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Chapter 15: Problem 47

One important reaction in the citric acid cycle is citrate(aq) \(\rightleftharpoons\) aconitate \((\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad K=0.031\) Write the equilibrium constant expression for the above reaction. Given that the concentrations of \([\text { citrate }(\mathrm{aq})]=0.00128 \mathrm{M},[\text { aconitate }(\mathrm{aq})]=4.0 \times\) \(10^{-5} \mathrm{M},\) and \(\left[\mathrm{H}_{2} \mathrm{O}\right]=55.5 \mathrm{M},\) calculate the reaction quotient. Is this reaction at equilibrium? If not, in which direction will it proceed?

Short Answer

Expert verified
The reaction quotient (\(Q\)) for the given reaction is calculated to be 1.738. As this is greater than the equilibrium constant (\(K = 0.031\)), the reaction is not at equilibrium and will proceed in the reverse direction.

Step by step solution

01

Write the equilibrium constant expression

The equilibrium constant expression for a chemical reaction is defined as the product of the concentrations of the products raised to their stoichiometric coefficients, divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. In the reaction given, citrate is the reactant and aconitate and water are the products. Therefore, the equilibrium constant expression \(K\) for the reaction is \(K = \frac{[aconitate][H_{2}O]}{[citrate]}\).
02

Calculate the reaction quotient

The reaction quotient \(Q\) is calculated the same way as the equilibrium constant: it is the ratio of the products' concentrations to the reactants' concentrations. Given the concentrations of citrate, aconitate, and water, we have \(Q = \frac{[aconitate][H_{2}O]}{[citrate]} = \frac{(4.0 \times 10^{-5})(55.5)}{0.00128}\). Solving this gives \(Q = 1.738\).
03

Determine if the reaction is at equilibrium

To determine if a reaction is at equilibrium, we compare the reaction quotient \(Q\) to the equilibrium constant \(K\). If \(Q = K\), the reaction is at equilibrium. If \(Q > K\), the reaction will proceed in the reverse direction to attain equilibrium. If \(Q < K\), the reaction will proceed in the forward direction to reach equilibrium. In this case, \(Q = 1.738\) and \(K = 0.031\). Since \(Q > K\), the reaction will proceed in the reverse direction to reach equilibrium.

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Most popular questions from this chapter

Write equilibrium constant expressions, \(K_{\mathrm{p}},\) for the reactions (a) \(\mathrm{CS}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) (b) \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaHCO}_{3}(\mathrm{s}) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

The decomposition of \(\mathrm{HI}(\mathrm{g})\) is represented by the equation $$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g})$$ \(\mathrm{HI}(\mathrm{g})\) is introduced into five identical \(400 \mathrm{cm}^{3}\) glass bulbs, and the five bulbs are maintained at \(623 \mathrm{K}\) Each bulb is opened after a period of time and analyzed for \(I_{2}\) by titration with \(0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq})\) $$\begin{array}{l} \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \longrightarrow \\ \quad \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(\mathrm{aq})+2 \mathrm{NaI}(\mathrm{aq}) \end{array}$$ Data for this experiment are provided in the table below. What is the value of \(K_{\mathrm{c}}\) at \(623 \mathrm{K} ?\) $$\begin{array}{llll} \hline & & & \text { Volume } \\ & \text { Initial } & \text { Time } & 0.0150 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \\ \text { Bulb } & \text { Mass of } & \text { Bulb } & \text { Required for } \\\ \text { Number } & \mathrm{Hl}(\mathrm{g}), \mathrm{g} & \text { Opened, } \mathrm{h} & \text { Titration, in } \mathrm{mL} \\ \hline 1 & 0.300 & 2 & 20.96 \\ 2 & 0.320 & 4 & 27.90 \\ 3 & 0.315 & 12 & 32.31 \\ 4 & 0.406 & 20 & 41.50 \\ 5 & 0.280 & 40 & 28.68 \\ \hline \end{array}$$

A sample of pure \(\mathrm{PCl}_{5}(\mathrm{g})\) is introduced into an evacuated flask and allowed to dissociate. $$ \mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ If the fraction of \(\mathrm{PCl}_{5}\) molecules that dissociate is denoted by \(\alpha,\) and if the total gas pressure is \(P\) show that $$ K_{\mathrm{p}}=\frac{\alpha^{2} P}{1-\alpha^{2}} $$

Starting with \(0.280 \mathrm{mol} \mathrm{SbCl}_{3}\) and \(0.160 \mathrm{mol} \mathrm{Cl}_{2},\) how many moles of \(\mathrm{SbCl}_{5}, \mathrm{SbCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present when equilibrium is established at \(248^{\circ} \mathrm{C}\) in a 2.50 L flask? $$\begin{aligned} \mathrm{SbCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 2.5 \times 10^{-2} \mathrm{at} \ 248^{\circ} \mathrm{C} \end{aligned}$$

Explain why the percent of molecules that dissociate into atoms in reactions of the type \(\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) always increases with an increase in temperature.
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