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Chapter 13: Problem 92

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. \(\operatorname{AgNO}_{3}(a q)\) is added? (AgSCN is insoluble.) c. \(\mathrm{NaOH}(a q)\) is added? [Fe(OH) \(_{3}\) is insoluble. \(]\) d. Fe(NO \(_{3} )_{3}(a q)\) is added?

Short Answer

Expert verified
a. The equilibrium position does not shift, but concentrations of each species will be halved. b. The equilibrium shifts to the left. c. The equilibrium shifts to the left. d. The equilibrium shifts to the right.

Step by step solution

01

a. Water is added, doubling the volume

According to Le Châtelier's principle, when the volume is increased, the equilibrium will shift towards the side with more moles of gas (or in this case, higher combined concentration of reactants or products). In this case, the number of moles on both sides of the reaction is the same, so the reaction quotient (Q) remains unchanged. The equilibrium position does not shift, but the concentrations of each species will be halved.
02

b. AgNO\(_{3}\)(aq) is added

When AgNO\(_{3}\)(aq) is added, it will react with SCN\(^-\) to form an insoluble AgSCN solid. This decreases the concentration of SCN\(^-\) in the solution. According to Le Châtelier's principle, the equilibrium will shift to counteract the change, and it will shift towards the reactant side to attempt to increase the concentration of SCN\(^-\) and maintain the equilibrium. So, the equilibrium will shift to the left.
03

c. NaOH(aq) is added

When NaOH(aq) is added, it will react with Fe\(^{3+}\) to form an insoluble Fe(OH)\(_{3}\) solid. This decreases the concentration of Fe\(^{3+}\) in the solution. According to Le Châtelier's principle, the equilibrium will try to counteract this change, so it will shift towards the reactant side attempt to increase the concentration of Fe\(^{3+}\). Thus, the equilibrium will shift to the left.
04

d. Fe(NO\(_{3}\))\(_{3}\)(aq) is added

By adding Fe(NO\(_{3}\))\(_{3}\)(aq) to the solution, the concentration of Fe\(^{3+}\) will increase. According to Le Châtelier's principle, the equilibrium will try to counteract this change, so it will shift towards the product side attempt to decrease the concentration of Fe\(^{3+}\). Therefore, the equilibrium will shift to the right.

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Most popular questions from this chapter

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ equilibrium constants $\left(K_{\mathrm{p}} \text { values ) as a function of temperature }\right.\( are\)\begin{array}{ll}{300^{\circ} \mathrm{C},} & {4.34 \times 10^{-3}} \\ {500^{\circ} \mathrm{C},} & {1.45 \times 10^{-5}} \\\ {600^{\circ} \mathrm{C},} & {2.25 \times 10^{-6}}\end{array}$ Is the reaction exothermic or endothermic?

At \(1100 \mathrm{K}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ What is the value of \(K\) at this temperature?

A 4.72 -g sample of methanol (CH_ 3 \(\mathrm{OH}\) ) was placed in an otherwise empty \(1.00-\mathrm{L}\) flask and heated to \(250 .^{\circ} \mathrm{C}\) to vaporize the methanol. Over time, the methanol vapor decomposed by the following reaction: $$\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)$$ After the system has reached equilibrium, a tiny hole is drilled in the side of the flask allowing gaseous compounds to effuse out of the flask. Measurements of the effusing gas show that it contains 33.0 times as much \(\mathrm{H}_{2}(g)\) as \(\mathrm{CH}_{3} \mathrm{OH}(g) .\) Calculate \(K\) for this reaction at \(250 . \mathrm{C} .\)

The partial pressures of an equilibrium mixture of $\mathrm{N}_{2} \mathrm{O}_{4}(g)\( and \)\mathrm{NO}_{2}(g)\( are \)P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.34\( atm and \)P_{\mathrm{NO}_{2}}=1.20 \mathrm{atm}$ at a certain temperature. The volume of the container is doubled. Calculate the partial pressures of the two gases when a new equilibrium is established.

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is $$\mathrm{HCO}_{3}^{-}(a q) \leftrightharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) K=5.6 \times 10^{-11}$$ If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into 1.00 \(\mathrm{L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}^{2-2}\) ?
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