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Chapter 15: Problem 78

An equilibrium mixture of \(\mathrm{SO}_{3}, \mathrm{SO}_{2}\), and \(\mathrm{O}_{2}\) at \(727^{\circ} \mathrm{C}\) is \(0.0160 \mathrm{MSO}_{3}, 0.0056 \mathrm{M} \mathrm{SO}_{2}\), and \(0.0021 \mathrm{M} \mathrm{O}_{2} .\) What is the value of \(K_{c}\) for the following reaction? $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$

Short Answer

Expert verified
The equilibrium constant, \( K_c \), is approximately 62.38.

Step by step solution

01

Write the Expression for Equilibrium Constant

First, we need to write the equilibrium expression for the given reaction. The reaction is: \( \mathrm{SO}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) \). The equilibrium constant \( K_c \) for this reaction is given by:\[ K_c = \frac{[\text{products}]}{[\text{reactants}]} = \frac{[\mathrm{SO}_3]}{[\mathrm{SO}_2][\mathrm{O}_2]^{1/2}} \]
02

Substitute the Equilibrium Concentrations into the Expression

Substitute the given equilibrium concentrations into the equilibrium constant expression:\[ [\mathrm{SO}_3] = 0.0160 \text{ M} \]\[ [\mathrm{SO}_2] = 0.0056 \text{ M} \]\[ [\mathrm{O}_2] = 0.0021 \text{ M} \]Now substitute these values into the expression for \( K_c \):\[ K_c = \frac{0.0160}{0.0056 \times (0.0021)^{1/2}} \]
03

Calculate the Value of Kc

Calculate the numerical value by solving the expression:First, find the square root of the \([\mathrm{O}_2]\) concentration:\[(0.0021)^{1/2} = 0.0458 \text{ (approximately)}\]Next, multiply this by \([\mathrm{SO}_2]\):\[0.0056 \times 0.0458 = 0.00025648 \]Finally, divide \([\mathrm{SO}_3]\) by this product to find \(K_c\):\[ K_c = \frac{0.0160}{0.00025648} \approx 62.38 \]
04

Final Answer Verification

Check each step to ensure correct arithmetic calculations and ensure the final answer makes sense. Here, re-evaluate steps if needed but in this scenario, our steps are correct, and \( K_c \) value calculated is reasonable for the reaction.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to the state in a reversible reaction where the rates of the forward and backward reactions are equal. At this point, the concentrations of reactants and products remain constant over time, although not necessarily equal. In simple terms, it means that the chemical process is occurring in both directions at the same speed.
For example, in the reaction between sulfur dioxide (\(\text{SO}_2\)) and oxygen (\(\text{O}_2\)) to form sulfur trioxide (\(\text{SO}_3\)), equilibrium is reached when the number of \(\text{SO}_3\) molecules being formed equals the number being decomposed back into \(\text{SO}_2\) and \(\text{O}_2\).
  • Concentrations remain constant, but bonds are continuously being broken and formed.
  • The system is dynamic, not static, representing a continuous interaction at the molecular level.
  • Conditions like pressure and temperature can affect the position of equilibrium.
Reaction Quotient
The reaction quotient, denoted as \(Q\), is a measure that helps predict which direction a reaction must go to reach equilibrium. Like the equilibrium constant \(K_c\), \(Q\) is calculated using the concentrations of the reactants and products. However, \(Q\) can be determined at any stage in a reaction, not just at equilibrium.
To compute \(Q\), you use the same expression as for \(K_c\): \[ Q = \frac{[\text{SO}_3]}{[\text{SO}_2][\text{O}_2]^{1/2}} \]where the concentrations employed are from any point in the reaction.
  • If \(Q < K_c\), the reaction will proceed forward, forming more products to reach equilibrium.
  • If \(Q > K_c\), the reaction moves backwards, producing more reactants until equilibrium is established.
  • When \(Q = K_c\), the system is at equilibrium, and no shift occurs.
Understanding \(Q\) allows chemists to predict reaction behaviors and adjust conditions accordingly.
Equilibrium Concentrations
Equilibrium concentrations refer to the stable concentrations of reactants and products measured when a reaction is at equilibrium. These concentrations can be used to calculate the equilibrium constant, \(K_c\). Notice that while concentrations may remain unchanged, the molecular interactions persist.
In the given exercise, the specific equilibrium concentrations were:
  • \([\text{SO}_3] = 0.0160\, \text{M}\)
  • \([\text{SO}_2] = 0.0056\, \text{M}\)
  • \([\text{O}_2] = 0.0021\, \text{M}\)
Using these values, we apply them into the \(K_c\) expression:\[ K_c = \frac{0.0160}{0.0056 \times (0.0021)^{1/2}} \]This calculation gives the equilibrium constant, which quantitatively describes the position of equilibrium in the reaction.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental rule that predicts how a change in conditions affects a chemical reaction at equilibrium. If a change in concentration, temperature, or pressure is introduced to a system at equilibrium, the system responds to counteract that change and establishes a new equilibrium.
Some practical ways this principle works include:
  • Increasing the concentration of a reactant will shift the equilibrium towards the products.
  • Raising the temperature typically shifts equilibrium in the endothermic direction. For an exothermic reaction, it pushes the equilibrium towards the reactants.
  • Changing the pressure affects gaseous reactions. Increasing pressure shifts equilibrium towards the side with fewer moles.
This principle is crucial in processes like the synthesis of chemicals, where maximizing product yield by manipulating conditions can be driven effectively using Le Chatelier's insight.

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Most popular questions from this chapter

(a) Predict the direction of reaction when chlorine gas is added to an equilibrium mixture of \(\mathrm{PCl}_{3}, \mathrm{PCl}_{5}\), and \(\mathrm{Cl}_{2}\). The reaction is $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ (b) What is the direction of reaction when chlorine gas is removed from an equilibrium mixture of these gases?

Explain why pure liquids and solids can be ignored when writing the equilibrium-constant expression.

Methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), is manufactured industrially by the reaction $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ A gaseous mixture at \(500 \mathrm{~K}\) is \(0.020 \mathrm{M} \mathrm{CH}_{3} \mathrm{OH}, 0.10 \mathrm{M} \mathrm{CO}\), and \(0.10 M \mathrm{H}_{2} .\) What will be the direction of reaction if this mixture goes to equilibrium? The equilibrium constant \(K_{c}\) equals \(10.5\) at \(500 \mathrm{~K}\).

At some temperature, a 100-L reaction vessel contains a mixture that is initially \(1.00 \mathrm{~mol} \mathrm{CO}\) and \(2.00 \mathrm{~mol} \mathrm{H}_{2}\). The vessel also contains a catalyst so that the following equilibrium is attained: $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ At equilibrium, the mixture contains \(0.100 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH} .\) In a later experiment in the same vessel, you start with \(1.00 \mathrm{~mol}\) \(\mathrm{CH}_{3} \mathrm{OH}\). How much methanol is there at equilibrium? Explain.

Write equilibrium-constant expressions \(K_{p}\) for each of the following reactions: a. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) b. \(2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\) c. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) d. \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)
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