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A certain reaction has \(\Delta H^{\circ}=-19.5 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+42.7 \mathrm{~J} / \mathrm{K} .\) (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 \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Short Answer

Expert verified
(a) The reaction is exothermic. (b) The reaction leads to an increase in the randomness or disorder of the system. (c) ΔG° for the reaction at 298 K is -32.2246 kJ. (d) The reaction is spontaneous at 298 K under standard conditions.

Step by step solution

01

Determine if the reaction is exothermic or endothermic

A reaction is exothermic if the enthalpy change, ΔH°, is negative (meaning heat is released), and endothermic if it's positive (meaning heat is absorbed). In this case, ΔH° is -19.5 kJ, which is negative, so the reaction is exothermic. Answer (a): The reaction is exothermic.
02

Determine the change in randomness or disorder of the system

A reaction leads to an increase in randomness or disorder if the entropy change, ΔS°, is positive, and a decrease if it's negative. In this case, ΔS° is +42.7 J/K, which is positive, so the reaction leads to an increase in the randomness or disorder of the system. Answer (b): The reaction leads to an increase in the randomness or disorder of the system.
03

Convert the units of ΔS° to kJ/K

To calculate the Gibbs free energy change, ΔG°, we need ΔS° to be in the same unit system as ΔH° which is in kJ. We need to convert J/K to kJ/K. 1 kJ = 1000 J So, ΔS° = 42.7 J/K * (1 kJ / 1000 J) = 0.0427 kJ/K.
04

Calculate ΔG° using the Gibbs free energy equation

Now we will use the Gibbs free energy equation to calculate ΔG°: ΔG° = ΔH° - (T * ΔS°) where T is the temperature in K (given as 298 K). ΔG° = -19.5 kJ - (298 K * 0.0427 kJ/K) = -19.5 kJ - 12.7246 kJ = -32.2246 kJ Answer (c): ΔG° for the reaction at 298 K is -32.2246 kJ.
05

Determine if the reaction is spontaneous at 298 K

A reaction is spontaneous at a given temperature if the Gibbs free energy change, ΔG°, is negative. In this case, ΔG° is -32.2246 kJ, which is negative, so the reaction is spontaneous at 298 K under standard conditions. Answer (d): The reaction 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.

Exothermic Reaction
An exothermic reaction is a chemical reaction that releases energy into the surroundings, usually in the form of heat. This happens because the energy required to break the bonds in the reactants is less than the energy released when new bonds form in the products.

This release of energy causes the overall enthalpy change (\(\Delta H^{\circ}\)) to be negative. A negative \(\Delta H^{\circ}\) means that the reactants lose energy, making the surroundings warmer.
This is what we observed in the exercise where the \(\Delta H^{\circ}\) was \(-19.5 \, \text{kJ}\).

If you've ever felt the warmth of a burning fire or the heat from your hands warming as you rub them together, you've experienced an exothermic reaction. In these reactions, not only is energy released, but the process also often increases the temperature of the surroundings.
  • Heat Released: It gives off heat, making it easy to feel.
  • Bond Energy: The energy to break existing bonds is less than the energy released making new bonds.
  • Example: Combustion is a common example of an exothermic process.
Entropy Change
Entropy is a measure of randomness or disorder within a system. When analyzing chemical reactions, entropy change (\(\Delta S^{\circ}\)) tells us if the disorder in a system is increasing or decreasing.

In the context of the exercise, the given \(\Delta S^{\circ}\) was \(+42.7 \, \text{J/K}\), which indicates an increase in disorder. This positive value suggests that the products of the reaction have more possible configurations or arrangements compared to the reactants.

Think of it like a messy room: initially, everything is neatly arranged (low entropy), but as you start to move things around, placing books randomly and leaving clothes on the floor, the disorder increases (high entropy).

In a chemical reaction, a positive \(\Delta S^{\circ}\) often means:
  • States of Matter: Mixing substances or a gas forming increases disorder.
  • Phase Change: Transition from solid to liquid or liquid to gas increases entropy.
  • Molecular Complexity: Larger molecules breaking into smaller ones causes more randomness.
Gibbs Free Energy
Gibbs free energy (\(\Delta G^{\circ}\)) links enthalpy (\(\Delta H^{\circ}\)), entropy (\(\Delta S^{\circ}\)), and temperature to predict if a reaction is spontaneous. The equation \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\) helps calculate this.

A spontaneous reaction at a given temperature is one that can proceed without needing continual energy input.
In the solution, introduced at 298 \, \text{K}, the reaction was confirmed to be spontaneous because \(\Delta G^{\circ} = -32.2246 \, \text{kJ}\). This negative value indicates it can occur naturally under these conditions.

Gibbs free energy helps us understand:
  • Spontaneity: Negative \(\Delta G^{\circ}\) means a reaction is spontaneous and can proceed on its own.
  • Temperature Influence: As temperatures (T) rise, the effect of \(\Delta S^{\circ}\) on \(\Delta G^{\circ}\) becomes more significant.
  • Feasibility: It predicts if a reaction pathway is feasible without inferring the speed of the reaction.

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Most popular questions from this chapter

Using data from Appendix \(C\), calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous under standard conditions. (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g)\) (c) \(6 \mathrm{Cl}_{2}(g)+2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{FeCl}_{3}(s)+3 \mathrm{O}_{2}(g)\) (d) \(\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

The oxidation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in body tissue produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2} .\) (a) Using data given in Appendix \(\mathrm{C}\), compare the equilibrium constants for the following reactions: $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) & \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) & \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

For the majority of the compounds listed in Appendix C, the value of \(\Delta G_{f}^{\circ}\) is more positive (or less negative) than the value of \(\Delta H_{f}^{\circ}\). (a) Explain this observation, using \(\mathrm{NH}_{3}(g), \mathrm{CCl}_{4}(l)\), and \(\mathrm{KNO}_{3}(s)\) as examples. (b) \(\mathrm{An}\) exception to this observation is \(\mathrm{CO}(g)\). Explain the trend in the \(\Delta H_{f}^{\circ}\) and \(\Delta G_{f}^{\circ}\) values for this molecule.

Use data from Appendix \(\mathrm{C}\) to calculate the equilibrium constant, \(K\), at \(298 \mathrm{~K}\) for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)\)

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) \(\mathrm{A}\) diatomic molecule dissociates into atoms. (e) A piece of charcoal is combusted to form \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

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