Question

Given that \(: \Delta \mathrm{G}_{1}^{\circ}(\mathrm{CuO})=-30.4 \mathrm{kcal} / \mathrm{mole}\) \(\Delta \mathrm{G}_{\mathrm{t}}^{\circ}\left(\mathrm{Cu}_{2} \mathrm{O}\right)=-34.98 \mathrm{kca} / \mathrm{mole} \quad \mathrm{T}=298 \mathrm{~K}\) Now on the basis of above data which of the following predictions will be most appropriate under the standard conditions and reversible reaction. (1) Finely divided form of CuO kept in excess \(\mathrm{O}_{2}\) would be completely converted to \(\mathrm{Cu}_{2} \mathrm{O}\) (2) Finely divided form of \(\mathrm{Cu}_{2} \mathrm{O}\) kept in excess \(\mathrm{O}_{2}\) would be completely converted to \(\mathrm{CuO}\) (3) Finely divided form of CuO kept in excess \(\mathrm{O}_{2}\) would be converted to a mixture of \(\mathrm{CuO}\) and \(\mathrm{Cu}_{2} \mathrm{O}\) (having more of \(\mathrm{CuO}\) ) (4) Finely divided form of CuO kept in excess \(\mathrm{O}_{2}\) would be converted to a mixture of \(\mathrm{CuO}\) and \(\mathrm{Cu}_{2} \mathrm{O}\) (having more of \(\mathrm{Cu}_{2} \mathrm{O}\) )

Short answer

Choice (2) is correct: Finely divided form of \( Cu_{2}O \) kept in excess \( O_{2} \) would be completely converted to \( CuO \).

Step-by-step solution

Write the Given Data

Given data:- \( \Delta G_{1}^{\circ}(\mathrm{CuO}) = -30.4 \mathrm{kcal/mol} \) - \(\Delta G_{t}^{\circ}(\mathrm{Cu}_{2}O) = -34.98 \mathrm{kcal/mol} \) - Temperature, T = 298 K

Define Gibbs Free Energy Change for Reaction

Consider the reversible reaction: \[2CuO(s) \rightarrow Cu_{2}O(s) + 1/2 O_{2}(g) \] The Gibbs free energy change for this reaction can be found using the given \( \Delta G \) values.

Calculate Net \( \Delta G^{\circ} \) for the Reaction

The net \( \Delta G^{\circ} \) can be calculated by subtracting the \(\Delta G^{\circ}(\mathrm{Cu}_{2}O)\) from \(\Delta G^{\circ}(\mathrm{CuO})\) converted to 2 moles:\[ \Delta G_{net}^{\circ} = \Delta G_{t}^{\circ}(Cu_{2}O) - 2 \Delta G_{1}^{\circ}(CuO) = -34.98 - 2(-30.4) = -34.98 + 60.8 = 25.82 \mathrm{kcal/mol} \]

Determine Equilibrium Based on \( \Delta G \)

Since the calculated \( \Delta G_{net}^{\circ} \) is positive (25.82 kcal/mol), the reaction is non-spontaneous in the forward direction under standard conditions. This indicates that the reaction will favor the formation of \( CuO \) rather than \( Cu_{2}O \).

Make the Correct Prediction

Given the fact that \( CuO \) formation is favored, the correct prediction is thus:Finely divided form of \( Cu_{2}O \) kept in excess \( O_{2} \) would be completely converted to \( CuO \)

Key concepts

Gibbs Free Energy Change

Gibbs Free Energy Change (∆G) is a critical concept in chemical thermodynamics.
It helps predict the spontaneity of a reaction under constant temperature and pressure.
The equation for Gibbs Free Energy is given by \( \Delta G = \Delta H - T\Delta S \), where \( \Delta H \) is the enthalpy change, \( T \) is the temperature, and \( \Delta S \) is the entropy change.
If \( \Delta G \) is negative, the reaction is spontaneous; if positive, the reaction is non-spontaneous.
In the context of the exercise, we calculated \( \Delta G_{net}^{\circ} \) for the reaction and found it to be 25.82 kcal/mol, indicating the reaction is non-spontaneous.
This tells us that the reaction does not proceed forward favorably under standard conditions.

Reversible Reactions

Reversible reactions are those that can occur in both the forward and reverse directions.
In these reactions, reactants are converted to products and vice versa.
The reaction does not go to completion, and both reactants and products are present at equilibrium.
For example, the reversible reaction in the exercise:
\[ 2CuO(s) \rightleftharpoons Cu_{2}O(s) + \frac{1}{2} O_{2}(g) \]
In reversible reactions, the Gibbs free energy change (\( \Delta G \)) determines which direction is favored.
A positive \( \Delta G \) (as we found in the exercise) means the reverse reaction is favored, leading to more of the reactants.

Chemical Thermodynamics

Chemical thermodynamics deals with the study of energy changes in chemical reactions.
It includes concepts such as enthalpy (\( \Delta H \)), entropy (\( \Delta S \)), and Gibbs free energy (\( \Delta G \)).
These parameters help predict the feasibility and extent of reactions.
Enthalpy is the heat content of a system, entropy is the measure of randomness, and Gibbs free energy combines these to predict reaction spontaneity.
In the exercise, we used these principles to find out that the reaction is not spontaneous and hence will favor the formation of CuO rather than Cu₂O.

Equilibrium Constants

Equilibrium constants (K) describe the ratio of concentrations of products to reactants at equilibrium.
It is a crucial parameter for understanding the extent of a reaction.
The relationship between \( K \) and Gibbs free energy change (\( \Delta G \)) is given by the equation: \[ \Delta G^{\circ} = -RT \ln K \]
When \( \Delta G^{\circ} = 0 \), the system is at equilibrium.
A positive \( \Delta G^{\circ} \) (as in the exercise) suggests a small \( K \), meaning the reactants are favored at equilibrium.
This aligns with our prediction that finely divided \( Cu_{2}O \) in excess \( O_{2} \) would be completely converted to \( CuO \).