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Chapter 17: Problem 88

Consider the following equilibrium constant versus temperature data for some reaction: $$\begin{array}{ll} {\boldsymbol{T}\left(^{\circ} \mathbf{C}\right)} & \quad {\text { K }} \\ \hline {109} & {2.54 \times 10^{4}} \\ {225} & {5.04 \times 10^{2}} \\ {303} & {6.33 \times 10^{1}} \\ {412} & {2.25 \times 10^{-1}} \\\ {539} & {3.03 \times 10^{-3}}\end{array}$$ Predict the signs for \(\Delta G^{\circ}, \Delta H^{\circ},\) and $\Delta S^{\circ}$ for this reaction at \(25^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Short Answer

Expert verified
In summary, for the given reaction at 25°C, we have the following signs: ∆G⁰ is negative, which indicates a spontaneous reaction. ∆H⁰ is negative, which indicates an exothermic reaction. ∆S° is negative, which indicates a decrease in the system's entropy.

Step by step solution

01

Convert the temperatures from Celsius to Kelvin

Since the van't Hoff equation uses temperature in Kelvin (K), we need to convert the given temperatures from Celsius to Kelvin. To do so, add 273.15 to each temperature in the table: \[T(K) = T(^\circ C) + 273.15\]
02

Choose two temperature-equilibrium constant pairs from the table

We will pick two pairs of data points (T and K) from the table to apply the van't Hoff equation. For the sake of this solution, let's use the first and second data points: T₁ = 109 °C = 382.15 K and K₁ = 2.54 × 10⁴ T₂ = 225 °C = 498.15 K and K₂ = 5.04 × 10²
03

Apply the van't Hoff equation

The van't Hoff equation relates the temperature and equilibrium constant to the standard Gibbs free energy change (∆G°), standard enthalpy change (∆H°), and standard entropy change (∆S°): \[\frac{d \ln K}{dT} = \frac{\Delta H^{\circ}}{RT^2}\] Since ∆H° and ∆S° are assumed to be constant with respect to temperature, we can express the van't Hoff equation for our two data points: \[\frac{\ln K_2 - \ln K_1}{T_2 - T_1} = \frac{\Delta H^{\circ}}{R} \left(\frac{1}{T_2}-\frac{1}{T_1}\right)\]
04

Calculate ∆H⁰

Solve the van't Hoff equation for ∆H⁰ by substituting the selected data points into the equation: \[\Delta H^{\circ} = R \frac{\ln K_2 - \ln K_1}{\frac{1}{T_2}-\frac{1}{T_1}}\] Where R is the gas constant, R = 8.314 J/(mol·K). \[\Delta H^{\circ} = 8.314 \frac{\ln(5.04 \times 10^{2}) - \ln(2.54 \times 10^{4})}{\frac{1}{498.15}-\frac{1}{382.15}}\] After calculation, we get: \[\Delta H^{\circ} = -32,230.53 \, \text{J/mol}\] Since ∆H⁰ is negative, the reaction is exothermic.
05

Calculate ∆G⁰

At 25°C (298.15 K), we will use the data point with the closest temperature to calculate ∆G⁰: T = 303 °C = 576.15 K and K = 6.33 × 10¹ Use the Gibbs free energy equation to calculate ∆G°: \[\Delta G^{\circ} = -RT \ln K\] \[\Delta G^{\circ} = -8.314 \times 298.15 \times \ln(6.33 \times 10^{1})\] After calculation, we get: \[\Delta G^{\circ} = -8,344.96 \, \text{J/mol}\] Since ∆G⁰ is negative, the reaction is spontaneous at 25°C.
06

Calculate ∆S⁰

Use the relationship between ∆G°, ∆H°, and ∆S°: \[\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}\] Rearrange to solve for ∆S°: \[\Delta S^{\circ} = \frac{\Delta H^{\circ}-\Delta G^{\circ}}{T}\] \[\Delta S^{\circ} = \frac{-32,230.53 \, \text{J/mol} - (-8,344.96 \, \text{J/mol})}{298.15 \, \text{K}}\] After calculation, we get: \[\Delta S^{\circ} = -80.30 \, \text{J/(mol·K)}\] Since ∆S⁰ is negative, the reaction leads to a decrease in the system's entropy.
07

Conclusion

In summary, for the given reaction at 25°C, we have the following signs: ∆G⁰ is negative, which indicates a spontaneous reaction. ∆H⁰ is negative, which indicates an exothermic reaction. ∆S° is negative, which indicates a decrease in the system's entropy.

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Most popular questions from this chapter

Consider the following reaction at \(25.0^{\circ} \mathrm{C} :\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03\) kJ/mol and \(-176.6 \mathrm{J} / \mathrm{K}\) . mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\) . Assuming \(\Delta H^{\circ}\) and $\Delta S^{\circ}$ are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\) .

What information can be determined from \(\Delta G\) for a reaction? Does one get the same information from \(\Delta G^{\circ},\) the standard free energy change? \(\Delta G^{\circ}\) allows determination of the equilibrium constant \(K\) for a reaction. How? How can one estimate the value of \(K\) at temperatures other than \(25^{\circ} \mathrm{C}\) for a reaction? How can one estimate the temperature where \(K=1\) for a reaction? Do all reactions have a specific temperature where \(K=1 ?\)

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a 100\(\%\) yield of pure $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}$ at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

Predict the sign of \(\Delta S\) for each of the following and explain. a. the evaporation of alcohol b. the freezing of water c. compressing an ideal gas at constant temperature d. dissolving NaCl in water

For a liquid, which would you expect to be larger, $\Delta S_{\text { fusion }}\( or \)\Delta S_{\text { evaporation }} ?$ Why?
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