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Chapter 16: Problem 20

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) \(K_{2}=-K_{1}^{1 / 2}\)

Short Answer

Expert verified
The correct relation is (a) \( K_2 = 1 /(K_1)^{1/2} \).

Step by step solution

01

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 \( K_1 \).
02

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.
03

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.
04

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).
05

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.

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Key Concepts

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

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:
  • \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)
This formula mirrors the ratio of the concentration of the products raised to their stoichiometric coefficients to that of the reactants.

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.
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:
  • \( ext{A} + ext{B} \rightleftharpoons 2 ext{C} \) 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.
By understanding stoichiometry, chemists can calculate how much product can form or how much reactant is needed to reach equilibrium.

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.

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Most popular questions from this chapter

Heating a metal carbonate leads to decomposition. $$ \mathrm{BaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{BaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ Predict the effect on the equilibrium of each change listed below. Answer by choosing (i) no change, (ii) shifts left, or (iii) shifts right. (a) add \(\mathrm{BaCO}_{3}\) (c) add BaO (b) add \(\mathrm{CO}_{2}\) (d) raise the temperature (e) increase the volume of the flask containing the reaction

The equilibrium constant for the butane \(\rightleftarrows\) isobutane isomerization reaction is 2.5 at \(25^{\circ} \mathrm{C}\). If 1.75 mol of butane and 1.25 mol of isobutane are mixed, is the system at equilibrium? If not, when it proceeds to equilibrium, which reagent increases in concentration? Calculate the concentrations of the two compounds when the system reaches equilibrium.

The total pressure for a mixture of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) is 1.5 atm. If \(K_{p}=6.75\) (at \(25^{\circ} \mathrm{C}\) ), calculate the partial pressure of each gas in the mixture. $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$

The equilibrium constant \(K\) for the reaction $$ \mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) $$ is \(6.66 \times 10^{-12}\) at \(1000 \mathrm{K}\). Calculate \(K\) for the reaction $$ 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}_{2}(\mathrm{g}) $$

Ammonium hydrogen sulfide decomposes on heating. $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftarrows \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) $$ If \(K_{p}\) for this reaction is 0.11 at \(25^{\circ} \mathrm{C}\) (when the partial pressures are measured in atmospheres), what is the total pressure in the flask at equilibrium?
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