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Chapter 15: Problem 64

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?

Short Answer

Expert verified
(a) The reaction is endothermic, as increasing the temperature favors product formation. (b) The balanced chemical equation has more molecules on the product side, as increasing the volume favors product formation.

Step by step solution

01

Examine the effect of temperature change

The reaction has an increased fraction of products when the temperature is increased. According to Le Chatelier's principle, increasing the temperature of an endothermic reaction will favor the formation of products, as heat is absorbed in the process. On the other hand, increasing the temperature of an exothermic reaction will favor the formation of reactants, as heat is released in the process. Therefore, the given reaction must be endothermic, as increasing the temperature favors product formation. Answer (a): The reaction is endothermic.
02

Examine the effect of volume change

The reaction also has an increased fraction of products when the volume of the reaction vessel is increased. According to Le Chatelier's principle, increasing the volume of a reaction vessel will favor the side of the reaction with more gas molecules, as the pressure is decreased. In order to determine which side of the balanced chemical equation has more molecules, we can apply this information to the given reaction. Since increasing the volume favors product formation, the product side must have more gas molecules than the reactant side. Answer (b): The balanced chemical equation has more molecules on the product side.

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Most popular questions from this chapter

For the equilibrium $$2 \operatorname{IBr}(g) \Longrightarrow \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C} .\) If 0.025 atm of IBr is placed in a 2.0 -L container, what is the partial pressure of all substances after equilibrium is reached?

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Consider the reaction $$\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)$$ At \(25^{\circ} \mathrm{C},\) the equilibrium constant is \(K_{c}=2.4 \times 10^{-5}\) for this reaction. (a) If excess CaSO \(_{4}(s)\) is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4},\) what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) ? (b) If the resulting solution has a volume of \(1.4 \mathrm{L},\) what is the minimum mass of \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium?

Consider the equilibrium $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at 298 \(\mathrm{K} )\) : \begin{equation} \begin{array}{l}{2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{NOBr}(g) \quad K_{c}=2.0} \\ {2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{c}=2.1 \times 10^{30}}\end{array} \end{equation}

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