Problem 102 - Chemistry Matter and Change

Question

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.

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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. \( \Delta S_{\text{rxn}}^{\circ} = -368 \mathrm{J} / \mathrm{K} = -0.368 \mathrm{kJ} / \mathrm{K} \).

Step 3: 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: \[ \Delta G_{\text{rxn}}^{\circ} = -718.3 \mathrm{kJ} - (298 \mathrm{K})(-0.368 \mathrm{kJ} / \mathrm{K}) \].

Step 4: Perform the Calculation

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

Step 5: 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.

Key Concepts

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°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.