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Consider the reaction $$ \mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{FeSCN}^{2+}(a q) $$ How will the equilibrium position shift if a. water is added, doubling the volume? b. \(\mathrm{AgNO}_{3}(a q)\) is added? (AgSCN is insoluble.) c. \(\mathrm{NaOH}(a q)\) is added? \(\left[\mathrm{Fe}(\mathrm{OH})_{3}\right.\) is insoluble.] d. \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) is added?

Short Answer

Expert verified
a. The equilibrium shifts to the reactants side due to the dilution of all species in the reaction. b. The equilibrium shifts to the left upon adding AgNO3(aq) to compensate for the loss of SCN⁻(aq) forming insoluble AgSCN. c. The equilibrium shifts to the left as NaOH(aq) reacts with Fe³⁺(aq) forming insoluble Fe(OH)₃. d. The equilibrium shifts to the right upon adding Fe(NO₃)₃(aq) as it increases the concentration of Fe³⁺(aq).

Step by step solution

01

Le Chatelier's Principle states that when the equilibrium system is disturbed by a change in concentration, pressure, or temperature, the system will counteract the change. In this case, the volume of the solution is doubled, causing a dilution of all the species in the reaction. The reaction will adjust by shifting to the side with more ions, which is the reactants side, since the total number of ions is higher on that side of the equation. #b. \(\mathrm{AgNO}_{3}(a q)\) is added. (AgSCN is insoluble.)

When \(\mathrm{AgNO}_{3}(a q)\) is added to the solution, it reacts with \(\mathrm{SCN}^{-}(a q)\) to form insoluble AgSCN(s), which removes \(\mathrm{SCN}^{-}(a q)\) from the equilibrium. According to Le Chatelier's Principle, the equilibrium will shift to counteract this change; in this case, the equilibrium will shift to the left to produce more \(\mathrm{SCN}^{-}(a q)\). #c. \(\mathrm{NaOH}(a q)\) is added. \(\left[\mathrm{Fe}(\mathrm{OH})_{3}\right.\) is insoluble.]
02

When \(\mathrm{NaOH}(a q)\) is added to the solution, it reacts with \(\mathrm{Fe}^{3+}(a q)\) to form insoluble \(\mathrm{Fe(OH)}_{3}(s)\), which removes \(\mathrm{Fe}^{3+}(a q)\) from the equilibrium. According to Le Chatelier's Principle, the equilibrium will shift to counteract this change; in this case, the equilibrium will shift to the left to produce more \(\mathrm{Fe}^{3+}(a q)\). #d. \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) is added.

When \(\mathrm{Fe(NO}_{3})_{3}(a q)\) is added to the solution, it increases the concentration of \(\mathrm{Fe}^{3+}(a q)\). According to Le Chatelier's Principle, the equilibrium will shift to counteract this change; in this case, the equilibrium will shift to the right to consume the added \(\mathrm{Fe}^{3+}(a q)\) and produce more \(\mathrm{FeSCN}^{2+}(a q)\).

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Most popular questions from this chapter

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$ \mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q) $$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10}\), what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

Explain the difference between \(K, K_{\mathrm{p}}\), and \(Q\).

At a particular temperature a 2.00-L flask at equilibrium contains \(2.80 \times 10^{-4} \mathrm{~mol} \mathrm{~N}_{2}, 2.50 \times 10^{-5} \mathrm{~mol} \mathrm{O}_{2}\), and \(2.00 \times 10^{-2}\) mol \(\mathrm{N}_{2} \mathrm{O} .\) Calculate \(K\) at this temperature for the reaction $$ 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2} \mathrm{O}(g) $$ If \(\left[\mathrm{N}_{2}\right]=2.00 \times 10^{-4} M,\left[\mathrm{~N}_{2} \mathrm{O}\right]=0.200 M\), and \(\left[\mathrm{O}_{2}\right]=0.00245\) \(M\), does this represent a system at equilibrium?

Consider the following reaction at a certain temperature: $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) $$ An equilibrium mixture contains \(1.0 \mathrm{~mol} \mathrm{Fe}, 1.0 \times 10^{-3} \mathrm{~mol} \mathrm{O}_{2}\), and \(2.0 \mathrm{~mol} \mathrm{Fe}_{2} \mathrm{O}_{3}\) all in a 2.0-L container. Calculate the value of \(K\) for this reaction.

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of \(4.5\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of \(9.0\) atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and \(\mathrm{b}\), does it matter from which direction an equilibrium position is reached?

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