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If \(K_{c}=1\) for the equilibrium \(2 \mathrm{A}(g) \rightleftharpoons \mathrm{B}(g),\) what is the relationship between \([\mathrm{A}]\) and \([\mathrm{B}]\) at equilibrium?

Short Answer

Expert verified
The relationship between the equilibrium concentrations of A and B can be expressed as: \([B] = [A]^2\).

Step by step solution

01

Write the equilibrium expression

For the given reaction, 2 A(g) ⇌ B(g), the equilibrium constant (Kc) can be expressed in terms of the equilibrium concentrations of the reactants and products. The general form of the expression for Kc is: Kc = \(\frac{[Products]}{[Reactants]}\) For this particular reaction, Kc will be: Kc = \(\frac{[B]}{[A]^2}\) We are given that Kc = 1.
02

Substitute Kc into the equilibrium expression and solve for the relationship

Now we can substitute the given Kc value (1) into the equilibrium expression: 1 = \(\frac{[B]}{[A]^2}\) To find the relationship, rearrange the equation to isolate [B] or [A] on one side. In this case, we can isolate [B]: [B] = [A]^2
03

State the relationship between [A] and [B]

At equilibrium, the concentration of B, [B], is equal to the square of the concentration of A, [A]^2. Therefore, the relationship between the equilibrium concentrations of A and B can be expressed as: [B] = [A]^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 refers to a state in a chemical reaction where the concentrations of reactants and products no longer change over time. This doesn't mean the reaction has stopped, but that the forward and reverse reactions occur at the same rate. As a result, the concentration of each substance remains constant, even though both reactions are still happening.
In simple terms, it's like a busy intersection where cars enter and leave at the same pace, so the number of cars in the intersection remains the same. Achieving equilibrium depends on the reaction conditions, such as temperature and pressure, but not on the initial concentrations of the reactants and products. It's a balance point that every reversible reaction strives to reach under set conditions. Understanding chemical equilibrium helps predict how changes in conditions will affect the concentrations of substances in a chemical reaction.
Concentration Relationship
In equilibrium calculations, understanding the concentration relationship between reactants and products is crucial. For the reaction given in the exercise, 2 A(g) ⇌ B(g), the concentrations of reactants and products have a specific mathematical relationship defined by the equilibrium constant, denoted as \( K_{c} \).
When the given \( K_{c} \) value is 1, it provides insight into the ratio of product concentration to its reactant concentration at equilibrium. In this reaction, it particularly means that at equilibrium, the concentration of \( B \) is equal to the square of the concentration of \( A \), expressed as
  • \( [B] = [A]^2 \)
This relationship highlights the importance of stoichiometry, the coefficients in the balanced equation, which show how reactants convert to products. It demonstrates how equilibrium constants help predict the concentrations of substances involved in a reaction at equilibrium.
Equilibrium Expression
An equilibrium expression mathematically represents the relationship between the concentrations of reactants and products for a reversible reaction at equilibrium. For the reaction 2 A(g) ⇌ B(g), the equilibrium expression is written using the equilibrium constant \( K_{c} \), where: \( K_{c} = \frac{[B]}{[A]^2} \).
Here, [B] represents the concentration of product \( B \), and [A] that of reactant \( A \). The coefficients from the balanced chemical equation become exponents in the expression. This expression shows us how to set up ratios that simplify to the value of \( K_{c} \), connecting it directly to the concentration values.
  • If \( K_{c} = 1 \), it indicates equal proportions reflected by the concentrations at equilibrium. This specific case demonstrates that the square of \( A's \) concentration must equate that of \( B \), or \([B] = [A]^2\).
Equilibrium expressions thus provide a powerful tool for understanding and predicting how chemical species will present themselves once a reaction reaches equilibrium. It helps chemists and anyone working with chemical reactions to predict and manipulate reaction conditions effectively.

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Most popular questions from this chapter

At \(2000^{\circ} \mathrm{C},\) the equilibrium constant for the reaction $$2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ is \(K_{c}=2.4 \times 10^{3} .\) If the initial concentration of \(\mathrm{NO}\) is \(0.175 \mathrm{M},\) what are the equilibrium concentrations of \(\mathrm{NO}\) \(\mathrm{N}_{2},\) and \(\mathrm{O}_{2} ?\)

Consider the following equilibrium, for which\(K_{p}=0.0752\) at \(480^{\circ} \mathrm{C} :\) $$2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g)$$ \begin{equation} \begin{array}{l}{\text { (a) What is the value of } K_{p} \text { for the reaction }} \\ {4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?} \\ {\text { (b) } \mathrm{What} \text { is the value of } K_{p} \text { for the reaction }} \\\ {\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?} \\ {\text { (c) What is the value of } K_{c} \text { for the reaction in part (b)? }}\end{array} \end{equation}

\(\mathrm{At} 900 \mathrm{K},\) the following reaction has \(K_{p}=0.345 :\) $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ In an equilibrium mixture the partial pressures of \(S \mathrm{O}_{2}\) and \(\mathrm{O}_{2}\) are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of \(\mathrm{SO}_{3}\) in the mixture?

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) .\) An equilibrium mixture in a 2.00 -L vessel is found to contain 0.0406 \(\mathrm{mol}\) \(\mathrm{CH}_{3} \mathrm{OH}, 0.170 \mathrm{mol} \mathrm{CO},\) and 0.302 \(\mathrm{mol} \mathrm{H}_{2}\) at 500 \(\mathrm{K}\) . Calculate \(K_{c}\) at this temperature.

When 2.00 \(\mathrm{mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 2.00 -L flask at 303 \(\mathrm{K}, 56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2} :\) $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{p}\) for this reaction at 303 \(\mathrm{K}\) . (c) According to Le Chatelier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture were transferred to a \(15.00-\mathrm{L}\) . vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when 2.00 mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(15.00-\mathrm{L}\) vessel at 303 \(\mathrm{K}\) .

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