Problem 54 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Show that the reaction \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) at \(300 \mathrm{~K}\), is spontaneous and exothermic, when the standard entropy change is \(-$$0.094 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\). The standard Gibbs free energies of formation for \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) are \(-394.4\) and \(-137.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\), respectively.

Step-by-Step Solution

Answer

The reaction is spontaneous and exothermic as ΔG° is -257.2 kJ/mol and ΔH° is -285.4 kJ/mol.

Step 1: Calculate standard Gibbs Free Energy Change (ΔG°)

The formula to calculate the change in Gibbs Free Energy, ΔG°, for the reaction is: \[ \Delta G^{\circ} = \sum \Delta G^{\circ}_{\text{products}} - \sum \Delta G^{\circ}_{\text{reactants}} \]For the given reaction: - The standard Gibbs energy of formation for CO₂(g) is -394.4 kJ/mol.- The standard Gibbs energy of formation for CO(g) is -137.2 kJ/mol, and for O₂(g) it is 0 kJ/mol since it is an element in its standard state.Substitute these values into the equation:\[ \Delta G^{\circ} = (-394.4) - \left((-137.2) + \frac{1}{2}(0) \right) \]\[ \Delta G^{\circ} = -394.4 + 137.2 \]\[ \Delta G^{\circ} = -257.2 \text{ kJ/mol} \]

Step 2: Check Spontaneity of the Reaction

A reaction is considered spontaneous if the standard Gibbs free energy change, ΔG°, is negative. We have calculated that \[ \Delta G^{\circ} = -257.2 \text{ kJ/mol} \]Since it is negative, the reaction is spontaneous.

Step 3: Calculate the Change in Enthalpy (ΔH°)

Use the Gibbs free energy equation:\[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \]Rearrange to solve for ΔH°:\[ \Delta H^{\circ} = \Delta G^{\circ} + T\Delta S^{\circ} \]Insert the given values:\[ T = 300\, \text{K}, \Delta S^{\circ} = -0.094 \text{ kJ/mol⋅K} \]Thus:\[ \Delta H^{\circ} = -257.2 + 300(-0.094) \]\[ \Delta H^{\circ} = -257.2 - 28.2 \]\[ \Delta H^{\circ} = -285.4 \text{kJ/mol} \]

Step 4: Check If Reaction is Exothermic

A reaction is exothermic if the change in enthalpy, ΔH°, is negative. From the previous step, we have calculated:\[ \Delta H^{\circ} = -285.4 \text{ kJ/mol} \]Thus, since ΔH° is negative, the reaction is exothermic.

Key Concepts

Enthalpy Change

Enthalpy change named as \( \Delta H \) measures the heat absorbed or released during a chemical reaction at constant pressure. It helps us understand whether a reaction is exothermic (releasing heat) or endothermic (absorbing heat).

In our reaction, carbon monoxide (CO) is converted to carbon dioxide (CO₂) with the addition of oxygen (O₂). The calculation of standard Gibbs free energy change, \( \Delta G^{\circ} \), involved figuring out if \( \Delta H^{\circ} \) was negative, meaning the reaction releases energy in the form of heat.

The formula that connects \( \Delta G^{\circ} \), \( \Delta H^{\circ} \), and \( \Delta S^{\circ} \) (standard entropy change) is:

\[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \]

We rearranged this to solve for \( \Delta H^{\circ} \) to find out if the reaction is exothermic. The reaction's \( \Delta H^{\circ} \) turned out to be \(-285.4 \text{kJ/mol}\), confirming it is indeed exothermic. This confirms that CO combustion releases energy, usually in the form of heat, making things warm.

Remember:

• Exothermic reactions result in a negative \( \Delta H \)
• Endothermic reactions result in a positive \( \Delta H \)

Reaction Spontaneity

The spontaneity of a reaction indicates whether it will occur without needing external energy. This is often determined by the Gibbs Free Energy Change, represented as \( \Delta G \).

For a reaction to be spontaneous, the value of \( \Delta G \) must be negative. Consider \[ \Delta G^{\circ} = \sum \Delta G^{\circ}_{\text{products}} - \sum \Delta G^{\circ}_{\text{reactants}} \]

With our example, the calculation shows a \( \Delta G^{\circ} \) of \(-257.2 \text{kJ/mol}\), indicating the reaction of CO to CO₂ is spontaneous!

Why is the spontaneity important? It helps us determine:

• If a reaction proceeds without outside intervention
• The feasibility of a reaction under supplied conditions

A key takeaway is that even though a reaction is spontaneous, it doesn't always proceed rapidly. Reaction rates derive from a separate principle of kinetics, meaning this spontaneity guarantees thermodynamically favorable conditions but not speed.

Standard Entropy Change

Standard Entropy Change, represented as \( \Delta S^{\circ} \), measures the disorder or randomness in a system as the reaction proceeds. The concept of entropy is linked with the second law of thermodynamics, which states that the total entropy of a closed system can never decrease over time.

For the examined reaction, \( \Delta S^{\circ} \) was given as \(-0.094 \text{kJ/mol⋅K}\). The negative sign suggests that the reaction leads to a decrease in entropy. This usually means the products are more ordered compared to the reactants.

The role of \( \Delta S^{\circ} \) can subtly affect whether a reaction is spontaneous, as linked by the equation:

\[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \]

The change in entropy, when paired with temperature \( T \) and enthalpy change \( \Delta H^{\circ} \), influences the Gibbs Free Energy Change \( \Delta G^{\circ} \) value.

• If \( \Delta S^{\circ} \) is positive, it generally favors spontaneity because more disorder contributes positively to the equation.
• If \( \Delta S^{\circ} \) is negative, it indicates more order and could oppose spontaneity if other factors (like \( \Delta H^{\circ} \)) do not counterbalance it.

Understanding entropy helps predict reaction behavior under different conditions and is crucial for applications where orderly transformation or creation of products is desired.