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Chapter 16: Problem 3

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.

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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.
04

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.
05

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.

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Most popular questions from this chapter

Consider the reaction: $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=-92.50 \mathrm{kJ}\) Which of the following statements is(are) true? a. This is an endothermic reaction. b. \(\Delta S^{\circ}\) for this reaction is negative. c. If the temperature is increased, the ratio \(\frac{\mathrm{PCl}_{5}}{\mathrm{PCl}_{3}}\) will increase. d. \(\Delta G^{\circ}\) for this reaction has to be negative at all temperatures. e. When \(\Delta G^{\circ}\) for this reaction is negative, then \(K_{\mathrm{p}}\) is greater than 1.00.

Given the following data: $$2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\Delta G^{\circ}=-6399 \mathrm{kJ}$$ $$\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta G^{\circ}=-394 \mathrm{kJ}$$ $$\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta G^{\circ}=-237 \mathrm{kJ}$$ calculate \(\Delta G^{\circ}\) for the reaction $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$

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 \(K=1 ?\)

The standard free energies of formation and the standard enthalpies of formation at \(298 \mathrm{K}\) for diffuoroacetylene \(\left(\mathrm{C}_{2} \mathrm{F}_{2}\right)\) and hexafluorobenzene \(\left(\mathrm{C}_{6} \mathrm{F}_{6}\right)\) are $$\begin{array}{ccc} & \Delta G_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & \Delta H_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) \\ \hline \mathrm{C}_{2} \mathrm{F}_{2}(g) & 191.2 & 241.3 \\\ \mathrm{C}_{6} \mathrm{F}_{6}(g) & 78.2 & 132.8 \end{array}$$ For the following reaction: $$\mathrm{C}_{6} \mathrm{F}_{6}(g) \rightleftharpoons 3 \mathrm{C}_{2} \mathrm{F}_{2}(g)$$ a. calculate \(\Delta S^{\circ}\) at \(298 \mathrm{K}\). b. calculate \(K\) at 298 K. c. estimate \(K\) at \(3000 .\) K, assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

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.)
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