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Chapter 19: Problem 57

For a certain chemical reaction, \(\Delta H^{\circ}=-35.4 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-85.5 \mathrm{~J} / \mathrm{K} .\) (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (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 is exothermic, as ΔH° is negative (-35.4 kJ). (b) The reaction leads to a decrease in the randomness or disorder of the system, as ΔS° is negative (-85.5 J/K). (c) The Gibbs free energy change (ΔG°) at 298 K is -9,941 J. (d) The reaction is spontaneous at 298 K under standard conditions, as ΔG° is negative.

Step by step solution

01

(a) Determine reaction type (Exothermic or Endothermic) based on ΔH°

In a reaction, if the enthalpy change (ΔH°) is negative, it indicates that the reaction releases heat and is therefore an exothermic reaction. If ΔH° is positive, the reaction absorbs heat and is endothermic. In this case, the value of the enthalpy change is ΔH° = -35.4 kJ (negative), so the reaction is exothermic.
02

(b) Determine change in randomness or disorder based on ΔS°

In a reaction, if the entropy change (ΔS°) is positive, it denotes that the reaction leads to an increase in randomness or disorder. If ΔS° is negative, the reaction leads to a decrease in randomness or disorder. In this case, the value of the entropy change is ΔS° = -85.5 J/K (negative), so the reaction leads to a decrease in the randomness or disorder of the system.
03

(c) Calculate ΔG° at 298 K

To calculate Gibbs free energy change (ΔG°), we need to use the following equation: ΔG° = ΔH° - TΔS° Given that ΔH° = -35.4 kJ and ΔS° = -85.5 J/K, and the temperature is 298 K, we can calculate the Gibbs free energy change as follows: First, convert ΔH° and ΔS° to the same unit. Let's use J: ΔH° = -35.4 kJ * 1000 J/kJ = -35,400 J ΔS° = -85.5 J/K Now, we can plug in the values into the equation: ΔG° = (-35,400 J) - (298 K * -85.5 J/K) = -35,400 J + 25,459 J = -9,941 J
04

(d) Determine reaction spontaneity based on ΔG°

The spontaneity of a reaction at a given temperature under standard conditions can be determined by examining the sign of ΔG°: - If ΔG° < 0, the reaction is spontaneous. - If ΔG° > 0, the reaction is non-spontaneous. - If ΔG° = 0, the reaction is at equilibrium. From part (c), we calculated ΔG° = -9,941 J, which is negative. Therefore, the reaction is spontaneous at 298 K under standard conditions.

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Most popular questions from this chapter

(a) If you are told that the entropy of a certain system is zero, what do you know about the system and the temperature? (b) The energy of a gas is increased by heating it. Using \(\mathrm{CO}_{2}\) as an example, illustrate the different ways in which additional energy can be distributed among the molecules of the gas. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(g)\) have nearly the same molar mass. At a given temperature, will they have the same number of microstates? Explain.

The normal freezing point of \(n\) -octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is \(-57{ }^{\circ} \mathrm{C}\). (a) Is the freezing of \(n\) -octane an endothermic or exothermic process? (b) In what temperature range is the freezing of \(n\) -octane a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid \(n\) -octane and solid \(n\) -octane are in equilibrium? Explain.

Indicate whether \(\Delta G\) increases, decreases, or does not change when the partial pressure of \(\mathrm{H}_{2}\) is increased in each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(2 \mathrm{HBr}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\)

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the controlled oxidation of methane: $$ \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. (b) How is \(\Delta G^{\circ}\) for the reaction expected to vary with increasing temperature? (c) Calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). Under standard conditions, is the reaction spontaneous at this temperature? (d) Is there a temperature at which the reaction would be at equilibrium under standard conditions and that is low enough so that the compounds involved are likely to be stable?

Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g),\) is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). (b) How much heat is produced in burning \(1 \mathrm{~mol}\) of \(\mathrm{C}_{2} \mathrm{H}_{2}\) under standard conditions if both reactants and products are brought to \(298 \mathrm{~K}\) ? (c) What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction?
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