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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掳C, we have the following signs: 鈭咷鈦� is negative, which indicates a spontaneous reaction. 鈭咹鈦� is negative, which indicates an exothermic reaction. 鈭哠掳 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: T鈧� = 109 掳C = 382.15 K and K鈧� = 2.54 脳 10鈦� T鈧� = 225 掳C = 498.15 K and K鈧� = 5.04 脳 10虏
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 (鈭咷掳), standard enthalpy change (鈭咹掳), and standard entropy change (鈭哠掳): \[\frac{d \ln K}{dT} = \frac{\Delta H^{\circ}}{RT^2}\] Since 鈭咹掳 and 鈭哠掳 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 鈭咹鈦�

Solve the van't Hoff equation for 鈭咹鈦� 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路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 鈭咹鈦� is negative, the reaction is exothermic.
05

Calculate 鈭咷鈦�

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

Calculate 鈭哠鈦�

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

Conclusion

In summary, for the given reaction at 25掳C, we have the following signs: 鈭咷鈦� is negative, which indicates a spontaneous reaction. 鈭咹鈦� is negative, which indicates an exothermic reaction. 鈭哠掳 is negative, which indicates a decrease in the system's entropy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs free energy
Gibbs free energy, often denoted as \(\Delta G^{\circ}\), is a vital concept in thermodynamics that determines the spontaneity of a reaction at constant pressure and temperature.
It is calculated using the equation: \[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \] where \( \Delta H^{\circ} \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S^{\circ} \) is the change in entropy.
- If \( \Delta G^{\circ} < 0 \), the reaction is spontaneous under the given conditions, meaning it can proceed without the input of additional energy.- If \( \Delta G^{\circ} > 0 \), the reaction is non-spontaneous, indicating that it would require energy to proceed.In the exercise, the calculated \( \Delta G^{\circ} \) value was negative, indicating that the reaction proceeds spontaneously at 25掳C. This means that under these conditions, the system has a natural tendency to proceed towards equilibrium.
enthalpy change
Enthalpy change, symbolized by \( \Delta H^{\circ} \), is a measure of the total heat content in a thermodynamic system.
It describes how much heat energy is absorbed or released during a chemical reaction at constant pressure.**Key Points about Enthalpy Change**:
  • **Exothermic Reactions**: If \( \Delta H^{\circ} < 0 \), the reaction releases heat to its surroundings, making it exothermic.
  • **Endothermic Reactions**: If \( \Delta H^{\circ} > 0 \), the reaction absorbs heat, making it endothermic.
In our provided exercise, the calculated \( \Delta H^{\circ} \) was negative, meaning the reaction is exothermic.
This indicates that energy is given off as heat when the reaction occurs, which is typical for reactions where bonds form, releasing energy.
entropy change
Entropy change, represented by \( \Delta S^{\circ} \), is a measure of the disorder or randomness in a thermodynamic system.
A system's entropy reflects how energy is distributed within it, often determining the feasibility of states or reactions.**Understanding Entropy Change**:
  • Positive \( \Delta S^{\circ} \): The system's disorder increases. In chemical reactions, this might happen when gases are produced from liquids or solids, or when a mixture is created from separate substances.
  • Negative \( \Delta S^{\circ} \): The system's disorder decreases. This frequently occurs when gases condense into liquids or solids, indicating energy focusing in order.
In the given exercise, the obtained \( \Delta S^{\circ} \) was negative, signifying a reduction in the system's entropy.
This usually implies a more ordered state due to the reaction process, reflecting a decrease in randomness as the products form.

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Most popular questions from this chapter

If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction $$\mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)$$ \(\Delta H^{\circ}\) for this reaction is 79.14 \(\mathrm{kJ} / \mathrm{mol}\) in the temperature range of 25 to \(125^{\circ} \mathrm{C}\) . Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C},\) calculate the partial pressure of \(\mathrm{CO}_{2}\) necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110 .^{\circ} \mathrm{C}\) (Hint: Manipulate the equation in Exercise 85.)

For the sublimation of iodine at \(25^{\circ} \mathrm{C}\) $$\mathrm{I}_{2}(s) \rightarrow \mathrm{I}_{2}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) are, respectively, 62 \(\mathrm{kJ}\) and 19 \(\mathrm{kJ}\) . Estimate the temperature at which iodine sublimes. Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Is \(\Delta S_{\text { surt }}\) favorable or unfavorable for exothermic reactions? Endothermic reactions? Explain.

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C},\) the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

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: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$
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