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Chapter 17: Problem 108

Consider the reaction: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ for which \(\Delta H\) is \(-233 \mathrm{kJ}\) and \(\Delta S\) is $-424 \mathrm{J} / \mathrm{K}$ . a. Calculate the free energy change for the reaction \((\Delta G)\) at 393 \(\mathrm{K} .\) b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

Short Answer

Expert verified
The free energy change for the reaction at 393 K is ΔG = -66,368 J. The reaction is spontaneous (ΔG < 0) at temperatures above 549.06 K.

Step by step solution

01

Recall the Gibbs free energy equation

We will use the Gibbs free energy equation to calculate the free energy change for the reaction and to find the conditions for spontaneity. The equation is given by: ΔG = ΔH - TΔS where ΔG is the change in free energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.
02

Calculate the free energy change at 393 K

We are given ΔH = -233 kJ and ΔS = -424 J/K. The temperature for part a is 393 K. We first need to convert ΔH to J since ΔS is in J/K. ΔH = -233 kJ × (1000 J/1 kJ) = -233,000 J Now we can plug in the values into the Gibbs free energy equation: ΔG = ΔH - TΔS = -233,000 J - (393 K × -424 J/K) = -233,000 J + 166,632 J = -66,368 J So, the free energy change for the reaction at 393 K is ΔG = -66,368 J.
03

Determine the conditions for spontaneity

To determine the temperatures at which this reaction is spontaneous, we need to analyze the criteria for spontaneity. A reaction is spontaneous when ΔG < 0. Therefore, we have: ΔG = ΔH - TΔS < 0 Now, we need to find the temperature boundaries that make this inequality true. We can rearrange the inequality as follows: T > ΔH / ΔS Note that the signs of both ΔH and ΔS are negative. So, we have: T > (-233,000 J) / (-424 J/K) = 549.06 K
04

Conclusion

The free energy change for the reaction at 393 K is ΔG = -66,368 J. The reaction is spontaneous (ΔG < 0) at temperatures above 549.06 K.

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Most popular questions from this chapter

For the reaction $$\mathrm{CS}_{2}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{SO}_{2}(g)$$ \(\Delta S^{\circ}\) is equal to \(-143 \mathrm{JK}\) . Use this value and data from Appendix 4 to calculate the value of \(S^{\circ}\) for $\mathrm{CS}_{2}(g) .$

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2} .\) Consider the following reactions and approximate standard free energy changes: $$\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} \quad \Delta G^{\circ}=-70 \mathrm{kJ}$$ $$\mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} \quad \Delta G^{\circ}=-80 \mathrm{kJ} $$ Using these data, estimate the equilibrium constant value at $25^{\circ} \mathrm{C}$ for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

Ethanethiol $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ; \text { also called ethyl mercaptan) is }\right.$ commonly added to natural gas to provide the “rotten egg” smell of a gas leak. The boiling point of ethanethiol is \(35^{\circ} \mathrm{C}\) and its heat of vaporization is 27.5 $\mathrm{kJ} / \mathrm{mol}$ . What is the entropy of vaporization for this substance?

The equilibrium constant for a certain reaction decreases from 8.84 to $3.25 \times 10^{-2}\( when the temperature increases from \)25^{\circ} \mathrm{C}$ to \(75^{\circ} \mathrm{C}\) . Estimate the temperature where \(K=1.00\) for this reaction. Estimate the value of \(\Delta S^{\circ}\) for this reaction. (Hint: Manipulate the equation in Exercise 85.)

Which of the following involve an increase in the entropy of the system? a. melting of a solid b. sublimation c. freezing d. mixing e. separation f. boiling
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