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Chapter 14: Problem 34

Explain why reactions with large equilibrium constants, such as the formation of rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right),\) may have very slow rates.

Short Answer

Expert verified
Large equilibrium constants indicate a favorable reaction, however, the rate of reactions is determined by kinetic factors such as activation energy. Therefore, reactions like rust formation, despite having a large equilibrium constant, can have a slow rate due to a high activation energy.

Step by step solution

01

Understanding the Nature of Equilibrium Constants

Equilibrium constants (K) measure the balance of product and reactant concentrations at equilibrium. A large K suggests that there are more products than reactants at equilibrium, indicating that the reaction is very favorable and goes almost to completion.
02

Understanding Reaction Rates

Reaction rates are a measure of the speed of a reaction. It is determined by how quickly the reactants are turned into products. The rate is influenced by factors such as the presence of a catalyst, temperature, pressure, and the concentrations of the reactants.
03

Connecting Equilibrium Constants and Reaction Rates

Although a large equilibrium constant means the reaction is favorable, it doesn't necessarily mean it will occur rapidly. The reaction rate can be slow if the reaction barrier or activation energy is high. In the case of rust formation, the reaction is energetically favorable (high K), but the activation energy is high, so the reaction rate is slow. This is why metal objects do not rust immediately in the presence of oxygen, but over time.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\), a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are \(\left[\mathrm{NO}_{2}\right]=0.0475 \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.487 \mathrm{M} .\) The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) is \(4.63 \times 10^{-3} . \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is colorless and \(\mathrm{NO}_{2}(g)\) has a brown color. \(]\)

When a gas was heated under atmospheric conditions, its color deepened. Heating above \(150^{\circ} \mathrm{C}\) caused the color to fade, and at \(550^{\circ} \mathrm{C}\) the color was barely detectable. However, at \(550^{\circ} \mathrm{C},\) the color was partially restored by increasing the pressure of the system. Which of the following best fits the above description? Justify your choice. (a) A mixture of hydrogen and bromine, (b) pure bromine, (c) a mixture of nitrogen dioxide and dinitrogen tetroxide. (Hint: Bromine has a reddish color and nitrogen dioxide is a brown gas. The other gases are colorless.)

Define equilibrium. Give two examples of a dynamic equilibrium.

The equilibrium constant \(K_{P}\) for the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ is 1.05 at \(250^{\circ} \mathrm{C}\). The reaction starts with a mixture of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) at pressures \(0.177 \mathrm{~atm},\) 0.223 atm, and 0.111 atm, respectively, at \(250^{\circ} \mathrm{C}\). When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why.

Consider the reaction $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ At \(430^{\circ} \mathrm{C},\) an equilibrium mixture consists of 0.020 mole of \(\mathrm{O}_{2}, 0.040\) mole of \(\mathrm{NO},\) and 0.96 mole of \(\mathrm{NO}_{2}\). Calculate \(K_{P}\) for the reaction, given that the total pressure is 0.20 atm.
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