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Chapter 17: Problem 60

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

Short Answer

Expert verified
The standard Gibbs free energy change (ΔG°) for the given reaction at 298 K is -25.45 kJ. The temperature at which ΔG° becomes zero is 328.3 K. ΔG° is negative below this temperature.

Step by step solution

01

Recall the Gibbs-Helmholtz equation

The Gibbs-Helmholtz equation calculates the relationship between the standard Gibbs free energy change (ΔG°), the standard enthalpy change (ΔH°), and the standard entropy change (ΔS°) for a given reaction: \[ΔG° = ΔH° - TΔS°\] where ΔG° is the standard Gibbs free energy change, ΔH° is the standard enthalpy change, T is the temperature in kelvin (K), and ΔS° is the standard entropy change.
02

Calculate ΔG° at 298 K

Using the Gibbs-Helmholtz equation, we can find ΔG° at 298 K given the values of ΔH° and ΔS°: ΔH° = -58.03 kJ = -58030 J (convert kJ to J) ΔS° = -176.6 J/K T = 298 K \[ΔG° = ΔH° - TΔS° = -58030 \,\text{J} - (298 \,\text{K} \times -176.6 \, \frac{\text{J}}{\text{K}})\]
03

Solve for ΔG° at 298 K

Calculate the value of ΔG° at 298 K by solving the equation: ΔG° = -58030 J + (298 K × 176.6 J/K) = -25,445 J = -25.45 kJ (convert J to kJ) So at 298 K, the standard Gibbs free energy change, ΔG°, is -25.45 kJ.
04

Determine the temperature at which ΔG° = 0

To find the temperature at which the standard Gibbs free energy change is zero, we need to set ΔG° = 0 in the Gibbs-Helmholtz equation and solve for T: \[0 = ΔH° - TΔS°\] Rearrange the equation to solve for T: \[T = \frac{ΔH°}{ΔS°} = \frac{-58030 \, \text{J}}{-176.6 \, \frac{\text{J}}{\text{K}}}\]
05

Calculate the temperature at which ΔG° = 0

Calculate the temperature at which ΔG° is 0: \(T = \frac{-58030 \, \text{J}}{-176.6 \, \frac{\text{J}}{\text{K}}} = 328.3 \, \text{K}\) So at 328.3 K, ΔG° becomes zero.
06

Determine whether ΔG° is negative above or below this temperature

Observe the sign of ΔS° to determine where ΔG° is negative: Since ΔS° is negative, ΔG° will be negative when the temperature is below 328.3 K. This is because a lower temperature will make the term TΔS° positive, resulting in a negative ΔG° due to the subtraction in the Gibbs-Helmholtz equation. In summary, at 298 K, the value of ΔG° for the given reaction is -25.45 kJ. The temperature at which ΔG° is zero is 328.3 K, and ΔG° is negative below this temperature.

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Most popular questions from this chapter

The melting point for carbon diselenide \(\left(\mathrm{CSe}_{2}\right)\) is \(-46^{\circ} \mathrm{C}\) . At a temperature of \(-75^{\circ} \mathrm{C},\) predict the signs for \(\Delta S_{\mathrm{surr}}\) and $\Delta S_{\mathrm{univ}}\( for the following process: \)\operatorname{CSe}_{2}(l) \rightarrow \operatorname{CSe}_{2}(s)$

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

For each of the following pairs, which substance has the greater value of S? a. \(\mathrm{N}_{2} \mathrm{O}(\text { at } 0 \mathrm{K})\) or He (at 10 \(\mathrm{K} )\) b. \(\mathrm{N}_{2} \mathrm{O}(g)\) (at $1 \mathrm{atm}, 25^{\circ} \mathrm{C} )\( or He(g) (at 1 atm, \)25^{\circ} \mathrm{C} )$ c. \(\mathrm{NH}_{3}(s)\) (at 196 $\mathrm{K} ) \longrightarrow \mathrm{NH}_{3}(l)(\text { at } 196 \mathrm{K})$

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

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?
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