Problem 107 - General Chemistry: Principles and Modern Applications

Question

Equilibrium is established in the reversible reaction \(2 \mathrm{A}+\mathrm{B} \rightleftharpoons 2 \mathrm{C} .\) The equilibrium concentrations are \([\mathrm{A}]=0.55 \mathrm{M},[\mathrm{B}]=0.33 \mathrm{M},[\mathrm{C}]=0.43 \mathrm{M}\) What is the value of \(K_{c}\) for this reaction?

Step-by-Step Solution

Answer

The value of \(K_{c}\) for this reaction is 0.209698.

Step 1: Write down the Mole Ratio of the Equation

From the equation, the mole ratio is as follows: \(2 \mathrm{A} + \mathrm{B} \rightleftharpoons 2 \mathrm{C}\). This implies that for every 2 moles of A and 1 mole of B reacting, they produce 2 moles of C in equilibrium.

Step 2: Write the Expression for the Equilibrium Constant[\(K_{c}\)]

The expression for the equilibrium constant \(K_c\) for reactions in general is given by: \(K_{c}\) = (concentration of products) / (concentration of reactants). Each concentration is raised to the power equivalent to the number of moles in the balanced reaction equation. In this case, the equilibrium constant \(K_c\) = \([C]^2\) / ([A]^2 [B]).

Step 3: Substitute the Given Concentrations

Substitute the given equilibrium concentration values into the \(K_c\) expression. Therefore, \(K_{c}\) = (0.43)^2 / ((0.55)^2 * 0.33).

Step 4: Perform the Calculation

Perform the calculation to arrive at the equilibrium constant. Using a calculator, this results in: \(K_c\) = 0.209698.

Key Concepts

Chemical Equilibrium

The concept of chemical equilibrium plays a pivotal role in understanding chemical reactions, especially reversible ones. In a system at equilibrium, the rate of the forward reaction, which consumes reactants to form products, is equal to the rate of the reverse reaction, where the products decompose back into reactants. This balance of rates does not imply that the reactants and products are in equal concentrations, but rather that their concentrations remain constant over time.

For students tackling this concept, it's crucial to digest that equilibrium does not mean that the reaction has stopped, but that it is dynamically proceeding at the same rate in both directions. This dynamism is why we can observe stable concentrations of both reactants and products in a reaction mixture at equilibrium.

Equilibrium Concentration Calculation

To calculate the equilibrium concentrations of substances in a chemical reaction, we must first understand the equilibrium constant expression, often denoted as \( K_c \). The equilibrium constant is a ratio of the concentration of products to the concentration of reactants, each raised to the power of their stoichiometric coefficients.

For example, when we have a balanced reaction equation such as \( 2A + B \rightleftharpoons 2C \), the equilibrium constant expression becomes \( K_c = \frac{[C]^2}{[A]^2[B]} \). Plugging in the concentrations into this expression allows us to solve for \( K_c \), which is a quantifiable measure of the position of equilibrium. Understanding this calculation is crucial for predicting the direction of the reaction and determining reaction yields.

Reversible Reactions

Reversible reactions are chemical processes that can proceed in both forward and reverse directions. To visualize this, consider that as reactants form products, some of these products can simultaneously revert to reactants. An important thing to remember about \reversible reactions\ is that they can reach a state of dynamic equilibrium where the concentrations of reactants and products remain unchanged.

To reinforce this learning point, emphasize that not all reactions are reversible, and those that are reversible may not always reach equilibrium. The extent to which a reaction is reversible and the ease with which equilibrium is achieved are indicated by \( K_c \), as it provides insight into whether the products or reactants are favored in the balanced reaction.