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For each of the following reactions, how will the \(\Delta G^{\circ}\) change as the temperature increases? (a) \(\begin{aligned} 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftarrows & 2 \mathrm{CO}_{2}(g) \\ \Delta H^{\circ}=&-565.97 \mathrm{~kJ} / \mathrm{mol}_{\mathrm{rxn}} \\ \Delta S^{\circ}=&-173.00 \mathrm{~J} / \mathrm{mol}_{\mathrm{rxn}} \cdot \mathrm{K} \end{aligned}\) (b) \(\begin{aligned} 2 \mathrm{H}_{2} \mathrm{O}(g) & \rightleftarrows 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ} &=483.64 \mathrm{~kJ} / \mathrm{mol}_{\mathrm{rxn}} \\ \Delta S^{\circ} &=90.01 \mathrm{~J} / \mathrm{mol}_{\mathrm{rxn}} \cdot \mathrm{K} \end{aligned}\) (c) \(\begin{aligned} 2 \mathrm{~N}_{2} \mathrm{O}(g) & \rightleftarrows 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ} &=-164.1 \mathrm{~kJ} / \mathrm{mol}_{\mathrm{rxn}} \\ \Delta S^{\circ} &=148.66 \mathrm{~J} / \mathrm{mol}_{\mathrm{rxn}} \cdot \mathrm{K} \end{aligned}\) (d) \(\begin{aligned} \mathrm{PbCl}_{2}(s) & \stackrel{H_{2} O}{\rightleftarrows} \mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \\ \Delta H^{\circ}=& 23.39 \mathrm{~kJ} / \mathrm{mol}_{\mathrm{rxn}} \\ \Delta S^{\circ}=&-12.5 \mathrm{~J} / \mathrm{mol}_{\mathrm{rxn}} \cdot \mathrm{K} \end{aligned}\).

Step by step solution

01

Understand the Gibbs Free Energy Equation

Gibbs Free Energy (\(\Delta G^{\circ}\) is determined using the equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] Where \(\Delta H^{\circ}\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S^{\circ}\) is the change in entropy. Determining how \(\Delta G^{\circ}\) changes with temperature involves analyzing the terms \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\).

02

Analyzing Reaction (a)

For the reaction \(2 \mathrm{CO}(g) + \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{CO}_{2}(g)\):- \(\Delta H^{\circ} = -565.97 \text{ kJ/mol}\)- \(\Delta S^{\circ} = -173.00 \text{ J/(mol} \cdot \text{K)}\)Both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are negative, so:- As temperature increases, the \(T\Delta S^{\circ}\) term becomes more negative.- \(\Delta G^{\circ}\) will increase (become less negative or more positive) with increasing temperature.

03

Analyzing Reaction (b)

For the reaction \(2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g)\):- \(\Delta H^{\circ} = 483.64 \text{ kJ/mol}\)- \(\Delta S^{\circ} = 90.01 \text{ J/(mol} \cdot \text{K)}\)Here, both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are positive:- As temperature increases, the positive \(T\Delta S^{\circ}\) term leads to a more negative effect on \(\Delta G^{\circ}\).- \(\Delta G^{\circ}\) decreases (becomes more negative) with increasing temperature.

04

Analyzing Reaction (c)

For the reaction \(2 \mathrm{~N}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{~N}_{2}(g) + \mathrm{O}_{2}(g)\):- \(\Delta H^{\circ} = -164.1 \text{ kJ/mol}\)- \(\Delta S^{\circ} = 148.66 \text{ J/(mol} \cdot \text{K)}\)Here, \(\Delta H^{\circ}\) is negative and \(\Delta S^{\circ}\) is positive:- As temperature increases, the \(T\Delta S^{\circ}\) term contributes positively.- \(\Delta G^{\circ}\) will decrease (become more negative) with increasing temperature.

05

Analyzing Reaction (d)

For the reaction \(\mathrm{PbCl}_{2}(s) \rightleftarrows \mathrm{Pb}^{2+}(aq) + 2 \mathrm{Cl}^{-}(aq)\):- \(\Delta H^{\circ} = 23.39 \text{ kJ/mol}\)- \(\Delta S^{\circ} = -12.5 \text{ J/(mol} \cdot \text{K)}\)Here, \(\Delta H^{\circ}\) is positive and \(\Delta S^{\circ}\) is negative:- As temperature increases, the \(T\Delta S^{\circ}\) term becomes more negative.- \(\Delta G^{\circ}\) increases (becomes more positive) with increasing temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy. It helps us understand how energy interacts within a system and how it can change forms. This field is crucial for studying chemical reactions as it lets us predict whether a reaction will occur spontaneously under given conditions by analyzing Gibbs Free Energy.

Thermodynamics consists of several laws that describe how energy moves:

• First Law: Energy cannot be created or destroyed, only transformed from one form to another. This is also known as the law of energy conservation.
• Second Law: Entropy in an isolated system always increases over time. This means systems naturally progress towards disorder.
• Third Law: As temperature approaches absolute zero, the entropy of a system approaches a constant minimum.

Gibbs Free Energy particularly focuses on the prediction and spontaneity of processes. For a reaction to be spontaneous, the change in Gibbs Free Energy (\(\Delta G^{\circ}\)) should be negative. By evaluating the balance between enthalpy (\(\Delta H^{\circ}\)) and entropy (\(\Delta S^{\circ}\)), we can determine whether a process will occur without external interference. Thus, thermodynamics provides a framework for understanding these spontaneous or non-spontaneous reactions.

Enthalpy

Enthalpy is a measure of the total heat content of a system and is symbolized by \(\Delta H^{\circ}\). It encompasses both the internal energy of the system and the energy required to displace its surroundings. Enthalpy changes help us evaluate the heat absorbed or released during a chemical reaction.

Reactions can be classified based on enthalpy changes:

• Exothermic Reactions: These reactions release heat to the surroundings, resulting in a negative \(\Delta H^{\circ}\). A burning candle is an example of an exothermic process.
• Endothermic Reactions: These absorb heat from the surroundings, leading to a positive \(\Delta H^{\circ}\). Melting ice is an endothermic process.

In the context of Gibbs Free Energy, enthalpy plays a crucial role. If a reaction shows a negative \(\Delta H^{\circ}\), it indicates an exothermic process, potentially contributing to a negative \(\Delta G^{\circ}\) and favoring spontaneity. Conversely, a positive \(\Delta H^{\circ}\) might need to be offset by a favorable entropy change to achieve a spontaneous reaction. Understanding enthalpy is vital in predicting how temperature changes will impact reaction feasibility.

Entropy

Entropy, which is represented by \(\Delta S^{\circ}\), is a measure of disorder or randomness in a system. It reflects the number of ways energy can be distributed among particles within a system. Higher entropy signifies greater disorder.

When a process results in increased randomness, \(\Delta S^{\circ}\) is positive. This can signify that a reaction is more likely to be spontaneous as it aligns with the second law of thermodynamics.

• Gases have higher entropy than liquids or solids because of their free-moving particles.
• Mixing and reactions generally increase entropy as they lead to greater dispersal of energy and particles.

In Gibbs Free Energy calculations, the term \(T\Delta S^{\circ}\) is crucial. As temperature increases, entropy's influence on spontaneity becomes more significant. If a reaction has a positive \(\Delta S^{\circ}\), increasing the temperature can favor the process's spontaneity by making \(T\Delta S^{\circ}\) larger. Conversely, a negative \(\Delta S^{\circ}\) becomes less favorable at higher temperatures, potentially making the process non-spontaneous. Thus, understanding entropy changes is essential to predicting reaction spontaneity at different temperatures.