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Gas \(\mathrm{A}_{2}\) reacts with gas \(\mathrm{B}_{2}\) to form gas \(\mathrm{AB}\) at a constant temperature. The bond energy of \(\mathrm{AB}\) 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.

Short Answer

Expert verified
In the given reaction, the change in enthalpy (ΔH) is negative, indicating an exothermic reaction. The change in entropy of the surroundings (ΔS_surr) is positive due to the release of heat, and the change in entropy (ΔS) for the reaction is negative as the number of gas particles decreases. The potential energy decreases as stronger bonds are formed in the product, while the random kinetic energy increases in the surroundings and reduces within the system.

Step by step solution

01

Determine the sign of ΔH

Since the bond energy of the product AB is much greater than that of the reactants A₂ and B₂, the formation of the product is associated with the release of energy. The energy released when the bonds in the products are formed is more significant than the energy required to break the bonds in the reactants. As a result, the change in enthalpy (ΔH) for the reaction will be negative, indicating an exothermic reaction.
02

Determine the sign of ΔS_surr

For an exothermic reaction, heat is released into the surroundings. This heat transfer increases the randomness or disorder of the surroundings. Therefore, the change in entropy of the surroundings (ΔS_surr) is positive.
03

Determine the sign of ΔS

During the reaction, two moles of gas (A₂ and B₂) combine to form one mole of gas (AB). This process decreases the number of gas particles and, thus, reduces the overall randomness or disorder in the system. Therefore, the change in entropy (ΔS) for the reaction is negative.
04

Explain the change in potential energy

As the reactants A₂ and B₂ form the product AB, new bonds are formed with greater bond energy. Potential energy is stored in these bonds. Since the bond energy of AB is much greater than that of A₂ and B₂, the overall potential energy of the system decreases during the reaction, as the reactants are transformed into a more stable product with stronger bonds.
05

Explain the change in random kinetic energy

During the reaction, energy is released due to the formation of stronger bonds in the product AB. This released energy increases the random kinetic energy of the particles in the surroundings by increasing their movement and making them more disordered. However, within the system, the reduction in the number of gas particles reduces the overall random kinetic energy.

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Key Concepts

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

Change in Enthalpy
Understanding the concept of enthalpy, denoted as \( \Delta H \) is crucial when analyzing chemical reactions. Enthalpy represents the total heat content of a system. A negative \( \Delta H \) implies that a reaction is exothermic—meaning it releases heat to its surroundings. In the reaction between \( \mathrm{A}_{2} \) and \( \mathrm{B}_{2} \) to form \( \mathrm{AB} \), the high bond energy of the product indicates that the overall heat content decreases during the reaction. Simply put, since forming \( \mathrm{AB} \) releases more energy than what’s needed to break the reactants' bonds, there’s a net energy release, associated with a negative change in enthalpy.

For students, imagine mixing ingredients to bake a cake. If mixing releases warmth, making the kitchen cozier, it’s akin to a negative \( \Delta H \) because energy is released during the mixing process. In chemical terms, the system gives off heat, resulting in a decrease in enthalpy.
Change in Entropy
Entropy, symbolized as \( \Delta S \), is a measure of disorder or randomness in a system. When assessing the change in entropy, important considerations include the nature and number of particles before and after the reaction. In our reaction, the conversion from two moles of gas to a single mole reduces the number of ways the gas particles can be arranged—thus reducing entropy, leading to a negative \( \Delta S \).

Imagine our previous cake analogy, but now focus on how the individual ingredients (like eggs and flour) are arranged. Initially, they are separate with many possible layouts, but once mixed, there’s only one form—the batter—signifying less randomness. The gas molecules in the system follow suit, with their randomness decreasing as they combine to form a less dispersed \( \mathrm{AB} \) molecule.
Exothermic Reactions
Exothermic reactions, such as our \( \mathrm{A}_{2} \) and \( \mathrm{B}_{2} \) combining to form \( \mathrm{AB} \), are characterized by the release of heat. This is an intuitive concept when you think of a bonfire: wood (the reactant) burns to produce ash and heat (the products), warming your hands. The same principle applies to chemical reactions. When products like \( \mathrm{AB} \) have stronger bonds than the reactants, energy is given off as these new bonds form. For learners, recognizing that exothermic reactions result in a temperature increase in the surroundings can help correlate the theoretical concept of \( \Delta H \) with a tangible warm sensation.
Bond Energy
Bond energy is a pivotal concept in understanding chemical reactions. It is the amount of energy required to break one mole of bonds in gaseous molecules under standard conditions. Stronger bonds have high bond energies and conversely, weaker bonds have lower bond energies. In our reaction, \( \mathrm{AB} \) has a higher bond energy than either \( \mathrm{A}_{2} \) or \( \mathrm{B}_{2} \), indicative of stronger bonds within the \( \mathrm{AB} \) molecule. For students, visualizing bond energy can be likened to the strength of a chain—in molecular terms, a more substantial chain (or stronger bond) requires more energy to break apart, thus having higher bond energy.

Think of each bond as a tiny spring holding atoms together; stronger springs (bonds) need more energy (force) to be stretched or snapped.
Random Kinetic Energy
Random kinetic energy refers to the energy that particles possess due to their random motion. It is an intrinsic part of a system’s total energy and is influenced by temperature and the number of particles. During our chemical reaction, when \( \mathrm{A}_{2} \) and \( \mathrm{B}_{2} \) form \( \mathrm{AB} \), there is a release of energy that increases the random kinetic energy of the surrounding particles. In the system itself, however, combining to form \( \mathrm{AB} \) reduces the number of particles and therefore the randomness of their motion. To visualize this, consider children playing in a playground: the more kids there are, and the faster they move, the more energy there is in random motion. In chemistry, particles in a substance behave similarly—their number and speed define the system's random kinetic energy.

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Most popular questions from this chapter

Given the following data: $$ \begin{aligned} 2 \mathrm{H}_{2}(g)+\mathrm{C}(s) & \longrightarrow \mathrm{CH}_{4}(g) & & \Delta G^{\circ}=-51 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & \Delta G^{\circ}=-474 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g) & & \Delta G^{\circ}=-394 \mathrm{~kJ} \end{aligned} $$ Calculate \(\Delta G^{\circ}\) for \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(l)\).

Consider a weak acid, HX. If a \(0.10-M\) solution of HX has a \(\mathrm{pH}\) of \(5.83\) at \(25^{\circ} \mathrm{C}\), what is \(\Delta G^{\circ}\) for the acid's dissociation reaction at \(25^{\circ} \mathrm{C} ?\)

Consider the following reaction: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of \(\pm 1\) in all quantities): a. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{~atm}\) b. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=200 \mathrm{~atm}, P_{\mathrm{H}_{2}}=600 \mathrm{~atm}\), \(P_{\mathrm{NH}_{3}}=200 \mathrm{~atm}\)

If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction $$ \mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g) $$ \(\Delta H^{\circ}\) for this reaction is \(79.14 \mathrm{~kJ} / \mathrm{mol}\) in the temperature range of 25 to \(125^{\circ} \mathrm{C}\). Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C}\), calculate the partial pressure of \(\mathrm{CO}_{2}\) necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110 .{ }^{\circ} \mathrm{C}\). (Hint: Manipulate the equation in Exercise 79.)

What information can be determined from \(\Delta G\) for a reaction? Does one get the same information from \(\Delta G^{\circ}\), the standard free energy change? \(\Delta G^{\circ}\) allows determination of the equilibrium constant \(K\) for a reaction. How? How can one estimate the value of \(K\) at temperatures other than \(25^{\circ} \mathrm{C}\) for a reaction? How can one estimate the temperature where \(K=1\) for a reaction? Do all reactions have a specific temperature where \(K=1\) ?

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