Problem 13 - Chemistry: Principles and Reactions

Question

Predict the sign of \(\Delta S^{\circ}\) for each of the following reactions. (a) \(\mathrm{O}_{3}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) (b) \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{5}(g)\) (c) \(\mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s)\)

Step-by-Step Solution

Answer

Question: Predict the sign of the entropy change (∆S°) for each of the following reactions: (a) O3(g) ⟶ O2(g)+O(g) (b) PCl3(g)+Cl2(g) ⟶ PCl5(g) (c) CuSO4(s)+5 H2O(l) ⟶ CuSO4 · 5 H2O(s) Answer: (a) The sign of ∆S° for this reaction is positive. (b) The sign of ∆S° for this reaction is negative. (c) The sign of ∆S° for this reaction is negative.

Step 1: Reaction (a)

For the reaction \(\mathrm{O}_{3}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\), we can see that there is 1 gaseous reactant and 2 gaseous products. The number of particles increases from the reactant to the products, and therefore, the entropy increases. This means the sign of \(\Delta S^\circ\) will be positive.

Step 2: Reaction (b)

For the reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g)\), there are 2 gaseous reactants and 1 gaseous product. The number of particles decreases from the reactants to the products, making the system more ordered. Thus, the sign of \(\Delta S^\circ\) will be negative.

Step 3: Reaction (c)

For the reaction \(\mathrm{CuSO}_{4}(s)+5 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s)\), there is a solid reactant and liquid reactants coming together to form a more ordered solid product. Since there is a change from less ordered states (solid and liquid) to a more ordered state (solid), the system becomes more ordered. Consequently, the sign of \(\Delta S^\circ\) will be negative.

Key Concepts

Chemical Thermodynamics

Chemical thermodynamics is the study of energy changes accompanying chemical reactions and physical processes. A key aspect of this field is understanding how energy and matter behave at the molecular level. This enables prediction of reaction spontaneity, equilibrium states, and energy flow.Some important topics in chemical thermodynamics include:

• Enthalpy (\( H \)) - the heat content of a system at constant pressure.
• Entropy (\( S \)) - the measure of disorder or randomness in a system.
• Gibbs Free Energy (\( G \)) - the energy associated with a chemical reaction that can be used to perform work.

In our exercise, predicting the sign of \( \Delta S^\circ \) helps us understand changes in disorder during reactions. For example, breaking ozone into oxygen gas and atomic oxygen increases entropy due to more particles resulting from the reaction. Thus, \( \Delta S^\circ \) is positive. In contrast, the formation of phosphorus pentachloride decreases particle number, leading to negative entropy change as the system becomes more ordered.

Reaction Mechanisms

Reaction mechanisms provide detailed steps involved in a chemical reaction, showing how reactants transform into products. Each step within a mechanism outlines collisions, bond breakages, and formations involved in the overall reaction. Understanding these mechanisms is crucial for predicting:

• Reaction rates and catalysis effects - which affect how quickly a chemical reaction occurs.
• Intermediates formed - temporary species appearing during the process.
• Reaction pathways - alternative routes a reaction may take.

In the given exercise, analyzing the reactions' mechanisms helps predict disorder changes. For the reaction where ozone transforms into oxygen and an oxygen atom, the mechanism involves breaking bonds, which results in increased particle number and thus higher entropy. On the other hand, combining chlorine gas and phosphorus trichloride to form phosphorus pentachloride creates fewer gaseous particles, thus indicating a more organized system.

Gibbs Free Energy

Gibbs Free Energy is an important concept in chemical thermodynamics, combining enthalpy and entropy to predict reaction spontaneity. It is represented by:\[ G = H - TS \]where \( H \) is enthalpy, \( T \) is temperature, and \( S \) is entropy.Key points about Gibbs Free Energy include:

• If \( \Delta G < 0 \) - the process is spontaneous.
• If \( \Delta G = 0 \) - the system is at equilibrium.
• If \( \Delta G > 0 \) - the process is non-spontaneous.

Examining \( \Delta G \) provides insight into the balance between enthalpy and entropy effects during a reaction.For the reactions in the exercise, video analysis of entropy change contributes to understanding the potential spontaneity of these reactions. A reaction resulting in increased entropy, such as the decomposition of ozone, may be spontaneous under certain conditions, as it potentially reduces \( \Delta G \). Conversely, a reaction decreasing entropy, like forming phosphorus pentachloride, may require energy input, making it non-spontaneous unless driven by other factors such as high enthalpy reduction.