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Chapter 15: Problem 61

Consider the following equilibrium for which \(\Delta H<0\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ How will each of the following changes affect an equilibrium mixture of the three gases: (a) O \(_{2}(g)\) is added to the system; (b) the reaction mixture is heated; (c) the volume of the reaction vessel is doubled; (d) a catalyst is added to the mixture; (e) the total pressure of the system is increased by adding a noble gas; (f) \(\mathrm{SO}_{3}(g)\) is removed from the system?

Short Answer

Expert verified
(a) Adding O\(_{2}\)(g) will shift the equilibrium to the right, increasing the concentration of SO\(_{3}\)(g) and decreasing the concentration of SO\(_{2}\)(g). (b) Heating the reaction mixture will shift the equilibrium to the left, increasing the concentrations of SO\(_{2}\)(g) and O\(_{2}\)(g) while decreasing the concentration of SO\(_{3}\)(g). (c) Doubling the volume of the reaction vessel will shift the equilibrium to the left, increasing the concentrations of SO\(_{2}\)(g) and O\(_{2}\)(g), and decreasing the concentration of SO\(_{3}\)(g). (d) Adding a catalyst will not affect the concentrations of any of the gases in the system. (e) Increasing the total pressure by adding a noble gas will not affect the position of the equilibrium, and the concentrations of the gases will remain the same. (f) Removing SO\(_{3}\)(g) will shift the equilibrium to the right, increasing the concentration of SO\(_{3}\)(g) and decreasing the concentrations of SO\(_{2}\)(g) and O\(_{2}\)(g).

Step by step solution

01

(a) Addition of O\(_{2}\)(g)

: When O\(_{2}\)(g) is added to the system, the concentration of O\(_{2}\) increases. According to Le Châtelier's principle, the equilibrium will shift to counteract this change and decrease the concentration of O\(_{2}\). The equilibrium will shift to the right, increasing the concentration of SO\(_{3}\)(g) and decreasing the concentration of SO\(_{2}\)(g).
02

(b) Heating the reaction mixture

: As the reaction is exothermic (\(\Delta H<0\)), increasing the temperature of the system will shift the equilibrium to the side that absorbs heat, which is the endothermic (reverse) direction. So, the equilibrium will shift to the left, increasing the concentrations of SO\(_{2}\)(g) and O\(_{2}\)(g), and decreasing the concentration of SO\(_{3}\)(g).
03

(c) Doubling the volume of the reaction vessel

: When the volume of the reaction vessel is doubled, the concentrations of all gases decrease. The equilibrium will shift to the side with more gas molecules, which, in this case, is the left side (3 molecules vs 2 molecules). As a result, the concentrations of SO\(_{2}\)(g) and O\(_{2}\)(g) will increase, while the concentration of SO\(_{3}\)(g) will decrease.
04

(d) Addition of a catalyst

: Adding a catalyst to the mixture speeds up the rate of both the forward and reverse reactions equally, without affecting the position of the equilibrium. Thus, adding a catalyst to the mixture will not change the concentrations of any of the gases in the system.
05

(e) Increasing the total pressure by adding a noble gas

: Increasing the total pressure of the system by adding a noble gas does not affect the partial pressures of the reactants or products, as the noble gas does not participate in the reaction. Therefore, according to Le Châtelier's principle, this change will not affect the position of the equilibrium, and the concentrations of the gases will remain the same.
06

(f) Removal of SO\(_{3}\)(g)

: When SO\(_{3}\)(g) is removed from the system, its concentration decreases. According to Le Châtelier's principle, the equilibrium will shift to counteract this change and increase the concentration of SO\(_{3}\). Hence, the equilibrium will shift to the right, increasing the concentration of SO\(_{3}\)(g) and decreasing the concentrations of SO\(_{2}\)(g) and O\(_{2}\)(g).

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Key Concepts

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

Le Chatelier's Principle
Understanding how a chemical reaction responds to changes in conditions is crucial in the world of chemistry. This is where Le Chatelier's Principle comes to the rescue. It predicts how a system at equilibrium reacts to disturbances, such as changes in concentration, temperature, volume, or pressure. If a chemical system at equilibrium experiences a disturbance, the system will adjust itself to partially counteract the change. It does this through a shift in the equilibrium position.

For instance, adding more of a reactant or product will shift the equilibrium to the opposite side to use up the added substance and re-establish equilibrium. Conversely, removing a substance from the system will cause a shift towards that substance's side to replace it. As a result, concentrations of substances within the system will change until equilibrium is reattained.
Equilibrium Shift
When discussing the impact of different changes on an equilibrium system, we often refer to an 'equilibrium shift'. This concept explains how the system adjusts to re-establish balance following a disturbance, as predicted by Le Chatelier's Principle. An equilibrium shift can go towards the products (right) or the reactants (left), and it is dependent on the type of disturbance.

Factors Influencing Equilibrium Shift:

  • Concentration changes: Increasing the concentration of reactants shifts the equilibrium to the right, while increasing the concentration of products shifts it to the left to restore balance.
  • Temperature variations: For exothermic reactions, raising the temperature shifts the equilibrium to the left as heat is considered a product. The opposite occurs for endothermic reactions.
  • Volume and pressure changes: According to the ideal gas law, reducing the volume increases pressure, which shifts the equilibrium towards the side with fewer gas molecules.
Reaction Quotient
A key tool for chemists to predict the direction of an equilibrium shift is the reaction quotient, denoted as Q. It's a measure that compares the relative amounts of products and reactants present in a reaction at any given moment.

To find Q, we take the same expression as the equilibrium constant (K) but use the current concentrations regardless of whether the system is at equilibrium. Comparing Q to K gives us valuable information:
  • If Q < K, the system will shift to the right, producing more products to reach equilibrium.
  • If Q > K, the system will shift to the left, producing more reactants to establish equilibrium.
  • If Q = K, the system is at equilibrium, and no shift is necessary.
Understanding how Q affects a system's move towards equilibrium is critical in predicting the impact of various changes on a reaction in progress.

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Most popular questions from this chapter

If \(K_{c}=0.042\) for \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{PCl}_{5}(g)\) at 500 \(\mathrm{K}\) , what is the value of \(K_{p}\) for this reaction at this temperature?

Consider the following equilibrium: \(2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{S}(g) \quad K_{c}=1.08 \times 10^{7} \mathrm{at} 700^{\circ} \mathrm{C}\) (a) Calculate \(K_{p}\) . (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{S} ?(\mathbf{c})\) Calculate the value of \(K_{c}\) if you rewrote the equation \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{S}(g)\)

For the equilibrium $$2 \operatorname{IBr}(g) \Longrightarrow \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C} .\) If 0.025 atm of IBr is placed in a 2.0 -L container, what is the partial pressure of all substances after equilibrium is reached?

An equilibrium mixture of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and \(\mathrm{HI}\) at \(458^{\circ} \mathrm{C}\) contains \(0.112 \mathrm{mol} \mathrm{H}_{2}, 0.112 \mathrm{mol} \mathrm{I}_{2},\) and 0.775 \(\mathrm{mol}\) HI in a 5.00 -L. vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of 0.200 \(\mathrm{mol}\) of \(\mathrm{HI}\) ?

Which of the following statements are true and which are false? (a) For the reaction \(2 \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{A}_{2} \mathrm{B}(g) K_{c}\) and \(K_{p}\) are numerically the same. (b) It is possible to distinguish \(K_{c}\) from \(K_{p}\) by comparing the units used to express the equilibrium constant. (c) For the equilibrium in (a), the value of \(K_{c}\) increases with increasing pressure.
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