Problem 90 - Chemistry Matter and Change

Question

The decomposition reaction \(\mathrm{A}_{2} \mathrm{B} \rightarrow 2 \mathrm{A}+\mathrm{B}\) proceeds to equilibrium at \(499^{\circ} \mathrm{C}\) . Analysis of the equilibrium mixture shows \(\left[\mathrm{A}_{2} \mathrm{B}\right]=0.855 \mathrm{mol} / \mathrm{L},[\mathrm{A}]=2.045 \mathrm{mol} / \mathrm{L},\) and \([\mathrm{B}]=1.026 \mathrm{mol} / \mathrm{L} .\) What is \(K_{\mathrm{eq}} ?(\text {Chapter } 17)\)

Step-by-Step Solution

Answer

The equilibrium constant (\(K_{eq}\)) for the decomposition reaction \(A_2B \rightarrow 2A + B\) is approximately \(4.93\), given the equilibrium concentrations \([A_2B] = 0.855\ mol/L\), \([A] = 2.045\ mol/L\), and \([B] = 1.026\ mol/L\).

Step 1: Write the equilibrium constant expression

In general, the equilibrium constant, \(K_{eq}\), for a reaction of the form \(aA + bB \rightleftharpoons cC + dD\) can be written as: \(K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}\) For the given decomposition reaction \(A_2B \rightarrow 2A + B\), the equilibrium constant expression will be: \(K_{eq} = \frac{[A]^2[B]}{[A_2B]}\)

Step 2: Plug in the given equilibrium concentrations

Now, we will substitute the given equilibrium concentrations into the equilibrium constant expression: \(K_{eq} = \frac{[2.045]^2[1.026]}{[0.855]}\)

Step 3: Calculate the equilibrium constant

Evaluate the expression to find the value of \(K_{eq}\): \(K_{eq} = \frac{(2.045)^2(1.026)}{0.855} \approx 4.93\) Thus, the equilibrium constant for the decomposition reaction \(A_2B \rightarrow 2A + B\) is \(K_{eq} \approx 4.93\).

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial for grasping how reactions occur and the conditions under which they can reverse. During a chemical reaction, reactants are converted into products, but as the reaction progresses, these products can start to transform back into reactants. Chemical equilibrium is attained when the rate of the forward reaction equals the rate of the reverse reaction.

This state does not imply that the reactants and products are present in equal amounts. Instead, it means that their concentrations remain constant over time, creating a dynamic balance where the concentration of neither reactants nor products is changing. It’s important to note the equilibrium does not halt chemical activity but maintains a consistent state of exchange.

Equilibrium Constant Expression

The equilibrium constant expression quantifies the concentrations of the products and reactants at chemical equilibrium. For a balanced chemical equation, this expression is derived by raising the concentration of each product to the power of its coefficient in the balanced equation, multiplying these together, and then doing the same for the reactants, placing them under the product term to form a ratio.

As an example, for a generic reaction where aA + bB \rightleftharpoons cC + dD, the equilibrium constant expression is: \(K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}\). The square brackets denote the concentration of substances (in mol/L), and the letters with subscripts correspond to the stoichiometric coefficients from the equation. It is essential to understand that the equilibrium constant expression depends on the balanced chemical equation of the reaction and is a reflection of how products and reactants are related at equilibrium.

Reaction Quotient

The reaction quotient, denoted as Q, serves as a predictor for the direction in which a reaction mixture will proceed to achieve equilibrium. It is calculated using the same expression as the equilibrium constant but using the actual concentrations of the reactants and products at any given moment, not necessarily at equilibrium.

Comparing Q to the equilibrium constant, K_eq, informs us about the reaction’s status:

• If Q < K_eq, the reaction will proceed forward to produce more products.
• If Q > K_eq, the reaction will go in reverse to yield more reactants.
• If Q = K_eq, the system is already at equilibrium, and no shift in the direction of the reaction is needed.

The role of Q is pivotal for chemists in predicting reaction behavior and in designing conditions that favor the formation of desired products.