Problem 14 - Chemistry in 30 days

Question

In a chemical reaction \(\Delta H=150 \mathrm{~kJ}\) and \(\Delta S=100 \mathrm{JK}^{-1}\) at \(300 \mathrm{~K}\). the \(\Delta G\) for the reaction is: (a) Zero (b) \(300 \mathrm{~kJ}\) (c) \(330 \mathrm{~kJ}\) (d) \(120 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

The \(\Delta G\) for the reaction is 120 kJ, so the correct answer is (d).

Step 1: Understand Gibbs Free Energy Equation

The Gibbs free energy change, \(\Delta G\), is calculated using the equation \(\Delta G = \Delta H - T \Delta S\). Here, \(\Delta H\) is the enthalpy change, \(\Delta S\) is the entropy change, and \(T\) is the temperature in Kelvin.

Step 2: Identify Given Values

From the problem, \(\Delta H = 150 \text{ kJ}\), \(\Delta S = 100 \text{ JK}^{-1}\), and \(T = 300 \text{ K}\). Since \(\Delta S\) is in \(\text{J K}^{-1}\), convert \(\Delta H\) into the same units: \(150 \text{ kJ} = 150,000 \text{ J}\).

Step 3: Convert Entropy Units

Ensure that \( \Delta S \) is consistent with the energy units of \( \Delta H \). No further conversion is needed since \( \Delta S \) is already in \( \text{ JK}^{-1} \).

Step 4: Plug Values into Gibbs Equation

Substitute the known values into the Gibbs equation: \(\Delta G = 150,000 \text{ J} - 300 \text{ K} \times 100 \text{ JK}^{-1}\).

Step 5: Perform the Calculation

Calculate the product of \(T\) and \(\Delta S\): \(300 \times 100 = 30,000 \text{ J}\). Then, compute \(\Delta G = 150,000 \text{ J} - 30,000 \text{ J} = 120,000 \text{ J}\).

Step 6: Convert Back to kJ

Since the answer choices are in \(\text{kJ}\), convert \(\Delta G\) back from \(\text{J}\) to \(\text{kJ}\): \(120,000 \text{ J} = 120 \text{ kJ}\).

Key Concepts

Enthalpy

Enthalpy is a key concept in understanding energy changes in thermodynamic systems. It represents the total heat content or energy of a system at constant pressure. We often denote it as \( \Delta H \) for changes in enthalpy during a chemical reaction.

When a reaction occurs, energy can either be absorbed or released. A positive \( \Delta H \) indicates that the reaction is endothermic, meaning it absorbs energy from the surroundings. Conversely, a negative \( \Delta H \) represents an exothermic reaction where energy is released.

• An increase in enthalpy usually results from the input of heat energy.
• Understanding whether a reaction is endothermic or exothermic can help predict how substances will interact.

In our exercise, we have an enthalpy change \( \Delta H = 150 \text{ kJ} \), meaning the system requires energy input as the reaction proceeds.

Entropy

Entropy is a measure of disorder or randomness within a system. In thermodynamics, it quantifies the energy in a system that is not available to do work. The symbol \( \Delta S \) represents change in entropy, and its units are typically \( \text{J K}^{-1} \).

Entropy plays a crucial role in determining the direction and spontaneity of a chemical reaction. A positive \( \Delta S \) implies an increase in disorder, while a negative \( \Delta S \) indicates a decrease.

• Systems naturally tend to move towards a state of higher entropy or disorder.
• Reactions with positive entropy changes are often favorable as they lead to greater chaos.

In the exercise, \( \Delta S = 100 \text{ JK}^{-1} \) suggests the system becomes more disordered with the reaction.

Thermodynamics

Thermodynamics is the study of energy, work, and heat, and how they interrelate in physical and chemical processes. The three laws of thermodynamics provide a framework for understanding these relationships.
The First Law, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed, only transferred or transformed. Therefore, in any process, the total energy within the system and its surroundings remains constant.
The Second Law deals with the direction of spontaneous processes and is closely associated with entropy. It states that the entropy of the universe always increases over time for any spontaneous process.

• Understanding these laws is essential to predicting and calculating changes in energy and disorder in reactions.
• It forms the basis for calculating other thermodynamic properties like Gibbs Free Energy.

Energy Conversion

Energy conversion refers to the transformation of energy from one form to another, which is vital in chemical reactions and thermodynamic processes. The Gibbs Free Energy equation, \( \Delta G = \Delta H - T \Delta S \), is pivotal in determining whether energy conversions in a reaction are feasible under constant temperature and pressure.

When calculating \( \Delta G \):

• If \( \Delta G \) is negative, the reaction is spontaneous, releasing free energy.
• If \( \Delta G \) is positive, the reaction is non-spontaneous, requiring energy input.
• If \( \Delta G \) is zero, the system is at equilibrium, with no net energy change.

Using the values from our problem, \( \Delta H = 150,000 \text{ J} \), \( \Delta S = 100 \text{ JK}^{-1} \), and \( T = 300 \text{ K} \), leads to \( \Delta G = 120,000 \text{ J} \) or \( 120 \text{ kJ} \), indicating the reaction needs an energy input to proceed.