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Using data from Appendix \(\mathrm{C}\), calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions. (a) \(2 \mathrm{Zn}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)\) (b) \(2 \mathrm{NaBr}(s) \longrightarrow 2 \mathrm{Na}(g)+\mathrm{Br}_{2}(g)\) (c) \(\mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)+\mathrm{H}_{2}(g)\)

Short Answer

Expert verified
For the given reactions, we calculated the Gibbs free energy changes (ΔG°) as follows: (a) ΔG° = -665.9 kJ/mol, reaction is spontaneous. (b) ΔG° = 483.4 kJ/mol, reaction is not spontaneous. (c) ΔG° = 36.12 kJ/mol, reaction is not spontaneous. Only reaction (a) is spontaneous at 298 K under standard conditions.

Step by step solution

01

Calculate ΔH° for each reaction

Calculate the enthalpy change (ΔH°) for each reaction by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products. Use the values from Appendix C. (a) ΔH° = [2 × (-348.0)] - [2 × 0 + 0] = -696.0 kJ/mol (b) ΔH° = [2 × 0 + 0] - [2 × (-362.1)] = 724.2 kJ/mol (c) ΔH° = [(-234.8) + 0] - [(-200.7) + (-50.45)] = 16.35 kJ/mol
02

Calculate ΔS° for each reaction

Calculate the entropy change (ΔS°) for each reaction by subtracting the sum of the standard entropies of the reactants from the sum of the standard entropies of the products. Use the values from Appendix C. (a) ΔS° = [2 × (43.6)] - [2 × (41.6) + 205.1] = 187.2 - 288.3 = -101.1 J/mol·K (b) ΔS° = [2 × (154.3) + 245.5] - [2 × (102.8)] = 808.1 J/mol·K (c) ΔS° = [(65.63) + 130.7] - [(188.8) + (74.87)] = -66.34 J/mol·K
03

Calculate ΔG° for each reaction

Calculate the Gibbs free energy change (ΔG°) for each reaction using the relationship: ΔG° = ΔH° - TΔS°. We are given the temperature T = 298 K. (a) ΔG° = -696.0 kJ/mol - 298 K × (-101.1 J/mol·K) / 1000 = -696.0 + 30.11 = -665.9 kJ/mol (b) ΔG° = 724.2 kJ/mol - 298 K × 808.1 J/mol·K / 1000 = 724.2 - 240.8 = 483.4 kJ/mol (c) ΔG° = 16.35 kJ/mol - 298 K × (-66.34 J/mol·K) / 1000 = 16.35 + 19.77 = 36.12 kJ/mol
04

Determine the spontaneity of the reactions

A reaction is spontaneous under the given conditions if ΔG° < 0. (a) ΔG° = -665.9 kJ/mol < 0, so the reaction is spontaneous. (b) ΔG° = 483.4 kJ/mol > 0, so the reaction is not spontaneous. (c) ΔG° = 36.12 kJ/mol > 0, so the reaction is not spontaneous. So, only reaction (a) is spontaneous at 298 K under standard conditions.

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Key Concepts

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

Enthalpy Change
Enthalpy change ( \( \Delta H^{\circ} \)) is a critical concept in understanding chemical reactions and their energy requirements. It represents the heat absorbed or released during a reaction under constant pressure. To find \( \Delta H^{\circ} \) for a reaction, one must subtract the total enthalpy of the reactants from that of the products. Here's how it works for our example reactions:
- For reaction (a), the formation of \( \text{ZnO} \) releases \(-696.0\) kJ/mol, indicating an exothermic process.- Reaction (b) absorbs \(724.2\) kJ/mol as \( \text{NaBr} \) breaks into gases, making it endothermic.- Lastly, reaction (c) has a small positive enthalpy change of \(16.35\) kJ/mol, suggesting slightly more energy is used than released.
The sign of \( \Delta H^{\circ} \)—negative for exothermic reactions and positive for endothermic ones—helps indicate whether the reaction gives off heat or requires added energy to proceed.
Entropy Change
Entropy change, \( \Delta S^{\circ} \), is a measure of disorder or randomness at the molecular level. In essence, it tells us whether a reaction results in more or less ordered products compared to the reactants. To calculate \( \Delta S^{\circ} \) for any given reaction, subtract the sum of the entropy values of the reactants from the sum of the entropy values of the products. Let's briefly see how this applies to our reactions:
- In reaction (a), \( \Delta S^{\circ} \) is \(-101.1\) J/mol·K, showing a significant decrease in randomness as \( \text{ZnO} \) solidifies from gases.- Reaction (b) drastically increases disorder, as seen by \( \Delta S^{\circ} = 808.1\) J/mol·K, from solid to gaseous stages.- Reaction (c) results in a decrease of \( \Delta S^{\circ} = -66.34\) J/mol·K since it goes from two gaseous reactants to a more ordered state.Consequently, a reaction increase in entropy (positive \( \Delta S^{\circ} \)) is generally favorable as it aligns with the natural tendency of systems to move towards disorder.
Reaction Spontaneity
Reaction spontaneity is determined by Gibbs Free Energy ( \( \Delta G^{\circ} \)), a central concept that combines enthalpy and entropy changes predicting if a reaction is self-proceeding. \( \Delta G^{\circ} \) is calculated with the formula: \[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \]where- \( \Delta H^{\circ} \) is the enthalpy change,- \( \Delta S^{\circ} \) is the entropy change, - and \( T\) is the temperature in Kelvin.A negative \( \Delta G^{\circ} \) indicates a spontaneous reaction ( \( \Delta G^{\circ} < 0 \)), while a positive value means non-spontaneity.
For example:- In reaction (a), the calculated \( \Delta G^{\circ} = -665.9\) kJ/mol suggests it spontaneously proceeds under standard conditions.- Conversely, reactions (b) and (c) are non-spontaneous, with \( \Delta G^{\circ} = 483.4\) and \( \Delta G^{\circ} = 36.12\) kJ/mol, respectively.
Understanding reaction spontaneity helps chemists predict reaction feasibility without external energy.

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Most popular questions from this chapter

In chemical kinetics, the entropy of activation is the entropy change for the process in which the reactants reach the activated complex. Predict whether the entropy of activation for a bimolecular process is usually positive or negative.

Consider a system that consists of two standard playing dice, with the state of the system defined by the sum of the values shown on the top faces. (a) The two arrangements of top faces shown here can be viewed as two possible microstates of the system. Explain. (b) To which state does each microstate correspond? (c) How many possible states are there for the system? (d) Which state or states have the highest entropy? Explain. (e) Which state or states have the lowest entropy? Explain. (f) Calculate the absolute entropy of the two-dice system.

Using data from Appendix \(\mathrm{C}\), write the equilibrium-constant expression and calculate the value of the equilibrium constant and the free- energy change for these reactions at \(298 \mathrm{~K}:\) (a) \(\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\)

Most liquids follow Trouton's rule (see Exercise 19.93 ), which states that the molar entropy of vaporization is approximately \(88 \pm 5 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\). The normal boiling points and enthalpies of vaporization of several organic liquids are as follows: \begin{tabular}{lcc} \hline & Normal Boiling & \\ Substance & Point \(\left({ }^{\circ} \mathrm{C}\right)\) & \(\Delta H_{\text {vap }}(\mathrm{k} / / \mathrm{mol})\) \\ \hline Acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\) & 56.1 & 29.1 \\\ Dimethyl ether, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}\) & -24.8 & 21.5 \\\ Ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) & 78.4 & 38.6 \\ Octane, \(\mathrm{C}_{\mathrm{s}} \mathrm{H}_{18}\) & 125.6 & 34.4 \\ Pyridine, \(\mathrm{C}_{5} \mathrm{H}_{\mathrm{S}} \mathrm{N}\) & 115.3 & 35.1 \\\ \hline \end{tabular} (a) Calculate \(\Delta S_{\text {vap }}\) for each of the liquids. Do all the liquids obey Trouton's rule? (b) With reference to intermolecular forces (Section 11.2), can you explain any exceptions to the rule? (c) Would you expect water to obey Trouton's rule? By using data in Appendix \(\mathrm{B}\), check the accuracy of your conclusion. (d) Chlorobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\right)\) boils at \(131.8^{\circ} \mathrm{C}\). Use Trouton's rule to estimate \(\Delta H_{\text {vap }}\) for this substance.

For each of the following processes, indicate whether the signs of \(\Delta S\) and \(\Delta H\) are expected to be positive, negative, or about zero. (a) A solid sublimes. (b) The temperature of a sample of \(\mathrm{Co}(s)\) is lowered from \(60^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C} .\) (c) Ethyl alcohol evaporates from a beaker. (d) A diatomic molecule dissociates into atoms. (e) A piece of charcoal is combusted to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\).

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