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Chapter 14: Problem 75

One way of preparing hydrogen is by decomposition of water: \(2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta E_{\mathrm{rxn}}=484 \mathrm{~kJ}\) (a) Would you expect the decomposition to be more complete at equilibrium when it is run at high temperature or when it is run at low temperature? Explain. (b) According to your answer to (a), would the reaction speed up, slow down, or occur at the same rate as before the temperature was changed? Explain.

Short Answer

Expert verified
(a) The decomposition of water will be more complete at equilibrium at high temperatures, as it is an endothermic reaction and the system favors the endothermic reaction to absorb the additional heat. (b) The reaction will speed up at higher temperatures, as the increased kinetic energy of reactant molecules leads to more frequent and energetic collisions, resulting in an increased likelihood of forming product molecules.

Step by step solution

01

(a) Identifying equilibrium shift

For an endothermic reaction, heat can be considered as a reactant. According to Le Chatelier's principle, if a stress is applied to a system at equilibrium, the system will adjust to counteract the applied stress. In this case, if the temperature increases, the system will respond by favoring the endothermic reaction to absorb this additional heat. In the given decomposition reaction, since it is endothermic (\( \Delta E_{\mathrm{rxn}}=484 \mathrm{~kJ}\)), an increase in temperature will shift the equilibrium towards the products side. Therefore, at high temperatures, the decomposition of water will be more complete at equilibrium.
02

(b) Effect of temperature on reaction speed

The rate of a chemical reaction depends on the temperature of the system, according to the collision theory and Arrhenius equation. As the temperature increases, the kinetic energy of reactant molecules increases, leading to more frequent and energetic collisions between them. This results in an increased likelihood of forming product molecules. Therefore, as we increase the temperature, the decomposition reaction will speed up.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\), the solubility of calcium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4}\), in water is \(6.1 \mathrm{mg} / \mathrm{L}\). (a) What are the equilibrium molar concentrations of \(\mathrm{Ca}^{2+}(a q)\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) ?\) (b) Calculate \(K_{\mathrm{sp}}\) for calcium oxalate.

Consider the gas-state reaction \(2 \mathrm{~A}(g)+3 \mathrm{~B}(g) \rightleftarrows \mathrm{C}(g)+\mathrm{D}(g)\) (a) Write the equilibrium constant expression for the reaction. (b) Write the reaction in reverse. (c) Write the equilibrium constant expression for the reverse reaction that you wrote for part (b). (d) Compare your answers to \((\mathrm{a})\) and \((\mathrm{c})\). What conclusion can you draw from the comparison? (e) Suppose \(K_{\mathrm{eq}}\) for a reaction \(=10.0 .\) What will the value of \(K_{\text {eq }}\) be for the reverse reaction?

Sometimes reactions are written with two arrows pointing in opposite directions instead of a single arrow going from reactants to products. What do the two arrows mean?

Write the expression for \(K_{\text {eq }}\) for the reaction \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

The saturation solubility of \(\mathrm{Ag}_{2} \mathrm{~S}\) at \(25^{\circ} \mathrm{C}\) is \(1.14 \times 10^{-17} \mathrm{M}\). What are the equilibrium concentrations of the cation and anion?
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