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The reaction \(3 \mathrm{~A}(g)+\mathrm{B}(s) \rightleftharpoons 2 \mathrm{C}(a q)+\mathrm{D}(a q)\) occurs at \(25^{\circ} \mathrm{C}\) in a flask, which has \(1.87\) L available for gas. After the reaction attains equilibrium, the amounts (mol) or concentrations \((M)\) of substances are as follows: \(2.48 \mathrm{~mol} \mathrm{~A} \quad 1.13 M \mathrm{C}\) \(2.41 \mathrm{~mol} \mathrm{~B} \quad 2.27 M \mathrm{D}\) What is the equilibrium constant \(K_{c}\) for this reaction at \(25^{\circ} \mathrm{C}\) ? a. \(1.24\) b. \(0.190\) c. \(0.516\) d. \(1.15\) e. \(0.939\)

Short Answer

Expert verified
The equilibrium constant \(K_{c}\) is approximately 1.24, so option (a) is correct.

Step by step solution

01

Understanding the Reaction and Equilibrium Constant

The reaction is given as \(3 \mathrm{~A}(g)+\mathrm{B}(s) \rightleftharpoons 2 \mathrm{C}(aq)+\mathrm{D}(aq)\). The equilibrium constant \(K_{c}\) is defined for aqueous concentrations in this reaction as: \[ K_{c} = \frac{[\mathrm{C}]^2[\mathrm{D}]}{[\mathrm{A}]^3} \] Since \(\mathrm{B}\) is a solid, it does not appear in the expression for \(K_{c}\). The concentrations of the products and reactants are needed to calculate \(K_{c}\).
02

Determining the Concentration of Reactant A

The concentration of a gas is given by its moles divided by the volume of the container. Given that there is \(2.48\) mol of \(\mathrm{A}\) in a \(1.87\) L flask, the concentration \([\mathrm{A}]\) is calculated as:\[ [\mathrm{A}] = \frac{2.48 \text{ moles}}{1.87 \text{ L}} \approx 1.326 \text{ M} \]
03

Substituting Values into the Equilibrium Constant Equation

Now we have the concentrations for all necessary components:- \([\mathrm{C}] = 1.13 \text{ M}\)- \([\mathrm{D}] = 2.27 \text{ M}\)- \([\mathrm{A}] = 1.326 \text{ M}\)Substitute into the expression for \(K_{c}\):\[ K_{c} = \frac{(1.13)^2 \times (2.27)}{(1.326)^3} \]
04

Calculating the Equilibrium Constant

Calculate each part of the equation separately:- \((1.13)^2 = 1.2769\)- The numerator \(1.2769 \times 2.27 = 2.899563\)- The denominator \((1.326)^3 = 2.3314936\)Now calculate \(K_{c}\):\[ K_{c} = \frac{2.899563}{2.3314936} \approx 1.24 \]
05

Conclusion and Answer Selection

The calculated equilibrium constant \(K_{c}\) is approximately \(1.24\). Therefore, the correct answer is option (a).

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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 of balance in a reversible chemical reaction. When a reaction reaches this state, the rate of the forward reaction equals the rate of the backward reaction. This does not necessarily mean that the concentrations of reactants and products are equal, but rather that they remain constant over time.
The equilibrium constant, denoted as \( K_c \), is a numerical value that characterizes the ratio of concentrations of products to reactants for a reaction at equilibrium. Each concentration is raised to the power of its coefficient in the balanced equation. It is important to note that this constant changes with temperature and pressure.
In our example reaction, the equilibrium constant expression is derived from the balanced chemical equation: \( 3 \mathrm{~A}(g)+\mathrm{B}(s) \rightleftharpoons 2 \mathrm{C}(aq)+\mathrm{D}(aq) \). Since \( \mathrm{B} \) is a solid, it is omitted from the expression for \( K_c \). Thus, \( K_c = \frac{[\mathrm{C}]^2[\mathrm{D}]}{[\mathrm{A}]^3} \), where the square brackets denote molarity or concentration in moles per liter.
Le Chatelier's Principle
Le Chatelier's Principle provides insight into how a system at equilibrium responds to external changes. It states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust itself to partially counteract the effect of the change.
For instance, if the concentration of reactants in a reaction is increased, the equilibrium will shift towards the products to reduce the impact of this change. Conversely, increasing the concentration of products will push the equilibrium towards the reactants.
In temperature changes, increasing the temperature of an exothermic reaction results in the equilibrium shifting towards the reactants as the system tries to absorb the extra heat. This principle not only helps in understanding the nature of equilibriums but also in predicting how equilibriums can be manipulated effectively.
Reversible Reactions
Reversible reactions are chemical reactions where the conversion of reactants to products and the conversion of products back to reactants occur simultaneously. These reactions are represented by the equilibrium symbol \( \rightleftharpoons \), indicating that reactions can proceed in both directions.
Reversibility is a common characteristic in many natural and industrial chemical processes. In reversible reactions, both reactants and products are present in the system, and the reaction reaches a state of equilibrium over time.
An example of a reversible reaction is the one mentioned in the exercise, \( 3 \mathrm{~A}(g)+\mathrm{B}(s) \rightleftharpoons 2 \mathrm{C}(aq)+\mathrm{D}(aq) \). Such reactions highlight the dynamic nature of equilibrium, where molecules change continuously, yet the overall concentrations remain constant, illustrating the balance that defines chemical equilibrium.

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Most popular questions from this chapter

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. \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

A researcher put \(0.400 \mathrm{~mol} \mathrm{PCl}_{3}\) and \(0.600 \mathrm{~mol} \mathrm{Cl}_{2}\) into a 5.00-L vessel at a given temperature to produce phosphorus pentachloride, \(\mathrm{PCl}_{5}\) : $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ What will be the composition of this gaseous mixture at equilibrium? \(K_{c}=25.6\) at the temperature of this experiment.

The following reaction has an equilibrium constant \(K_{c}\) equal to \(3.07 \times 10^{-4}\) at \(24^{\circ} \mathrm{C}\) $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ For each of the following compositions, decide whether the reaction mixture is at equilibrium. If it is not, decide which direction the reaction should go. a. \([\mathrm{NOBr}]=0.0720 M,[\mathrm{NO}]=0.0162 M,\left[\mathrm{Br}_{2}\right]=0.0123 M\) b. \([\mathrm{NOBr}]=0.121 M,[\mathrm{NO}]=0.0159 M,\left[\mathrm{Br}_{2}\right]=0.0139 M\) c. \([\mathrm{NOBr}]=0.103 M,[\mathrm{NO}]=0.0134 M,\left[\mathrm{Br}_{2}\right]=0.0181 M\) d. \([\mathrm{NOBr}]=0.0472 M,[\mathrm{NO}]=0.0121 M,\left[\mathrm{Br}_{2}\right]=0.0105 M\)

When a continuous stream of hydrogen gas, \(\mathrm{H}_{2}\), passes over hot magnetic iron oxide, \(\mathrm{Fe}_{3} \mathrm{O}_{4}\), metallic iron and water vapor form. When a continuous stream of water vapor passes over hot metallic iron, the oxide \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) and \(\mathrm{H}_{2}\) form. Explain why the reaction goes in one direction in one case but in the reverse direction in the other.

The following reaction is important in the manufacture of sulfuric acid. $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ At \(900 \mathrm{~K}, 0.0216 \mathrm{~mol}\) of \(\mathrm{SO}_{2}\) and \(0.0148 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) are sealed in a 1.00-L reaction vessel. When equilibrium is reached, the concentration of \(\mathrm{SO}_{3}\) is determined to be \(0.0175 \mathrm{M}\). Calculate \(K_{c}\) for this reaction.

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