Question

Tell whether reactions with the following values of \(\Delta H\) and \(\Delta S\) are spontaneous or nonspontaneous and whether they are exothermic or endothermic: (a) \(\Delta H=-48 \mathrm{~kJ} ; \Delta S=+135 \mathrm{~J} / \mathrm{K}\) at \(400 \mathrm{~K}\) (b) \(\Delta H--48 \mathrm{~kJ} ; \Delta S--135 \mathrm{~J} / \mathrm{K}\) al \(400 \mathrm{~K}\) (c) \(\Delta H=+48 \mathrm{~kJ} ; \Delta S=+135 \mathrm{~J} / \mathrm{K}\) at \(400 \mathrm{~K}\) (d) \(\Delta H=+48 \mathrm{~kJ} ; \Delta S=-135 \mathrm{~J} / \mathrm{K}\) at \(400 \mathrm{~K}\)

Short answer

(a) Spontaneous, Exothermic; (b) Nonspontaneous, Exothermic; (c) Spontaneous, Endothermic; (d) Nonspontaneous, Endothermic.

Step-by-step solution

Understanding Spontaneity Criteria

A reaction is spontaneous if the Gibbs free energy change, \( \Delta G \), is negative. This is calculated using the formula:\[ \Delta G = \Delta H - T \Delta S \]Where, \( \Delta H \) is enthalpy change, \( \Delta S \) is entropy change, and \( T \) is temperature in Kelvin.

Classifying Exothermic and Endothermic

Reactions are classified based on \( \Delta H \):- **Exothermic**: \( \Delta H < 0 \)- **Endothermic**: \( \Delta H > 0 \)

Solving for (a)

Given: \( \Delta H = -48 \text{ kJ}, \Delta S = +135 \text{ J/K}, T = 400 \text{ K} \)Convert \( \Delta S \) to \( \text{kJ/K} \): \[ \Delta S = 0.135 \text{ kJ/K} \]Calculate \( \Delta G \):\[ \Delta G = (-48) - (400 \times 0.135) \]\[ \Delta G = -48 - 54 = -102 \text{ kJ} \]Since \( \Delta G < 0 \), the reaction is spontaneous. It's exothermic as \( \Delta H < 0 \).

Solving for (b)

Given: \( \Delta H = -48 \text{ kJ}, \Delta S = -135 \text{ J/K}, T = 400 \text{ K} \)Convert \( \Delta S \) to \( \text{kJ/K} \): \[ \Delta S = -0.135 \text{ kJ/K} \]Calculate \( \Delta G \):\[ \Delta G = (-48) - (400 \times -0.135) \]\[ \Delta G = -48 + 54 = 6 \text{ kJ} \]Since \( \Delta G > 0 \), the reaction is nonspontaneous. It's exothermic as \( \Delta H < 0 \).

Solving for (c)

Given: \( \Delta H = +48 \text{ kJ}, \Delta S = +135 \text{ J/K}, T = 400 \text{ K} \)Convert \( \Delta S \) to \( \text{kJ/K} \): \[ \Delta S = 0.135 \text{ kJ/K} \]Calculate \( \Delta G \):\[ \Delta G = (+48) - (400 \times 0.135) \]\[ \Delta G = 48 - 54 = -6 \text{ kJ} \]Since \( \Delta G < 0 \), the reaction is spontaneous. It's endothermic as \( \Delta H > 0 \).

Solving for (d)

Given: \( \Delta H = +48 \text{ kJ}, \Delta S = -135 \text{ J/K}, T = 400 \text{ K} \)Convert \( \Delta S \) to \( \text{kJ/K} \): \[ \Delta S = -0.135 \text{ kJ/K} \]Calculate \( \Delta G \):\[ \Delta G = (+48) - (400 \times -0.135) \]\[ \Delta G = 48 + 54 = 102 \text{ kJ} \]Since \( \Delta G > 0 \), the reaction is nonspontaneous. It's endothermic as \( \Delta H > 0 \).

Key concepts

Spontaneous Reactions

In chemistry, the concept of spontaneity refers to whether a reaction occurs naturally without needing to be driven by an external force. The key to determining spontaneity is the Gibbs Free Energy change, denoted as \( \Delta G \). The formula for calculating \( \Delta G \) is: \[ \Delta G = \Delta H - T \Delta S \] where \( \Delta H \) represents the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) symbolizes the change in entropy. A reaction is deemed spontaneous when \( \Delta G < 0 \). If \( \Delta G > 0 \), the reaction is nonspontaneous, meaning it will not occur under the given conditions without additional energy input.
Here are some tips to remember:

• \( \Delta G < 0 \): Reaction is spontaneous
• \( \Delta G > 0 \): Reaction is nonspontaneous
• \( \Delta G = 0 \): Reaction is at equilibrium

By applying this knowledge to the given conditions, you can determine whether a reaction is likely to occur on its own.

Exothermic and Endothermic Reactions

Reactions can release or absorb energy, and this is characterized by the term enthalpy change, \( \Delta H \). Depending on whether energy is released or absorbed, reactions can be classified as either exothermic or endothermic.

• Exothermic Reactions: These reactions release heat, hence \( \Delta H < 0 \) (negative). This means that the final energy of the system is less than the initial.
• Endothermic Reactions: These processes absorb heat, meaning \( \Delta H > 0 \) (positive). The system's energy increases as it proceeds.

Understanding whether a reaction is exothermic or endothermic helps predict temperature effects and feasibility under different conditions. For instance, a spontaneous, exothermic reaction at a particular temperature may transfer heat to its surroundings, while an endothermic one might require a heat source to proceed.

Enthalpy and Entropy

Enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) are pivotal in understanding the spontaneity of a chemical reaction. Enthalpy measures the heat content of a system and indicates whether heat is absorbed or released during a reaction.
On the other hand, entropy is a measure of disorder or randomness in a system. Generally, systems tend to move towards higher entropy, or more disorder. When analyzing a reaction's spontaneity using \( \Delta G = \Delta H - T \Delta S \), both \( \Delta H \) and \( \Delta S \) play critical roles.
The interaction between enthalpy and entropy determines whether a reaction favors progress or equilibrium.

• High entropy change (positive \( \Delta S \)) means increased disorder and can drive a reaction towards spontaneity even if \( \Delta H \) is positive.
• A negative \( \Delta H \) could overcome a small negative entropy to still result in a spontaneous process.

Examining both enthalpy and entropy changes alongside temperature gives a holistic view of a chemical reaction's likelihood under specific conditions.