Question

A certain reaction has \(\Delta H^{\circ}=+20.0 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+100.0 \mathrm{~J} / \mathrm{K} .\) (a) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K} .(\mathbf{d})\) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Short answer

(a) The reaction leads to an increase in the randomness or disorder of the system, as ΔS⁰ is positive. (b) The reaction leads to a decrease in the randomness or disorder of the surroundings, as ΔH⁰ is positive (endothermic reaction). (c) At 298 K, ΔG⁰ = -9,800 J. (d) The reaction is spontaneous at 298 K under standard conditions, as ΔG⁰ is negative.

Step-by-step solution

Analyzing the change in randomness or disorder of the system

We are given the entropy change ΔS⁰ as +100.0 J/K. Since the value is positive, it implies that the reaction leads to an increase in the randomness or disorder of the system.

Analyzing the change in randomness or disorder of the surroundings

We are given the enthalpy change ΔH⁰ as +20.0 kJ. Since the value is positive, it means that the reaction is endothermic. Therefore, the reaction absorbs heat from its surroundings, leading to a decrease in the randomness or disorder of the surroundings.

Calculating Gibbs free energy change (ΔG⁰) at 298 K

We can use the relation between Gibbs free energy change, enthalpy change, and entropy change (ΔG⁰ = ΔH⁰ - TΔS⁰) to calculate ΔG⁰ at 298 K. Given : ΔH⁰ = +20.0 kJ ΔS⁰ = +100.0 J/K T = 298 K First, we need to convert ΔH⁰ and ΔS⁰ to the same units. Here, we will convert ΔH⁰ to J by multiplying it by 1000: ΔH⁰ = 20.0 kJ × 1000 J/kJ = 20,000 J Now we can use the formula: ΔG⁰ = ΔH⁰ - TΔS⁰ ΔG⁰ = 20,000 J - (298 K × 100 J/K)

Calculate ΔG⁰

By substituting the values, we get: ΔG⁰ = 20,000 J - 29,800 J ΔG⁰ = -9,800 J

Determine if the reaction is spontaneous at 298 K

Since the Gibbs free energy change ΔG⁰ is negative (-9,800 J), the reaction is spontaneous at 298 K under standard conditions.

Key concepts

Entropy Change

Entropy change refers to the variation in the randomness or disorder of a system during a chemical reaction. In thermodynamics, it is represented as \( \Delta S \). When \( \Delta S \) is positive, it indicates an increase in disorder. Conversely, a negative \( \Delta S \) implies a decrease in disorder.
In this exercise, the provided \( \Delta S^{\circ} = +100.0 \ \mathrm{J/K} \) suggests a positive entropy change. This tells us that the products of the reaction are more random than the reactants. As a result, the system's entropy increases, making the molecules more disordered.
This information can be vital when predicting the behavior of chemical reactions.

• A positive \( \Delta S \) could mean gases are being formed, increasing randomness compared to liquids or solids.
• It helps in assessing spontaneity together with enthalpy and temperature.

Enthalpy Change

Enthalpy change involves the heat change of a system during a reaction and is denoted as \( \Delta H \). It signifies whether heat is absorbed or released.
A positive \( \Delta H \) indicates that the reaction is endothermic, absorbing heat from the surroundings. A negative \( \Delta H \) means it is exothermic, releasing heat. In this exercise, the enthalpy change \( \Delta H^{\circ} = +20.0 \ \mathrm{kJ} \) suggests the reaction absorbs energy.
This absorption can cause the surroundings to cool down due to reduced thermal motion.

• This reaction's positive \( \Delta H \) leads to a decrease in the surroundings' disorder, as energy absorption often decreases molecular motion.
• Understanding \( \Delta H \) helps in determining reaction feasibility and conditions required for the reaction to proceed.

Spontaneity of Reaction

The spontaneity of a reaction determines if it can proceed without external influence. It's assessed using Gibbs free energy change \( \Delta G \).
The expression \( \Delta G = \Delta H - T \Delta S \) helps us evaluate this. If \( \Delta G \) is negative, the reaction is spontaneous; if positive, it's non-spontaneous.
For this reaction, calculating \( \Delta G^{\circ} \) at \( 298 \ \mathrm{K} \): \[ \Delta G^{\circ} = +20,000 \ \mathrm{J} - (298 \ \mathrm{K} \times 100 \ \mathrm{J/K}) = -9,800 \ \mathrm{J} \]
The result indicates that the reaction is spontaneous under standard conditions. This is because:

• A negative \( \Delta G \) suggests the process is thermodynamically favorable.
• The reaction can progress without needing additional energy inputs.

Understanding spontaneity is crucial for determining if a reaction can occur naturally and how conditions like temperature influence it.