Write the reaction quotients \(Q_{c}\) for (a) \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{~s}) \rightarrow \mathrm{Fe}^{3}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq})\) (b) \(\mathrm{CuSO}_{4}(\mathrm{~s})+5 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (s) (c) \(\mathrm{ZnO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{Zn}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\)

Short Answer

Expert verified
\(Q_{c}\) for (a) is \([\mathrm{Fe}^{3+}][\mathrm{OH}^{-}]^3\). There is no \(Q_{c}\) defined for (b) as there are no aqueous or gas phase reactants/products. \(Q_{c}\) for (c) is \(\frac{P_{\mathrm{CO}_2}}{P_{\mathrm{CO}}}\).

Step by step solution

01

Identify Relevant Species

For reaction quotient calculations, include only the concentration of aqueous solutions and the partial pressure of gases. Solid and pure liquid concentrations remain constant and are not included.
02

Write Reaction Quotient for (a)

Using the reaction for (a), the reaction quotient, \(Q_c\), is defined as the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the reactants raised to their stoichiometric coefficients. Since \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) is a solid, it is not included in \(Q_c\). The expression is \[Q_c = [\mathrm{Fe}^{3+}] [\mathrm{OH}^{-}]^3\]
03

Write Reaction Quotient for (b)

For (b), \(CuSO_4(\mathrm{s})\) and \(CuSO_4 \cdot 5H_2O(\mathrm{s})\) are both solids, so the reaction quotient is not defined. The reaction does not contain any species in aqueous solutions or gases, hence \(Q_c\) does not apply.
04

Write Reaction Quotient for (c)

For (c), \(ZnO(\mathrm{s})\) and \(Zn(\mathrm{s})\) are solids and are not included. Only \(CO(\mathrm{g})\) and \(CO_2(\mathrm{g})\) are gases so they will be included. The expression for the reaction quotient is \[Q_c = \frac{P_{\mathrm{CO}_2}}{P_{\mathrm{CO}}}\]

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. In simpler terms, at equilibrium, the amounts of reactants and products remain constant over time, even though the reactions continue to occur. This balance does not mean that the reactants and products are present in equal amounts but rather that their concentrations have stabilized and are not changing with time.

Understanding chemical equilibrium is fundamental when studying reactions, as it can influence the outcome of a chemical process. In an equilibrium state, the reaction quotient (\(Q\textsubscript{c}\) or \(Q\textsubscript{p}\) for pressure) will equal the equilibrium constant (\(K\textsubscript{c}\) or \(K\textsubscript{p}\)) of the reaction. This equilibrium constant is a reflection of the ratio of the concentrations of products to reactants at equilibrium. Every different reaction at given conditions has a unique equilibrium constant that can indicate the position of the equilibrium and the extent to which reactants are converted to products.
Stoichiometry
Stoichiometry is a section of chemistry that involves the calculation of relative quantities of reactants and products in chemical reactions. It's grounded in the law of conservation of mass, where the total mass of reactants equals the total mass of products. In a balanced chemical equation, the stoichiometric coefficients represent the number of molecules (or moles) involved in the reaction for each substance.

To accurately calculate the reaction quotient (\(Q\textsubscript{c}\)) or the equilibrium constant (\(K\textsubscript{c}\)), one needs to apply the principles of stoichiometry. Each concentration (in mol/L) of a substance in the reaction quotient expression is raised to the power of its stoichiometric coefficient. It is crucial in computing the correct values for both equations and predictions regarding the direction of the reaction. Knowing the proper stoichiometry is essential for students in predicting product formation and for industries to optimize the production and yield of chemical processes.
Equilibrium Constant
The equilibrium constant, represented as (\(K\textsubscript{c}\) for concentrations or \(K\textsubscript{p}\) for pressures), is a number that expresses the relationship between the amounts of products and reactants at the point of chemical equilibrium for a reversible reaction at a constant temperature. The equilibrium constant provides insight into the position of equilibrium and the extent of the reaction: a large value of (\(K\textsubscript{c}\)) or (\(K\textsubscript{p}\)) indicates that, at equilibrium, a greater concentration of products is present compared to reactants, suggesting the reaction favors the production of products. Conversely, a small value points to a higher concentration of reactants.

Understanding how to calculate and apply the equilibrium constant is invaluable when predicting the outcome of a reaction. When you calculate the reaction quotient (\(Q\textsubscript{c}\)) during a reaction that hasn't yet reached equilibrium, you can predict which way the reaction will proceed to reach equilibrium; if \(Q\textsubscript{c} < K\textsubscript{c}\), the reaction will proceed forward to produce more products, and if \(Q\textsubscript{c} > K\textsubscript{c}\), the reaction will proceed in the reverse direction to produce more reactants.

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Most popular questions from this chapter

A mixture consisting of \(1.1 \mathrm{mmol} \mathrm{SO}_{2}\) and \(2.2 \mathrm{mmol} \mathrm{O}_{2}\) in a \(250-\mathrm{mL}\) container was heated to \(500 \mathrm{~K}\) and allowed to reach equilibrium. Will more sulfur trioxide be formed if that equilibrium mixture is cooled to \(25^{\circ} \mathrm{C}\) ? For the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\) \(\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), \mathrm{K}=2.5 \times 10^{10}\) at \(500 \mathrm{~K}\) and \(4.0 \times 10^{24}\) at \(25^{\circ} \mathrm{C}\).

Write the equilibrium expressions \(K\) for the following reactions. (a) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{NO}_{2}(\mathrm{~g})\) (b) \(\mathrm{SbCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g}) \Rightarrow \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})\)

Consider the equilibrium \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g})=\) \(\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). (a) If the partial pressure of \(\mathrm{CO}_{2}\) is increased, what happens to the partial pressure of \(\mathrm{CH}_{4}\) ? (b) If the partial pressure of \(\mathrm{CH}_{4}\) is decreased, what happens to the partial pressure of \(\mathrm{CO}_{2}\) ? (c) If the concentration of \(\mathrm{CH}_{4}\) is increased, what happens to the equilibrium constant for the reaction? (d) If the concentration of \(\mathrm{H}_{2} \mathrm{O}\) is decreased, what happens to the concentration of \(\mathrm{CO}_{2}\) ?

Explain why the following equilibria are heterogencous and write the reaction quoticnt \(Q\) for each one. (a) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})=\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})\) (b) \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{OH}_{2} \mathrm{O}\) (s) \(\rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{~s})+10 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(2 \mathrm{KCO}_{3}\) (s) \(\Rightarrow 2 \mathrm{KCl}\) (s) \(+3 \mathrm{O}_{2}\) (g)

The equilibrium constant \(K=3.5 \times 10^{4}\) for the reaction \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})=\mathrm{PCl}_{5}(\mathrm{~g})\) at \(760^{\circ} \mathrm{C}\). At equilibrium, the partial pressure of \(\mathrm{PCl}_{5}\) was \(2.2 \times\) \(10^{4}\) bar and that of \(\mathrm{PCl}_{3}\) was \(1.33\) bar. What was the equilibrium partial pressure of \(\mathrm{Cl}_{2}\) ?

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