Problem 3 - Chemistry An Atoms First Approach

Question

Gas \(\mathrm{A}_{2}\) reacts with gas \(\mathrm{B}_{2}\) to form gas \(\mathrm{AB}\) at a constant temperature. The bond energy of AB is much greater than that of either reactant. What can be said about the sign of \(\Delta H ? \Delta S_{\text {surr }}\) ? \(\Delta S ?\) Explain how potential energy changes for this process. Explain how random kinetic energy changes during the process.

Step-by-Step Solution

Answer

The change in enthalpy (∆H) for the reaction A2 + B2 → AB is negative, as more stable, lower-energy AB bonds are formed, releasing energy. The entropy of surroundings (∆S_surr) is positive, as heat is absorbed by the surroundings, increasing their randomness. The system entropy (∆S) is negative, as the number of gaseous species decreases, reducing randomness. During the process, potential energy decreases due to the formation of stable bonds, releasing heat, while random kinetic energy decreases within the reaction but increases in the surroundings.

Step 1: 1. Determine the sign of ∆H

Since the bond energy of AB is greater than those of A2 and B2, the formation of AB is favored over its breakdown into A2 and B2. This means that the energy is released as the reaction progresses and more stable, lower-energy AB bonds are formed. Given that energy is released during the reaction, the change in enthalpy (∆H) will be negative.

Step 2: 2. Determine the sign of ∆S_surr

Given that the reaction releases heat (as evident by the negative ∆H), the heat will be absorbed by the surroundings. As heat is transferred to the surroundings, the entropy (∆S_surr) will increase. The increase in entropy of the surroundings results in a positive ∆S_surr.

Step 3: 3. Determine the sign of ∆S

For the reaction A2 + B2 → AB, the number of gaseous species decreases from two to one. This decrease in the number of species results in a decrease in randomness or disorder, which corresponds to a decrease in entropy (∆S) of the system. Therefore, the sign of ∆S for this reaction will be negative.

Step 4: 4. Potential energy changes during the process

The potential energy changes during this reaction can be described by the change in bond energy. As A2 and B2 react to form AB, the higher-energy bonds in A2 and B2 break while lower-energy, more stable bonds form in the AB molecule. The stability of the product molecule means that the potential energy of the system decreases. This energy decrease is released as heat, contributing to the negative ∆H.

Step 5: 5. Random kinetic energy changes during the process

As the reaction progresses, the number of gaseous species decreases, causing a decrease in the randomness or disorder within the system. This means that the random kinetic energy of the molecules is reduced as they form the more stable product. Although the energy from the bonds is released as heat, this energy is absorbed by the surroundings, leading to an increase in random kinetic energy in the surroundings rather than within the reaction itself.

Key Concepts

Enthalpy (ΔH)

In chemistry, enthalpy a measure of the total energy of a thermodynamic system, is a vital concept.During chemical reactions, enthalpy changes captured as \( \Delta H \) reveal whether a process absorbs or releases heat.In the reaction between gas \( \mathrm{A}_{2} \) and gas \( \mathrm{B}_{2} \)forming gas \( \mathrm{AB} \), the bond energy of the resulting AB bond far exceeds that of the original bonds.This stronger bond formation releases energy, which is why \( \Delta H \) is negative.A negative \( \Delta H \) indicates an exothermic reaction.Simply put, the reaction liberates energy as it forms more stable bonds.The concept of enthalpy helps us understand how energy conservation plays out during chemical processes. It's the direct measure of heat transfer involved in breaking and forming chemical bonds.

Entropy (ΔS)

Entropy, signified as \( \Delta S \), revolves around the degree of randomness or disorder within a system.When gas \( \mathrm{A}_{2} \) and gas \( \mathrm{B}_{2} \)combine to produce gas \( \mathrm{AB} \), there's a reduction in the number of gaseous molecules.This reduction from two molecules to one signifies a decrease in the system's disorder.Thus, \( \Delta S \) for the reaction is negative.In this context, we also discuss the surroundings.The energy released boosts the surroundings' entropy, leading to a positive \( \Delta S_{\text{surr}} \).Key points to remember:

• Reduction in system randomness results in negative \( \Delta S \).
• Increased randomness in surroundings leads to positive \( \Delta S_{\text{surr}} \).

Entropy offers insight into the spontaneous nature of processes and how energy distribution affects both system and environment.

Potential Energy changes in reactions

Potential energy in a chemical context relates to the stored energy within bonds.As the reaction between \( \mathrm{A}_{2} \) and \( \mathrm{B}_{2} \)transforms into \( \mathrm{AB} \), there’s a decrease in the potential energy.This occurs because weaker, high-energy bonds in the reactants are replaced by the stronger, low-energy bond in \( \mathrm{AB} \).The decrease in potential energy signifies a move to a more stable state, with the excess energy released as heat.Understanding potential energy shifts helps reveal why some reactions are spontaneous and others require added energy.Remember:

• Breaking high-energy bonds in \( \mathrm{A}_{2} \) and \( \mathrm{B}_{2} \)lowers the system's potential energy.
• Forming a strong \( \mathrm{AB} \) bond stabilizes and releases energy.

These dynamics illustrate why certain reactions are more energetically favorable and how potential energy maintains chemical stability.