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If \(K_{c}=1\) for the equilibrium \(3 \mathrm{~A}(g) \rightleftharpoons 2 \mathrm{~B}(g)\), what is the relationship between [A] and [B] at equilibrium?

Short Answer

Expert verified
The relationship between the concentrations of A and B at equilibrium is: \[[B] = \sqrt{[A]^3}\]

Step by step solution

01

Write down the equilibrium constant expression

For the given equilibrium, we can write the Kc expression as follows: \[K_c = \frac{[B]^2}{[A]^3}\] Since we're given that Kc = 1 for this reaction, the equation becomes: \[1 = \frac{[B]^2}{[A]^3}\]
02

Solve for the relationship between [A] and [B]

To find the relationship between the concentrations of A and B, we simply solve the equation for one of the variables in terms of the other. In this case, let's solve for [B]^2: \[[B]^2 = [A]^3\] Now, we can find the relationship between [A] and [B] by taking the square root of both sides of the equation: \[[B] = \sqrt{[A]^3}\] This equation shows the relationship between the concentrations of A and B at equilibrium. That is, the concentration of B is equal to the square root of the cube of the concentration of 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 in a chemical reaction where the concentrations of reactants and products remain constant over time. At this point, the rate at which the reactants convert to products equals the rate of the products converting back to reactants. This does not mean the quantities of reactants and products are equal, but rather that their rates are balanced, resulting in no net change.

In the context of our given reaction, \(3A(g) \rightleftharpoons 2B(g)\), the system reaches equilibrium when the forward reaction rate (\(3A \rightarrow 2B\)) equals the reverse rate (\(2B \rightarrow 3A\)). Recognizing equilibrium in a chemical reaction helps students understand how different conditions can affect reaction dynamics.
Concentration Relationship
The relationship between the concentrations of the reactants and products at equilibrium is important for understanding how equilibrium is maintained. In our example, we're interested in the relationship between \([A]\) and \([B]\).

This relationship can be derived from the expression of the equilibrium constant \(K_c\) which is provided for the reaction. Using the expression \(K_c = \frac{[B]^2}{[A]^3}\), we can explore how changing the concentration of one component will impact the other while maintaining \([K_c] = 1\). In essence, the concentration relationship at equilibrium can reveal how reagents distribute themselves in a reaction.

For the exercise, the equation \([B] = \sqrt{[A]^3}\) shows that the concentration of \([B]\) is linked directly to the concentration of \([A]\) raised to the power of three and then square rooted.
Kc Expression
The equilibrium constant, represented as \(K_c\), is a vital part of understanding chemical equilibria. It quantifies the ratio of concentration of products to reactants, each raised to the power of their respective coefficients from the balanced equation.

For the reaction \(3A(g) \rightleftharpoons 2B(g)\), the \(K_c\) expression is written as \(K_c = \frac{[B]^2}{[A]^3}\). This formula helps determine how the concentrations of \([A]\) and \([B]\) relate to each other at equilibrium, given a specific \(K_c\) value.

When \(K_c = 1\), it shows a state of balance where neither reactants nor products dominate. This allows us to derive the relationship \([B]^2 = [A]^3\), simplifying to \([B] = \sqrt{[A]^3}\), a useful format to express how product concentration depends on reactant concentration under equilibrium conditions. Understanding \(K_c\) expressions equips students with the tools to interpret and predict the behavior of chemical systems.

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Most popular questions from this chapter

At \(700 \mathrm{~K}\), the equilibrium constant for the reaction $$\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ is \(K_{p}=77\). A flask is charged with \(202.7 \mathrm{kPa}\) of \(\mathrm{CCl}_{4}\), which then reaches equilibrium at \(700 \mathrm{~K}\). (a) What fraction of the \(\mathrm{CCl}_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?\) (b) What are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

At \(900^{\circ} \mathrm{C}, K_{p}=51.2\) for the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressure of \(\mathrm{NO}(g)\) is half the pressure of \(\mathrm{NOBr}(g)\), what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g)+\mathrm{Cl}(g)$$ At \(25^{\circ} \mathrm{C}\), the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?(\mathbf{b})\) Are reactants or products more plentiful at equilibrium?

True or false: When the temperature of an exothermic reaction increases, the rate constant of the forward reaction decreases, which leads to a decrease in the equilibrium constant, \(K_{c}\)

Bromine and hydrogen react in the gas phase to form hydrogen bromide: \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) .\) The reaction enthalpy is \(\Delta H^{\circ}=-6 \mathrm{~kJ} .\) (a) To increase the equilibrium yield of hydrogen bromide would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen bromide by controlling the pressure of this reaction? If so, would high or low pressure favor formation of \(\mathrm{HBr}(g) ?\)

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