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Chapter 19: Problem 56

A certain reaction has \(\Delta H^{\circ}=+20.0 \mathrm{~kJ}\) and $\Delta S^{\circ}=\( \)+100.0 \mathrm{~J} / \mathrm{K} .$ (a) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at $298 \mathrm{~K} .(\mathbf{d})\( Is the reaction spontaneous at \)298 \mathrm{~K}$ under standard conditions?

Short Answer

Expert verified
(a) The reaction leads to an increase in the randomness or disorder of the system, as ΔS⁰ is positive. (b) The reaction leads to a decrease in the randomness or disorder of the surroundings, as ΔH⁰ is positive (endothermic reaction). (c) At 298 K, ΔG⁰ = -9,800 J. (d) The reaction is spontaneous at 298 K under standard conditions, as ΔG⁰ is negative.

Step by step solution

01

Analyzing the change in randomness or disorder of the system

We are given the entropy change ΔS⁰ as +100.0 J/K. Since the value is positive, it implies that the reaction leads to an increase in the randomness or disorder of the system.
02

Analyzing the change in randomness or disorder of the surroundings

We are given the enthalpy change ΔH⁰ as +20.0 kJ. Since the value is positive, it means that the reaction is endothermic. Therefore, the reaction absorbs heat from its surroundings, leading to a decrease in the randomness or disorder of the surroundings.
03

Calculating Gibbs free energy change (ΔG⁰) at 298 K

We can use the relation between Gibbs free energy change, enthalpy change, and entropy change (ΔG⁰ = ΔH⁰ - TΔS⁰) to calculate ΔG⁰ at 298 K. Given : ΔH⁰ = +20.0 kJ ΔS⁰ = +100.0 J/K T = 298 K First, we need to convert ΔH⁰ and ΔS⁰ to the same units. Here, we will convert ΔH⁰ to J by multiplying it by 1000: ΔH⁰ = 20.0 kJ × 1000 J/kJ = 20,000 J Now we can use the formula: ΔG⁰ = ΔH⁰ - TΔS⁰ ΔG⁰ = 20,000 J - (298 K × 100 J/K)
04

Calculate ΔG⁰

By substituting the values, we get: ΔG⁰ = 20,000 J - 29,800 J ΔG⁰ = -9,800 J
05

Determine if the reaction is spontaneous at 298 K

Since the Gibbs free energy change ΔG⁰ is negative (-9,800 J), the reaction is spontaneous at 298 K under standard conditions.

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Most popular questions from this chapter

The reaction $2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)$ is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for \(\Delta S^{\circ}\). Did your classmate make a mistake in the calculation? Explain.

Consider what happens when a sample of the explosive TNT is detonated under atmospheric pressure. (a) Is the detonation a reversible process? (b) What is the sign of \(q\) for this process? (c) Is w positive, negative, or zero for the process?

Indicate whether each of the following statements is trueor false. If it is false, correct it. (a) The feasibility of manufacturing \(\mathrm{NH}_{3}\) from \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) depends entirely on the value of $\Delta H\( for the process \)\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) .$ (b) The reaction of \(\mathrm{Na}(s)\) with \(\mathrm{Cl}_{2}(g)\) to form \(\mathrm{NaCl}(s)\) is a spontaneous process. (c) A spontaneous process can in principle be conducted reversibly. (d) Spontaneous processes in general require that work be done to force them to proceed. (e) Spontaneous processes are those that are exothermic and that lead to a higher degree of order in the system.

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$, is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix \(C\), calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S)\) ? (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7. (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

The normal boiling point of the element mercury (Hg) is $356.7{ }^{\circ} \mathrm{C},\( and its molar enthalpy of vaporization is \)\Delta H_{\text {vap }}=59.11 \mathrm{~kJ} / \mathrm{mol} .$ (a) When Hg boils at its nor- mal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(2.00 \mathrm{~mol}\) of \(\mathrm{Hg}\) is vaporized at \(356.7^{\circ} \mathrm{C}\).
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