Question

For a certain chemical reaction, \(\Delta H^{\circ}=-35.4 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-85.5 \mathrm{~J} / \mathrm{K} .\) (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (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 is exothermic, as ΔH° is negative (-35.4 kJ). (b) The reaction leads to a decrease in the randomness or disorder of the system, as ΔS° is negative (-85.5 J/K). (c) The Gibbs free energy change (ΔG°) at 298 K is -9,941 J. (d) The reaction is spontaneous at 298 K under standard conditions, as ΔG° is negative.

Step-by-step solution

(a) Determine reaction type (Exothermic or Endothermic) based on ΔH°

In a reaction, if the enthalpy change (ΔH°) is negative, it indicates that the reaction releases heat and is therefore an exothermic reaction. If ΔH° is positive, the reaction absorbs heat and is endothermic. In this case, the value of the enthalpy change is ΔH° = -35.4 kJ (negative), so the reaction is exothermic.

(b) Determine change in randomness or disorder based on ΔS°

In a reaction, if the entropy change (ΔS°) is positive, it denotes that the reaction leads to an increase in randomness or disorder. If ΔS° is negative, the reaction leads to a decrease in randomness or disorder. In this case, the value of the entropy change is ΔS° = -85.5 J/K (negative), so the reaction leads to a decrease in the randomness or disorder of the system.

(c) Calculate ΔG° at 298 K

To calculate Gibbs free energy change (ΔG°), we need to use the following equation: ΔG° = ΔH° - TΔS° Given that ΔH° = -35.4 kJ and ΔS° = -85.5 J/K, and the temperature is 298 K, we can calculate the Gibbs free energy change as follows: First, convert ΔH° and ΔS° to the same unit. Let's use J: ΔH° = -35.4 kJ * 1000 J/kJ = -35,400 J ΔS° = -85.5 J/K Now, we can plug in the values into the equation: ΔG° = (-35,400 J) - (298 K * -85.5 J/K) = -35,400 J + 25,459 J = -9,941 J

(d) Determine reaction spontaneity based on ΔG°

The spontaneity of a reaction at a given temperature under standard conditions can be determined by examining the sign of ΔG°: - If ΔG° < 0, the reaction is spontaneous. - If ΔG° > 0, the reaction is non-spontaneous. - If ΔG° = 0, the reaction is at equilibrium. From part (c), we calculated ΔG° = -9,941 J, which is negative. Therefore, the reaction is spontaneous at 298 K under standard conditions.

Key concepts

Thermodynamics

Thermodynamics plays a crucial role in understanding chemical reactions, especially when analyzing Gibbs Free Energy. It is a branch of physics that deals with the relationships between heat, work, temperature, and energy.
In the context of a chemical reaction, several thermodynamic parameters help us predict reaction behaviors under different conditions.
- **Enthalpy (\( \Delta H^{\circ} \))**: This represents the heat change at constant pressure. A positive value indicates an endothermic reaction, while a negative value implies an exothermic reaction.- **Entropy (\( \Delta S^{\circ} \))**: It measures the disorder or randomness in a system. Positive entropy change suggests increased randomness, whereas negative entropy change signifies decreased randomness.- **Gibbs Free Energy (\( \Delta G^{\circ} \))**: This parameter predicts if a reaction will occur spontaneously under constant temperature and pressure. Calculated using the equation \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \), where \( T \) is the temperature in Kelvin.By examining these parameters, one can determine not only the energy changes involved in a reaction but also the feasibility and spontaneity of the reaction.

Exothermic Reaction

An exothermic reaction releases heat into its surroundings. This heat release occurs because the energy required to break the bonds of the reactants is less than the energy released when the products are formed.

Key characteristics of exothermic reactions include:

• **Negative Enthalpy Change ( \(\Delta H^{\circ} < 0 \))**: This indicates that the reaction results in a net release of energy.
• **Temperature Increase**: Often, the temperature of the environment increases, as energy disperses into the surroundings.

Consider the original exercise: \( \Delta H^{\circ} = -35.4 \space \text{kJ} \). Since this value is negative, the reaction is exothermic and releases energy.
This release of energy can influence other properties of the reaction, such as its spontaneity and speed.

Entropy Change

Entropy is a measure of the randomness or disorder of a system, and its change (\( \Delta S^{\circ} \)) provides insights into how disorder fluctuates during a reaction.
If \( \Delta S^{\circ} > 0 \), the system's disorder increases, which is often favorable for spontaneity. Conversely, \( \Delta S^{\circ} < 0 \) indicates a decrease in disorder.Examining reactions from the standpoint of entropy:

• Processes that involve gases or the breaking down of solid structures into more disordered states often have positive entropy changes.
• Reactions forming more ordered structures typically exhibit negative entropy changes.

For the exercise mentioned, \( \Delta S^{\circ}= -85.5 \space \text{J/K} \) shows a decline in disorder as the reaction proceeds. This is an influential factor in the calculation of Gibbs Free Energy, impacting whether the reaction will be spontaneous or not.Understanding entropy changes helps in determining the overall tendency of the reaction, aligning with the principle that nature favors processes leading to higher disorder.