Problem 30 - Chemistry and Chemical Reactivity

Question

Determine whether each of the reactions listed below is entropy-favored or -disfavored under standard conditions. Predict how an increase in temperature will affect the value of \(\Delta G_{\text {ren }}^{\circ}\) (a) \(\mathrm{I}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{I}(\mathrm{g})\) (b) \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{SO}_{3}(\mathrm{g})\) (c) \(\operatorname{sicl}_{1}(g)+2 H_{2} O(\ell) \longrightarrow S i O_{2}(s)+4 H C l(g)\) (d) \(\mathrm{P}_{4}(\mathrm{s}, \text { white })+6 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{PH}_{3}(\mathrm{g})\)

Step-by-Step Solution

Answer

(a) Entropy-favored, decreases \( \Delta G^{\circ} \); (b) Entropy-disfavored, increases \( \Delta G^{\circ} \); (c) Entropy-favored, decreases \( \Delta G^{\circ} \); (d) Entropy-disfavored, increases \( \Delta G^{\circ} \).

Step 1: Analyze Reaction (a)

The reaction is \( \mathrm{I}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{I}(\mathrm{g}) \). Here, one molecule of \( \mathrm{I}_{2} \) gas is dissociating into two atoms of \( \mathrm{I} \) gas. This leads to an increase in the number of gas particles and therefore increases the disorder or entropy (\( \Delta S^{\circ} > 0 \)). The reaction is entropy-favored. An increase in temperature decreases \( \Delta G^{\circ} \) since \( \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \) and \( \Delta S^{\circ} \) is positive.

Step 2: Analyze Reaction (b)

The reaction is \( 2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{SO}_{3}(\mathrm{g}) \). Here, three molecules of gas combine to form two molecules of gas. This results in a decrease in the number of gas particles, leading to a decrease in entropy (\( \Delta S^{\circ} < 0 \)). The reaction is entropy-disfavored. An increase in temperature increases \( \Delta G^{\circ} \) since \( \Delta S^{\circ} \) is negative.

Step 3: Analyze Reaction (c)

The reaction is \( \operatorname{sicl}_{1}(g) + 2 H_{2} O(\ell) \longrightarrow S i O_{2}(s) + 4 H C l(g) \). The reaction involves going from one gaseous molecule and two liquids to a solid and four gaseous molecules. There is an increase in entropy because gas formation leads to greater disorder (\( \Delta S^{\circ} > 0 \)). The reaction is entropy-favored. An increase in temperature will decrease \( \Delta G^{\circ} \) as \( \Delta S^{\circ} \) is positive.

Step 4: Analyze Reaction (d)

The reaction is \( \mathrm{P}_{4}(\mathrm{s}, \text{ white }) + 6 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{PH}_{3}(\mathrm{g}) \). In this reaction, there is a transformation from 6 molecule moles of gas to 4 molecule moles of gas. There is a decrease in entropy due to decreased particle mobility (\( \Delta S^{\circ} < 0 \)). This makes the reaction entropy-disfavored. An increase in temperature increases \( \Delta G^{\circ} \).

Step 5: Conclusion

Reactions (a) and (c) are entropy-favored due to an increase in disorder. Reactions (b) and (d) are entropy-disfavored because the number of gas molecules decreases.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, often symbolized as \( \Delta G \), is a crucial concept in understanding chemical reactions. It helps us predict whether a reaction will occur spontaneously under constant pressure and temperature. The equation \( \Delta G = \Delta H - T\Delta S \) summarizes this relationship. Here, \( \Delta H \) represents the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy.
When \( \Delta G < 0 \), a reaction is spontaneous, meaning it can proceed without adding energy. Conversely, a positive \( \Delta G \) suggests non-spontaneity, and the reaction won’t occur on its own. The interplay between enthalpy and entropy changes can shift \( \Delta G \) values, dictating the likelihood of a reaction occurring. Thus, understanding Gibbs Free Energy is key to discerning the spontaneity and feasibility of chemical reactions.

Temperature Effect on Reactions

Temperature plays a pivotal role in altering the course and rate of chemical reactions. It primarily affects the reaction's entropy component. According to the Gibbs Free Energy equation, increasing the temperature can significantly influence \( \Delta G \) because \( T\Delta S \) becomes more substantial.

• For reactions with a positive \( \Delta S \), higher temperatures often enhance spontaneity, as the \( T\Delta S \) term becomes more significant, making \( \Delta G \) less than zero.
• For reactions with a negative \( \Delta S \), increased temperature can make the process less favorable, potentially turning spontaneous reactions into non-spontaneous ones, as \( \Delta G \) may increase.

Temperature changes can also affect reaction equilibria and rates, thus influencing how fast or in what direction a reaction proceeds. Therefore, understanding the temperature's impact is vital for predicting changes in chemical reaction behaviors.

Entropy Change in Reactions

Entropy is a measure of disorder or randomness in a system. Entropy change \( \Delta S \) in a chemical reaction indicates how the disorder of the system changes from reactants to products.

• An increase in the number of gas molecules typically results in a positive \( \Delta S \), as observed in Reaction (a), where \( I_2(g) \rightarrow 2I(g) \).
• A decrease, like in Reaction (b), where \( 2SO_2(g) + O_2(g) \rightarrow 2SO_3(g) \), leads to a negative \( \Delta S \).

When reactions produce more disorder, they are often entropy-favored or spontaneous at higher temperatures, as entropy contributes positively to reducing \( \Delta G \). Conversely, reactions that result in more order tend to be entropy-disfavored, especially noticeable at higher temperatures where \( T\Delta S \) plays a larger role in determining spontaneity.

Chemical Equilibrium

Chemical equilibrium is the state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction, resulting in no net change in the concentration of reactants and products. At equilibrium, \( \Delta G \) reaches zero, denoting a balanced system.

• When chemical reactions reach equilibrium, they exist in the lowest possible energy state, allowing for energy conservation.
• Le Chatelier's principle helps predict changes, stipulating that a system at equilibrium will adjust to counteract any alterations in concentration, pressure, or temperature.

Understanding how to manipulate conditions to achieve equilibrium or how external changes shift equilibrium remains critical for managing industrial chemical processes and biological systems. Thus, equilibrium concepts are foundational in leveraging reaction conditions for optimal outcomes.