Problem 20
Question
Which of the following correctly relates the equilibrium constants for the two reactions shown? \(\mathrm{A}+\mathrm{B} \rightleftarrows 2 \mathrm{C} \quad K_{1}\) \(\begin{array}{ccc}C & \leftrightarrows_{2} A+\frac{1}{2} B & K_{2}\end{array}\) (a) \(K_{2}=1 /\left(K_{1}\right)^{1 / 2}\) (c) \(K_{2}=K_{1}^{2}\) (b) \(K_{2}=1 / K_{1}\) (d) \(K_{2}=-K_{1}^{1 / 2}\)
Step-by-Step Solution
Verified Answer
The correct relation is (a) \( K_2 = 1 /(K_1)^{1/2} \).
1Step 1: Understanding the Reactions
We have two chemical equations involving equilibrium constants. The first reaction is \( \text{A} + \text{B} \rightleftarrows 2 \text{C} \) with equilibrium constant \( K_1 \), and the second reaction is \( \text{C} \leftrightarrows \text{A} + \frac{1}{2} \text{B} \) with equilibrium constant \( K_2 \). Our task is to find the correct relationship between \( K_2 \) and \( K_1 \).
2Step 2: Writing the General Expression for K_1
For the reaction \( \text{A} + \text{B} \rightleftarrows 2\text{C} \), the equilibrium constant \( K_1 \) is given by:\[K_1 = \frac{[\text{C}]^2}{[\text{A}][\text{B}]}\]This formula reflects the concentration of the products over reactants at equilibrium.
3Step 3: Writing the General Expression for K_2
For the reaction \( \text{C} \leftrightarrows \text{A} + \frac{1}{2}\text{B} \), the equilibrium constant \( K_2 \) is:\[K_2 = \frac{[\text{A}][\text{B}]^{1/2}}{[\text{C}]}\]This reflects the formula for the reverse reaction as analyzed from its components.
4Step 4: Relating K_2 to K_1
To relate \( K_2 \) and \( K_1 \), notice that the second reaction can be seen as the reverse of the first reaction with coefficients halved. When a reaction is reversed, the equilibrium constant for the reverse reaction is \( 1/K \) of the forward reaction. Hence, the equilibrium constant for the given version of the reversed reaction is:\[K_2 = \frac{1}{K_1^{1/2}}\]because we also have to consider the stoichiometric change in the reaction (which involves squaring the original equation).
5Step 5: Selecting the Correct Relationship
Evaluating the relationships provided in the options, (a) precisely represents the relationship derived:(a) \( K_2 = 1 /(K_1)^{1/2} \). The other options do not match the mathematical analysis of the equilibrium constant relationships.
Key Concepts
Equilibrium ConstantsReversible ReactionsStoichiometry
Equilibrium Constants
In the study of chemical equilibrium, equilibrium constants play a crucial role in describing the balance of reversible reactions. An equilibrium constant, represented as \( K \), quantifies the concentrations of reactants and products at equilibrium for a particular reaction. In essence, it reflects how far the reaction proceeds before reaching a state of balance.
For a general reaction, such as \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression would be:
Understanding equilibrium constants allows chemists to predict the position of equilibrium under different conditions and how shifts in concentration or pressure affect the system.
For a general reaction, such as \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression would be:
- \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)
Understanding equilibrium constants allows chemists to predict the position of equilibrium under different conditions and how shifts in concentration or pressure affect the system.
Reversible Reactions
Reversible reactions are fascinating in chemistry because they can proceed in both forward and backward directions. These reactions reach a state of dynamic equilibrium, where the rates of the forward and reverse reactions are equal, resulting in stable concentrations of reactants and products.
This dynamic nature means that reversible reactions never "really" stop. Instead, they continuously convert reactants to products and vice versa, albeit without any net change in their concentrations over time.
An important aspect of reversible reactions is that if you reverse a reaction, you take the reciprocal of the equilibrium constant for the original direction. Moreover, if you alter the stoichiometry, such as by halving the number of molecules involved, you adjust the equilibrium constant accordingly.
This dynamic nature means that reversible reactions never "really" stop. Instead, they continuously convert reactants to products and vice versa, albeit without any net change in their concentrations over time.
An important aspect of reversible reactions is that if you reverse a reaction, you take the reciprocal of the equilibrium constant for the original direction. Moreover, if you alter the stoichiometry, such as by halving the number of molecules involved, you adjust the equilibrium constant accordingly.
Stoichiometry
Stoichiometry in the context of reversible reactions involves the quantitative relationship between reactants and products in a chemical reaction. It provides insight into how changes in these quantities affect the system's equilibrium.
The coefficients in a balanced chemical equation represent the mole ratio of the substances involved. For example, in our reactions:
When stoichiometric coefficients change, as in the reversal and alteration of reactions, this directly influences the equilibrium constant, emphasizing the importance of a clear understanding of these relationships.
The coefficients in a balanced chemical equation represent the mole ratio of the substances involved. For example, in our reactions:
- \( ext{A} + ext{B} \rightleftharpoons 2 ext{C} \) implies that 1 mole of \( ext{A} \) and 1 mole of \( ext{B} \) generate 2 moles of \( ext{C} \).
- Reversing the reaction so \( ext{C} \rightleftharpoons ext{A} + rac{1}{2} ext{B} \) alters this stoichiometric relationship.
When stoichiometric coefficients change, as in the reversal and alteration of reactions, this directly influences the equilibrium constant, emphasizing the importance of a clear understanding of these relationships.
Other exercises in this chapter
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