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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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Key Concepts

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

Equilibrium Shift
Le Châtelier's principle helps predict how the position of equilibrium will shift when there is a change in concentration, pressure, or temperature. An equilibrium shift refers to the reaction's tendency to restore balance after being disturbed. When the system at equilibrium experiences stress, it will adjust either left or right to counteract the change and restore equilibrium. When the concentration of reactants or products is altered, the system shifts in a direction that offsets this imbalance.
  • If the concentration of a reactant is decreased or a product is increased, the equilibrium shifts to the left (towards reactants).
  • Conversely, if a reactant concentration is increased or a product decreased, the equilibrium shifts to the right (towards products).
Therefore, understanding equilibrium shifts is crucial for predicting the outcome of changes in reaction conditions.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products. In equilibrium reactions, both the forward and reverse reactions occur simultaneously. As a result, reactants turn into products and vice versa, until the rates of the forward and reverse reactions equalize.In a reversible reaction like \[ ext{Fe}^{3+}(aq) + ext{SCN}^-(aq) ightleftharpoons ext{FeSCN}^{2+}(aq) \]both the forward and backward reactions are significant. Thus, the system aims to reach a state where the concentrations of reactants and products remain constant over time.
When conditions change, the position of equilibrium can shift, altering the concentration balance in the solution.
Reaction Quotient
The reaction quotient, denoted as \(Q\), is a measure of the relative amounts of products and reactants present during a reaction at any given point in time. It is similar to the equilibrium constant \(K\) but differs by considering concentrations at states other than equilibrium.The mathematical expression for \(Q\) is given as:\[Q = \frac{[ ext{Products}]}{[ ext{Reactants}]} \]Where the concentrations of each component are raised to the power of their stoichiometric coefficients in the balanced chemical equation. By comparing \(Q\) to \(K\), chemists can predict the direction of an equilibrium shift:
  • If \(Q < K\), the reaction shifts right, favoring products to reach equilibrium.
  • If \(Q > K\), the reaction shifts left, favoring reactants.
  • If \(Q = K\), the reaction is already at equilibrium.
This measure helps understand and predict how changes in conditions can affect the system.
Concentration Changes
In chemical reactions, concentration changes significantly impact the position of equilibrium. When a component in the reaction experiences an increase or decrease in concentration, the system reacts according to Le Châtelier's principle to counterbalance the change.Adding a compound like \( ext{AgNO}_{3}(aq)\) or \( ext{NaOH}(aq)\) can result in precipitation of an insoluble compound, thereby reducing the concentration of one of the components. This prompts a shift in equilibrium to restore lost ions.For instance:
  • Adding \( ext{AgNO}_{3}(aq)\) decreases \( ext{SCN}^-\) concentration by forming \( ext{AgSCN}\) solid, shifting the equilibrium left.
  • Adding \( ext{NaOH}(aq)\) decreases \( ext{Fe}^{3+}\) concentration by forming \( ext{Fe(OH)}_3\) solid, also shifting the equilibrium left.
Understanding these shifts allows chemists to manipulate reaction conditions for desired outcomes.

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Most popular questions from this chapter

In a given experiment, 5.2 moles of pure NOCl were placed in an otherwise empty \(2.0-\mathrm{L}\) container. Equilibrium was established by the following reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K=1.6 \times 10^{-5}$$ a. Using numerical values for the concentrations in the Initial row and expressions containing the variable \(x\) in both the Change and Equilibrium rows, complete the following table summarizing what happens as this reaction reaches equilibrium. Let \(x=\) the concentration of \(\mathrm{Cl}_{2}\) that is present at equilibrium. b. Calculate the equilibrium concentrations for all species.

Calculate a value for the equilibrium constant for the reaction $$\mathrm{O}_{2}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)$$ given $$\mathrm{NO}_{2}(g) \stackrel{h \nu}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) \quad K=6.8 \times 10^{-49}$$ $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad K=5.8 \times 10^{-34}$$ (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

Le Chatelier's principle is stated (Section 13.7\()\) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

At a particular temperature, \(K_{\mathrm{p}}=1.00 \times 10^{2}\) for the reaction $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \leftrightharpoons 2 \mathrm{HI}(g)$$ If 2.00 atm of \(\mathrm{H}_{2}(g)\) and 2.00 atm of \(\mathrm{I}_{2}(g)\) are introduced into a \(1.00-\mathrm{L}\) container, calculate the equilibrium partial pressures of all species.

Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at \(300 .\) K. At equilibrium the total pressure was 110.5 torr. The reaction is $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ a. Calculate the value of \(K_{\mathrm{p}}\) . b. What would be the partial pressures of all species if NO and \(\mathrm{Br}_{2},\) both at an initial partial pressure of \(0.30 \mathrm{atm},\) were allowed to come to equilibrium at this temperature?
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