Problem 102 Is the following reaction to con... [FREE SOLUTION]

Chapter 15: Problem 102

Is the following reaction to convert copper(II) sulfide to copper(II) sulfate spontaneous under standard conditions $$ \operatorname{Cus}(\mathrm{s})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CuSO}_{4}(\mathrm{s}) . \Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=-718.3 $$ $$ \mathrm{k} \mathrm{J}, \text { and } \Delta S_{\mathrm{rxn}}^{\circ}=-368 \mathrm{J} / \mathrm{K} $$. Explain.

Short Answer

Expert verified

The reaction is spontaneous under standard conditions due to a negative Gibbs free energy change.

Step by step solution

Understanding Reaction Spontaneity

To determine if a reaction is spontaneous under standard conditions, we need to evaluate the Gibbs free energy change, denoted as $ \Delta G_{\text{rxn}}^{\circ} $. A reaction is spontaneous if $ \Delta G_{\text{rxn}}^{\circ} < 0 $. The formula for $ \Delta G $ is: $ \Delta G_{\text{rxn}}^{\circ} = \Delta H_{\text{rxn}}^{\circ} - T\Delta S_{\text{rxn}}^{\circ} $, where $ T $ is the temperature in Kelvin, typically 298 K for standard conditions.

Convert Units of Entropy

Convert the entropy change from $ \mathrm{J} / \mathrm{K} $ to $ \mathrm{kJ} / \mathrm{K} $ to match the units of the enthalpy change. $ \Delta S_{\text{rxn}}^{\circ} = -368 \mathrm{J} / \mathrm{K} = -0.368 \mathrm{kJ} / \mathrm{K} $.

Calculate Gibbs Free Energy Change

Using the formula $ \Delta G_{\text{rxn}}^{\circ} = \Delta H_{\text{rxn}}^{\circ} - T\Delta S_{\text{rxn}}^{\circ} $, substitute $ \Delta H_{\text{rxn}}^{\circ} = -718.3 \mathrm{kJ} $, $ T = 298 \mathrm{K} $, and $ \Delta S_{\text{rxn}}^{\circ} = -0.368 \mathrm{kJ} / \mathrm{K} $. Calculate: \[ \Delta G_{\text{rxn}}^{\circ} = -718.3 \mathrm{kJ} - (298 \mathrm{K})(-0.368 \mathrm{kJ} / \mathrm{K}) \].

Perform the Calculation

Follow the calculation: \[ \Delta G_{\text{rxn}}^{\circ} = -718.3 \mathrm{kJ} + 109.664 \mathrm{kJ} = -608.636 \mathrm{kJ} \].

Analyze the Result

The calculated $ \Delta G_{\text{rxn}}^{\circ} = -608.636 \mathrm{kJ} $ is negative, which indicates that the reaction is spontaneous under standard conditions.

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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, denoted as $ \Delta G $, plays a crucial role in determining the spontaneity of a chemical reaction. To decide if a reaction will proceed without outside intervention, we look at $ \Delta G $. If $ \Delta G < 0 $, the reaction is spontaneous. This means it can occur naturally without any added energy input. To compute $ \Delta G $, we use the formula:

$ \Delta G_{\text{rxn}}^{\circ} = \Delta H_{\text{rxn}}^{\circ} - T \Delta S_{\text{rxn}}^{\circ} $

Where:

$ \Delta H $ is the enthalpy change of the reaction
$ T $ is the temperature in Kelvin, often 298 K under standard conditions
$ \Delta S $ is the entropy change of the reaction

Breaking down these components helps us understand how energy dispersal impacts reaction spontaneity.

Enthalpy

Enthalpy, represented by $ \Delta H $, refers to the heat content of a system. It helps us understand how heat is absorbed or released during a chemical reaction. When examining reactions like the conversion of copper(II) sulfide to copper(II) sulfate, $ \Delta H_{\text{rxn}}^{\circ} $ indicates whether the reaction releases or absorbs heat.

If $ \Delta H < 0 $, the reaction is exothermic, meaning it releases heat to the surroundings. This usually favors spontaneity because the system is losing energy.
If $ \Delta H > 0 $, the reaction is endothermic, implying it absorbs heat. These reactions are often non-spontaneous unless counteracted by other factors like increased entropy.

In our exercise, $ \Delta H_{\text{rxn}}^{\circ} = -718.3 \text{kJ} $, marking the reaction as strongly exothermic.

Entropy Change

Entropy, denoted as $ \Delta S $, measures the disorder or randomness of a system. In chemistry, entropy change provides insight into the degree of chaos or energy dispersal in a reaction.

$ \Delta S > 0 $ implies an increase in randomness, generally favoring spontaneity as the universe tends towards more disorder.
Conversely, $ \Delta S < 0 $ suggests a decrease in randomness, which may oppose spontaneity.

For our reaction, $ \Delta S_{\text{rxn}}^{\circ} = -368 \text{J/K} $ (or $-0.368 \text{kJ/K}$), indicates a reduction in entropy. Despite this, the significant negative enthalpy still results in a negative $ \Delta G $, making the reaction spontaneous under standard conditions.

Standard Conditions

In chemical thermodynamics, the term 'standard conditions' offers a baseline for understanding reactions consistently. Standard conditions imply:

Pressure at 1 atmosphere
Temperature at 298 K (25掳C)
Concentrations of any solutions at 1 M

These conditions allow chemists to compare different reactions' behaviors under a uniform set of parameters. When analyzing reaction spontaneity, using standard conditions helps simplify calculations and predictions. For example, our given $ \Delta H_{\text{rxn}}^{\circ} $ and $ \Delta S_{\text{rxn}}^{\circ} $ are measured under these standards, ensuring the computed $ \Delta G_{\text{rxn}}^{\circ} $ accurately reflects spontaneous nature without external influences like changing temperature or pressure.

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Short Answer

Step by step solution

Understanding Reaction Spontaneity

Convert Units of Entropy

Calculate Gibbs Free Energy Change

Perform the Calculation

Analyze the Result

Key Concepts

Gibbs Free Energy

Enthalpy

Entropy Change

Standard Conditions

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