG° $$ For reaction 1: $$ W_{max1}=-ΔG°_1 $$ For reaction 2: $$ W_{max2}=-ΔG°_2 $$
06

Compare the maximum work

Finally, compare the maximum works (W_max1 and W_max2) calculated for both reactions. The reaction with higher maximum work represents the process that can produce more work under standard conditions.

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Most popular questions from this chapter

Indicate whether each statement is true or false. (a) Unlike enthalpy, where we can only ever know changes in \(H,\) we can know absolute values of $S .(\mathbf{b})\( If you heat a gas such as \)\mathrm{CO}_{2}$, you will increase its degrees of translational, rotational and vibrational motions. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(g)\) have nearly the same molar mass. At a given temperature, they will have the same number of microstates.

Indicate whether \(\Delta G\) increases, decreases, or does not change when the partial pressure of \(\mathrm{H}_{2}\) is increased in each of the following reactions: (a) $\mathrm{H}_{2}(g)+\mathrm{NiO}(s) \longrightarrow \mathrm{Ni}(s)+\mathrm{H}_{2} \mathrm{O}(g)$ (b) $\mathrm{H}_{2}(g)+\mathrm{S}(s) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)$ (c) $\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$

For a certain chemical reaction, $\Delta H^{\circ}=-40.0 \mathrm{k} \mathrm{J}\( and \)\Delta S^{\circ}=-150.0 \mathrm{~J} / \mathrm{K}$. (a) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at $298 \mathrm{~K}$ under standard conditions?

For a particular reaction, \(\Delta H=30.0 \mathrm{~kJ}\) and $\Delta S=90.0 \mathrm{~J} / \mathrm{K}\(. Assume that \)\Delta H\( and \)\Delta S$ do not vary with temperature. (a) At what temperature will the reaction have \(\Delta G=0 ?\) (b) If \(\mathrm{T}\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

The value of \(K_{a}\) for nitrous acid \(\left(\mathrm{HNO}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{a}\). (b) By using the value of \(K_{a}\) calculate \(\Delta G^{\circ}\) for the dissociation of nitrous acid in aqueous solution. (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when $\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} \mathrm{M}\(, \)\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} \mathrm{M},\( and \)\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M} ?$
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