Problem 39 - Chemistry: Principles and Reactions

Question

Discuss the effect of temperature change on the spontaneity of the following reactions at 1 atm. $$ \text { (a) } \begin{aligned} 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) \longrightarrow 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \\ \Delta H^{\circ}=+830.8 \mathrm{~kJ} ; \Delta S^{\circ}=+168 \mathrm{~J} / \mathrm{K} \end{aligned} $$ (b) \(2 \mathrm{As}(s)+3 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{AsF}_{3}(l)\) \(\Delta H^{\circ}=-1643 \mathrm{~kJ} ; \Delta S^{\circ}=-0.316 \mathrm{~kJ} / \mathrm{K}\) (c) \(\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{C}(s)+\frac{1}{2} \mathrm{O}_{2}(g)\) \(\Delta H^{\circ}=110.5 \mathrm{~kJ} ; \Delta S^{\circ}=-89.4 \mathrm{~J} / \mathrm{K}\)

Step-by-Step Solution

Answer

Answer: For reaction (a), an increase in temperature makes it more spontaneous. The spontaneity of reaction (b) is not affected by the change in temperature, and it remains spontaneous at all temperatures. For reaction (c), the spontaneity is not affected by the change in temperature, and it remains non-spontaneous at all temperatures.

Step 1: Reaction (a)

In this reaction, we have: $$\Delta H^{\circ} = +830.8 \mathrm{~kJ}$$ $$\Delta S^{\circ} = +168 \mathrm{~J}/\mathrm{K} = +0.168 \mathrm{~kJ}/\mathrm{K}$$ 1. If the temperature is low: - The term \(T\Delta S^{\circ}\) will be smaller than \(\Delta H^{\circ}\) - \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} > 0\) - The reaction is non-spontaneous. 2. If the temperature is high: - The term \(T\Delta S^{\circ}\) will be larger than \(\Delta H^{\circ}\) - \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} < 0\) - The reaction is spontaneous. Therefore, an increase in temperature makes reaction (a) more spontaneous.

Step 2: Reaction (b)

In this reaction, we have: $$\Delta H^{\circ} = -1643 \mathrm{~kJ}$$ $$\Delta S^{\circ} = -0.316 \mathrm{~kJ}/\mathrm{K}$$ For all temperatures: - \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} < 0\) - The reaction is spontaneous. Therefore, the spontaneity of reaction (b) is not affected by the change in temperature.

Step 3: Reaction (c)

In this reaction, we have: $$\Delta H^{\circ} = +110.5 \mathrm{~kJ}$$ $$\Delta S^{\circ} = -89.4 \mathrm{~J}/\mathrm{K} = -0.0894 \mathrm{~kJ}/\mathrm{K}$$ 1. If the temperature is low: - The term \(-T\Delta S^{\circ}\) will be smaller than \(\Delta H^{\circ}\) - \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} > 0\) - The reaction is non-spontaneous. 2. If the temperature is high: - The term \(-T\Delta S^{\circ}\) will still be smaller than \(\Delta H^{\circ}\) - \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} > 0\) - The reaction is still non-spontaneous. Therefore, the spontaneity of reaction (c) is not affected by the change in temperature, and the reaction always remains non-spontaneous.

Key Concepts

Gibbs Free Energy

The Gibbs free energy, represented by \( G \), is a thermodynamic function that predicts the spontaneity of chemical reactions. It is calculated using the equation \( \Delta G = \Delta H - T\Delta S \), where \( \Delta G \) is the change in Gibbs free energy, \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy. A reaction is spontaneous if \( \Delta G \) is negative and non-spontaneous if \( \Delta G \) is positive.

Understanding \( \Delta G \) is crucial for predicting whether a reaction will occur without additional input of energy. This calculation allows us to assess the feasibility of reactions in various industries, such as pharmaceuticals and materials science. When the exercise refers to \( \Delta G^\circ \) being greater than or less than zero, it signifies if a reaction is spontaneous under standard conditions.

Enthalpy Change

Enthalpy change, denoted by \( \Delta H \), is the total heat content change during a chemical reaction at constant pressure. It is a measure of the energy absorbed or released in a reaction. If \( \Delta H \) is negative, the reaction is exothermic, releasing heat to the surroundings. Conversely, a positive \( \Delta H \) indicates an endothermic reaction, where heat is absorbed.

In the exercise, enthalpy changes play a vital role in determining spontaneity in conjunction with entropy changes and temperature. Knowing the \( \Delta H \) value, along with the \( \Delta S \) and temperature, helps in calculating the Gibbs free energy to predict if a reaction is spontaneous.

Entropy Change

Entropy, symbolized by \( S \), relates to the disorder or randomness in a system. The change in entropy, \( \Delta S \) for a reaction, indicates whether the system becomes more or less disordered. An increase in entropy \( (+\Delta S) \) means the system has more disorder, while a decrease \( (-\Delta S) \) means it has less disorder.

The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. Therefore, the universe favors processes that increase its overall entropy. In the provided example, the entropy change, whether positive or negative, influences the spontaneity of reactions when combined with enthalpy change and temperature.

Temperature Effects on Reaction Spontaneity

Temperature is a critical factor influencing the spontaneity of reactions. It affects both the enthalpy \( (\Delta H) \) and entropy \( (\Delta S) \) terms in the Gibbs free energy equation. As seen in the exercise, a low temperature may result in a non-spontaneous reaction due to the small \( T\Delta S \) term, whereas a high temperature can increase \( T\Delta S \) and potentially make the reaction spontaneous.

This indicates that reactions that are non-spontaneous at one temperature may become spontaneous at a higher temperature. That's why in some industrial processes, controlling the temperature is a way to drive reactions that would not occur under standard conditions.