A reaction is represented by this equation: \(\mathrm{A}(a q)+2 \mathrm{B}(a q) \rightleftharpoons 2 \mathrm{C}(a q) \quad K_{c}=1 \times 10^{3}\) (a) Write the mathematical expression for the equilibrium constant. (b) Using concentrations \(\leq 1\) M, identify two sets of concentrations that describe a mixture of \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) at equilibrium.

Short Answer

Expert verified
The expression for the equilibrium constant is \( K_{c} = \frac{[C]^2}{[A][B]^2} \). Two sets of concentrations at equilibrium are [A] = 1 M, [B] = 1 M, [C] = 10 M and [A] = 0.01 M, [B] = 0.01 M, [C] = 0.3162 M.

Step by step solution

01

Write the expression for the equilibrium constant

Use the general equation for the equilibrium constant (Kc) for a reaction aA + bB \rightleftharpoons cC + dD: Kc = [C]^c [D]^d / [A]^a [B]^b. Apply this to the given reaction: \( K_{c} = \frac{[C]^2}{[A][B]^2} \).
02

Choose the first set of concentrations

Let's choose concentrations such that the ratio equals the given equilibrium constant, Kc (1 x 10^3). Assume a concentration of 1 M for A and 1 M for B. Calculate the concentration of C: \( K_{c} = \frac{[C]^2}{1 \times 1^2} = 1 \times 10^{3} \), which gives us \( [C] = \sqrt{1 \times 10^{3}} = 10^{\frac{3}{2}} / 10^{\frac{1}{2}} = 10 \) M.
03

Choose the second set of concentrations

For the second set, let's assume a concentration of 0.01 M for A and B. Using the equilibrium expression: \( K_{c} = \frac{[C]^2}{0.01 \times 0.01^2} = 1 \times 10^{3} \), calculate [C]: \( 1 \times 10^{3} = \frac{[C]^2}{10^{-4}} \), which gives us \( [C] = \sqrt{1 \times 10^{-1}} = 10^{-\frac{1}{2}} = 0.3162 \) M.

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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 rates of the forward and reverse reactions are equal, leading to no net change in the concentrations of the reactants and products over time. It is important to note that while the concentrations remain constant, the reactions are still occurring, but at the same rate in both directions.

For the reaction \(\mathrm{A}(aq) + 2\mathrm{B}(aq) \rightleftharpoons 2\mathrm{C}(aq)\), we reach equilibrium when the rate at which \(\mathrm{A}\) and \(\mathrm{B}\) combine to form \(\mathrm{C}\) is exactly balanced by the rate at which \(\mathrm{C}\) breaks down into \(\mathrm{A}\) and \(\mathrm{B}\). The equilibrium constant expression, derived from the law of mass action, quantitatively describes this balance.
Reaction Quotient
The reaction quotient, \(Q\), is a measure that tells us whether a system is at equilibrium, and if not, in which direction it needs to shift to reach equilibrium. It has the same form as the equilibrium constant expression, but it uses the current concentrations of the reactants and products, which may or may not be at equilibrium.

For our reaction \(\mathrm{A} + 2\mathrm{B} \rightleftharpoons 2\mathrm{C}\), the reaction quotient is \(Q = \frac{[C]^2}{[A][B]^2}\). If \(Q < K_c\), the reaction will proceed in the forward direction to produce more products, and if \(Q > K_c\), the reaction will proceed in the reverse direction to produce more reactants until \(Q\) equals \(K_c\) when equilibrium is achieved.
Le Chatelier's Principle
Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change. This principle is vital for predicting how changes in concentration, temperature, and pressure can affect the composition of the equilibrium mixture.

Applying Le Chatelier's principle to our reaction, if more \(\mathrm{A}\) or \(\mathrm{B}\) is added to the system, the equilibrium will shift to the right, increasing the production of \(\mathrm{C}\) until a new equilibrium is established. Conversely, if some \(\mathrm{C}\) is removed, the reaction will shift to produce more \(\mathrm{C}\) in response. The changes occur in a way that partially offsets the disturbance, aligning with Le Chatelier's principle.

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