Write equilibrium constant expressions, \(K_{\mathrm{c}},\) for the reactions (a) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) (b) \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})\) (c) \(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{CO}_{3}^{2-}(\mathrm{aq}) \rightleftharpoons\) \(\mathrm{MgCO}_{3}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq})\)

Short Answer

Expert verified
The equilibrium constant expressions for the provided reactions are (a) \( K_c = \frac{{[\mathrm{NO_{2}}]^2}}{{[\mathrm{NO}]^2 [\mathrm{O_2}]}} \), (b) \( K_c = \frac{{[\mathrm{Zn}^{2+}]}}{{[\mathrm{Ag^{+}}]^2}} \), and (c) \( K_c = \frac{{[\mathrm{OH^{-}}]^2}}{{[\mathrm{CO_{3}^{2-}]}} \).

Step by step solution

01

Writing the Equilibrium Constant Expression for Reaction (a)

For the reaction \(2 \mathrm{NO(g)} + \mathrm{O_{2}(g)} \rightleftharpoons 2 \mathrm{NO_{2}(g)}\), the equilibrium constant expression would be \[ K_c = \frac{{[\mathrm{NO_{2}}]^2}}{{[\mathrm{NO}]^2 [\mathrm{O_2}]}} \] The concentrations of \(NO_{2}\), \(NO\) and \(O_{2}\) are raised to their stoichiometric coefficients.
02

Writing the Equilibrium Constant Expression for Reaction (b)

For the reaction \(\mathrm{Zn(s)} + 2 \mathrm{Ag^{+}(aq)} \rightleftharpoons \mathrm{Zn^{2+}(aq)} + 2 \mathrm{Ag(s)}\), write the equilibrium constant expression as \[ K_c = \frac{{[\mathrm{Zn}^{2+}]}}{{[\mathrm{Ag^{+}}]^2}} \] In this reaction, \(Zn\) is a pure solid and \(Ag\) is a pure metal; therefore, they are not included in the equilibrium expression.
03

Writing the Equilibrium Constant Expression for Reaction (c)

For the reaction \(\mathrm{Mg(OH)_{2}(s)} + \mathrm{CO_{3}^{2-}(aq)} \rightleftharpoons \mathrm{MgCO_{3}(s)} + 2 \mathrm{OH^{-}(aq)}\), the equilibrium constant expression would be \[ K_c = \frac{{[\mathrm{OH^{-}}]^2}}{{[\mathrm{CO_{3}^{2-}]}} \] Here, \(\mathrm{Mg(OH)_{2}}\) and \(\mathrm{MgCO_{3}}\) are both solids so they aren't included in the equilibrium expression.

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Most popular questions from this chapter

In your own words, define or explain the following terms or symbols: (a) \(K_{\mathrm{p}} ;\) (b) \(Q_{\mathrm{c}} ;\) (c) \(\Delta n_{\text {gas }}\)

A classic experiment in equilibrium studies dating from 1862 involved the reaction in solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) to produce ethyl acetate and water. $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}$$ The reaction can be followed by analyzing the equilibrium mixture for its acetic acid content. $$\begin{array}{r} 2 \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightleftharpoons \\ \mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In one experiment, a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol is brought to equilibrium. A sample containing exactly one-hundredth of the equilibrium mixture requires \(28.85 \mathrm{mL} 0.1000 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. Calculate the equilibrium constant, \(K_{c}\), for the ethanol-acetic acid reaction based on this experiment.

Can a mixture of \(2.2 \mathrm{mol} \mathrm{O}_{2}, 3.6 \mathrm{mol} \mathrm{SO}_{2},\) and \(1.8 \mathrm{mol}\) \(\mathrm{SO}_{3}\) be maintained indefinitely in a \(7.2 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=100\) in this reaction? Explain. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) $$

\(1.00 \times 10^{-3} \mathrm{mol} \mathrm{PCl}_{5}\) is introduced into a \(250.0 \mathrm{mL}\) flask, and equilibrium is established at \(284^{\circ} \mathrm{C}\) : \(\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) .\) The quantity of \(\mathrm{Cl}_{2}(\mathrm{g})\) present at equilibrium is found to be \(9.65 \times 10^{-4} \mathrm{mol}\) What is the value of \(K_{c}\) for the dissociation reaction at \(284^{\circ} \mathrm{C} ?\)

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