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Chapter 12: Problem 14

WEB Consider the following reaction at \(122^{\circ} \mathrm{C}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ (a) Write an equilibrium constant expression for the reaction and call the constant \(K_{1}\). (b) Write an equilibrium constant expression for the decomposition of one mole of \(\mathrm{SO}_{3}\) to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) and call the constant \(K_{2}\). (c) Relate \(K_{1}\) and \(K_{2}\).

Short Answer

Expert verified
Question: Derive the relationship between the equilibrium constants \(K_{1}\) and \(K_{2}\) for a given chemical reaction. Solution: For a chemical reaction at equilibrium that can be represented by the sum of two smaller reactions, the relationship between the equilibrium constants can be obtained by multiplying the constants of the smaller reactions. In this case, if the original reaction has the equilibrium constant \(K_{1}\) and the two smaller reactions have constants \(K_{2}\), the relationship is \(K_{1} = K_{2}^{2}\).

Step by step solution

01

(a) Writing the equilibrium constant expression for the given reaction

For the given reaction at equilibrium: $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ To write the equilibrium constant expression, we need to use the concentrations of the reactants and products at equilibrium. The equilibrium constant, \(K_{1}\), for this reaction can be written as: $$K_{1} = \frac{[\mathrm{SO}_{2}]^{2}[\mathrm{O}_{2}]}{[\mathrm{SO}_{3}]^{2}}$$ This formula represents the relationship between the equilibrium concentrations of the reactants and products.
02

(b) Writing the equilibrium constant expression for the decomposition of one mole of \(\mathrm{SO}_{3}\)

We can rewrite the original reaction as the decomposition of one mole of \(\mathrm{SO}_{3}(g)\): $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+0.5\mathrm{O}_{2}(g)$$ The equilibrium constant expression for this reaction, \(K_{2}\), can be written as: $$K_{2} = \frac{[\mathrm{SO}_{2}][\mathrm{O}_{2}]^{0.5}}{[\mathrm{SO}_{3}]}$$ This expression represents the relationship between the equilibrium concentrations of the reactants and products for the decomposition of one mole of \(\mathrm{SO}_{3}\).
03

(c) Relating \(K_{1}\) and \(K_{2}\)

To relate \(K_{1}\) and \(K_{2}\), let's consider the original reaction in terms of the decomposition of one mole of \(\mathrm{SO}_{3}\). The original reaction can be written as the sum of two decomposition reactions, each involving one mole of \(\mathrm{SO}_{3}\): $$\begin{aligned} \mathrm{SO}_{3}(g) &\rightleftharpoons \mathrm{SO}_{2}(g)+0.5\mathrm{O}_{2}(g) \\ \mathrm{SO}_{3}(g) &\rightleftharpoons \mathrm{SO}_{2}(g)+0.5\mathrm{O}_{2}(g) \\ \hline 2 \mathrm{SO}_{3}(g) &\rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \end{aligned}$$ Since the original reaction is the sum of the two decomposition reactions, the equilibrium constant for the original reaction, \(K_{1}\), can be obtained by multiplying the equilibrium constants of the two smaller reactions, \(K_{2}\): $$K_{1} = K_{2} \cdot K_{2} = K_{2}^{2}$$ In conclusion, the relationship between \(K_{1}\) and \(K_{2}\) can be represented as \(K_{1} = K_{2}^{2}\).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, leading to no net change in the concentration of reactants and products over time. It is essential to note that this does not mean the reactants and products are present in equal amounts, but rather that their concentrations have stabilized in a particular ratio that is specific to the reaction's conditions, such as temperature and pressure.

For example, in the provided exercise we consider the reversible reaction of sulfur trioxide decomposing into sulfur dioxide and oxygen. At equilibrium, the system reaches a state where the decomposition of sulfur trioxide and the formation of sulfur trioxide from sulfur dioxide and oxygen occur at the same rate.
Reaction Quotient
The reaction quotient, Q, helps determine the direction in which a reaction will proceed to reach equilibrium. Calculated using the same formula as the equilibrium constant (K), the reaction quotient uses the concentrations of the reactants and products at any point in time, not just at equilibrium.

For the reaction involving sulfur trioxide, if Q < K, the reaction will proceed forward, turning more reactants into products. Conversely, if Q > K, the reaction will go in the reverse direction to form more reactants. Only when Q = K is the system at equilibrium. Understanding Q, along with the equilibrium constant expression, can predict the shift in a reaction under altered conditions.
Le Châtelier's Principle
Le Châtelier's principle provides insight into how a system at equilibrium reacts to disturbances. It states that if an external change is applied to a system at equilibrium, the system will adjust itself to minimize that change.

These disturbances can include changes in concentration, pressure, or temperature. For instance, if the concentration of a reactant is increased, the system will respond by producing more products, and vice versa. If the exercise's reaction were subjected to increased pressure, the system would shift to the side with fewer moles of gas to relieve some of the pressure.
Dynamic Equilibrium
Dynamic equilibrium is the actual nature of the state achieved when a chemical reaction reaches equilibrium. Unlike static equilibrium, where no reactions occur, dynamic equilibrium is characterized by the continuous and simultaneous occurrence of both the forward and reverse reactions. The rates of these opposing reactions are equal, so there's no net change in the concentrations of reactants and products over time.

In the context of our sulfur trioxide reaction, even when it has reached equilibrium, sulfur trioxide molecules continue to form sulfur dioxide and oxygen, while sulfur dioxide and oxygen molecules also continue to form sulfur trioxide at exactly the same rate. This incessant activity underlines the 'dynamic' aspect of the equilibrium.

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Most popular questions from this chapter

For the decomposition of \(\mathrm{CaCO}_{3}\) at \(900^{\circ} \mathrm{C}, K=1.04\). $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{O}_{2}(g)$$ Find the smallest mass of \(\mathrm{CaCO}_{3}\) needed to reach equilibrium in a 5.00-L vessel at \(900^{\circ} \mathrm{C}\).

At a certain temperature, \(K\) is \(1.3 \times 10^{5}\) for the reaction $$2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g)$$ What is the equilibrium pressure of hydrogen sulfide if those of hydrogen and sulfur gases are \(0.103\) atm and \(0.417\) atm, respectively?

The following data are for the system $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ $$\begin{array}{ccccccc}\hline \text { Time (s) } & 0 & 20 & 40 & 60 & 80 & 100 \\ P_{\mathrm{A}} \text { (atm) } & 1.00 & 0.83 & 0.72 & 0.65 & 0.62 & 0.62 \\ P_{\mathrm{B}} \text { (atm) } & 0.00 & 0.34 & 0.56 & 0.70 & 0.76 & 0.76 \\ \hline\end{array}$$ Prepare a graph of \(P_{\Lambda}\) and \(P_{\mathrm{B}}\) versus time and use it to answer the following questions: (a) Estimate \(P_{\mathrm{A}}\) and \(P_{\mathrm{g}}\) after \(30 \mathrm{~s}\). (b) Estimate \(P_{\mathrm{A}}\) after \(150 \mathrm{~s}\). (c) Estimate \(P_{\mathrm{B}}\) when \(P_{\mathrm{A}}=0.700 \mathrm{~atm}\).

A sealed flask has \(0.541\) atm of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\). The following equilibrium is established. $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium, the partial pressure of oxygen is measured to be \(0.216\) atm. Calculate \(K\) for the decomposition of \(\mathrm{SO}_{3}\) at \(1000 \mathrm{~K}\).

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is \(0.45\). If \(20.0 \mathrm{~mL}\) \((d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00\) -L vessel and heated to \(180^{\circ} \mathrm{C}\), what percentage remains undissociated at equilibrium?
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