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Chapter 16: Problem 55

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 \(-176.6 \mathrm{J} / \mathrm{K},\) respectively. What is the value of \(\Delta G^{\circ}\) at 298 K? Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

Short Answer

Expert verified
The value of ΔG° at 298 K is -5373.2 J. The temperature at which ΔG° = 0 is approximately 328.6 K. ΔG° is negative below this temperature (T < 328.6 K).

Step by step solution

01

Calculate ΔG° at 298 K

To calculate the change in Gibbs free energy (ΔG°) at 298 K, we can use the given values of ΔH° and ΔS° with the Gibbs-Helmholtz equation: ΔG° = ΔH° - TΔS° ΔG° = -58.03 kJ - (298 K)(-176.6 J/K) Note that the units for ΔH° are in kJ and for ΔS° are in J/K. We need to convert these units to the same base before calculation. Therefore, we will convert ΔH° into J by multiplying by 1000: ΔH° = -58.03 kJ × 1000 J/kJ = -58030 J Now, we can substitute the converted values into the equation: ΔG° = -58030 J - (298 K)(-176.6 J/K)
02

Solve for ΔG°

Now, we can solve the equation for ΔG°: ΔG° = -58030 J + (298 K)(176.6 J/K) ΔG° = -58030 J + 52656.8 J ΔG° = -5373.2 J So, the value of ΔG° at 298 K is -5373.2 J.
03

Find the temperature at which ΔG° = 0

To find the temperature at which ΔG° = 0, we'll set the Gibbs-Helmholtz equation equal to zero: 0 = ΔH° - TΔS° To solve for T, we can rearrange the equation: T = ΔH°/ΔS° Now we can substitute the values of ΔH° and ΔS°: T = (-58030 J)/(-176.6 J/K)
04

Solve for T

Now, we can solve for T: T = 328.6 K So, the temperature at which ΔG° = 0 is approximately 328.6 K.
05

Determine if ΔG° is negative above or below this temperature

To determine if ΔG° is negative above or below this temperature, we can look at the signs of ΔH° and ΔS°. If both have the same sign, ΔG° will be negative below the calculated temperature, while if they have opposite signs, ΔG° will be negative above the calculated temperature. In this case, both ΔH° and ΔS° are negative (-58030 J and -176.6 J/K, respectively). Therefore, ΔG° is negative below the temperature at which it is equal to 0: ΔG° < 0 when T < 328.6 K ΔG° becomes negative below 328.6 K.

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Most popular questions from this chapter

The value of \(\Delta G^{\circ}\) for the reaction $$2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(l)$$ is \(-5490 .\) kJ. Use this value and data from Appendix 4 to calculate the standard free energy of formation for \(\mathrm{C}_{4} \mathrm{H}_{10}(g).\)

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ} ?\) a. \(\mathrm{C}_{\text {graphite }}(s)\) or \(\mathrm{C}_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

a. Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{\mathrm{f}} / k_{\mathrm{r}},\) where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{\mathrm{a}} / R T}\right).\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction.”

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}_{\text {rhombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{SO}_{3}(g) \longrightarrow 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\) c. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

Some water is placed in a coffee-cup calorimeter. When \(1.0 \mathrm{g}\) of an ionic solid is added, the temperature of the solution increases from \(21.5^{\circ} \mathrm{C}\) to \(24.2^{\circ} \mathrm{C}\) as the solid dissolves. For the dissolving process, what are the signs for \(\Delta S_{\mathrm{sys}}, \Delta S_{\text {surr, and }}\) \(\Delta S_{\text {univ }} ?\)
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