Question

For a reaction at \(300 \mathrm{~K}\), enthalpy and entropy changes are \(-11.5 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\) and \(-105 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively. What is the change in Gibbs free energy? (a) \(25 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(15 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(20 \mathrm{kJmol}^{-1}\)

Short answer

The change in Gibbs free energy is \( 20 \mathrm{kJmol}^{-1} \), option (d).

Step-by-step solution

Understand the Gibbs Free Energy Equation

The change in Gibbs free energy for a reaction at constant temperature and pressure is given by the equation \( \Delta G = \Delta H - T\Delta S \), where \( \Delta G \) is the change in Gibbs free energy, \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy.

Identify Given Values

From the problem, we are given the enthalpy change \( \Delta H = -11.5 \times 10^3 \mathrm{~J~mol}^{-1} \) and the entropy change \( \Delta S = -105 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1} \) at a temperature \( T = 300 \mathrm{~K} \).

Insert Values into the Equation

Substitute the given values into the Gibbs free energy equation: \( \Delta G = (-11.5 \times 10^3) - (300)(-105) \).

Calculate the Product of Temperature and Entropy Change

Calculate the product of temperature \( T = 300 \mathrm{~K} \) and entropy change \( \Delta S = -105 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1} \): \( T\Delta S = 300 \times (-105) = -31500 \mathrm{~J~mol}^{-1} \).

Calculate the Change in Gibbs Free Energy

Using the values: \( \Delta G = (-11.5 \times 10^3) + 31500 \), we have \( \Delta G = -11500 + 31500 = 20000 \mathrm{~J~mol}^{-1} \). Convert it to kilojoules: \( \Delta G = 20 \mathrm{~kJ~mol}^{-1} \).

Select the Correct Answer

Compare the calculated \( \Delta G \) value with the options provided. The correct answer is (d) \( 20 \mathrm{kJmol}^{-1} \).

Key concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is the measure of heat content change in a system during a chemical reaction. Imagine it as the energy absorbed or released, reflecting the bond-making or bond-breaking nature in a reaction.
When \( \Delta H \) is negative, the reaction is exothermic, meaning it releases heat. Conversely, a positive \( \Delta H \) signifies an endothermic reaction, where the system absorbs heat. In the exercise, the enthalpy change is given as \(-11.5 \times 10^3 \text{ J mol}^{-1}\), indicating an exothermic reaction.
This concept is crucial because it helps predict how much heat will need to be added or removed from a system to sustain a reaction at constant pressure. It’s one part of what we need to understand the overall spontaneity of the process when combined with entropy change.

Entropy Change

Entropy change, represented as \( \Delta S \), describes the degree of disorder or randomness in a system. A simple way to grasp this is thinking about how messy your room gets over time. Processes where the disorder increases are often more favored. Thus, a positive \( \Delta S \) suggests an increase in disorder.
In the exercise, \( \Delta S = -105 \text{ JK}^{-1} \text{ mol}^{-1} \) means the reaction leads to a decrease in disorder, or makes the system more ordered. This is key because it affects the overall Gibbs free energy change, which determines if a reaction is spontaneous.

• A positive \( \Delta S \) typically means products of a reaction are more disordered compared to reactants.
• A negative \( \Delta S \) implies products are more ordered.

Understanding entropy helps predict whether a reaction will proceed without external input, especially alongside changes in enthalpy.

Thermodynamics Equation

The thermodynamics equation for Gibbs free energy, given by \( \Delta G = \Delta H - T \Delta S \), is a cornerstone in chemistry, blending enthalpy and entropy changes to determine spontaneity. This equation allows us to calculate the available energy in a system to do work at a constant temperature and pressure.
Each part of the equation plays a pivotal role:

• \( \Delta H \): Provides insight into heat changes in the reaction.
• \( T \Delta S \): Determines how temperature and entropy affect system randomness.

In the exercise, the values \( \Delta H = -11.5 \times 10^3 \text{ J mol}^{-1} \), \( T = 300 \text{ K} \), and \( \Delta S = -105 \text{ JK}^{-1} \text{ mol}^{-1} \) are substituted into the equation. Calculating \( T \Delta S \) gives \(-31500 \text{ J mol}^{-1}\), which when added to \( \Delta H \) results in \( \Delta G = 20000 \text{ J mol}^{-1} \), or \( 20 \text{ kJ mol}^{-1} \).
This teaches us that despite the reaction having an enthalpy release, the increase in order due to negative entropy change impacts the Gibbs free energy, providing a full picture of spontaneity.