Problem 19 - Chemistry and Chemical Reactivity

Question

Which of the following correctly relates the equilibrium constants for the two reactions shown? \(A+B \rightleftarrows 2 C \quad K_{1}\) \(2 \mathrm{A}+2 \mathrm{B} \rightleftarrows 4 \mathrm{C} \quad K_{2}\) (a) \(K_{2}=2 K_{1}\) (c) \(K_{2}=1 / K_{1}\) (b) \(K_{2}=K_{1}^{2}\) (d) \(K_{2}=1 / K_{1}^{2}\)

Step-by-Step Solution

Answer

(b) \(K_{2} = K_{1}^{2}\)

Step 1: Understanding the Problem

We are given two chemical equilibrium reactions: \(A + B \rightleftarrows 2C\) with equilibrium constant \(K_1\) and \(2A + 2B \rightleftarrows 4C\) with equilibrium constant \(K_2\). The task is to determine the relationship between \(K_1\) and \(K_2\).

Step 2: Writing the Expression for K1

For the reaction \(A + B \rightleftarrows 2C\), the equilibrium constant \(K_1\) can be written as: \[ K_1 = \frac{[C]^2}{[A][B]} \]

Step 3: Writing the Expression for K2

For the reaction \(2A + 2B \rightleftarrows 4C\), the equilibrium constant \(K_2\) can be expressed as: \[ K_2 = \frac{[C]^4}{[A]^2[B]^2} \]

Step 4: Relating K2 to K1 by Reaction Doubling

Notice that the second reaction is exactly double the first in stoichiometry. When a chemical equation is doubled, the equilibrium constant is squared. Thus, \[ K_2 = (K_1)^2 \]

Step 5: Checking Choices Given

The correct relationship from the given answer choices is (b): \(K_2 = K_1^2\).

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs in a reversible chemical reaction when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of reactants and products remain constant. This doesn't mean that reactants and products are equal, but that their rates of change match over time, establishing a state of balance. In the equation context, consider the reaction: \( A + B \rightleftharpoons 2C \). When this reaction reaches equilibrium, the amount of \( A \), \( B \), and \( C \) in the system does not change.
A few important points about chemical equilibrium are:

• It is dynamic, meaning that reactions don't stop; rather, they occur at the same rate in both directions.
• Different conditions like temperature and pressure can shift the equilibrium point.
• The equilibrium state is predictable through the equilibrium constant \( K \), which helps determine the ratio of products to reactants at equilibrium.

Understanding this concept aids in determining how equilibrium constraints affect chemical reactions and their constants, as seen with \( K_1 \) and \( K_2 \) in reversible reactions.

Reaction Stoichiometry

Stoichiometry is a key component in understanding chemical reactions, as it involves the calculation of reactants and products in chemical equations. It's like a recipe that shows how much of each ingredient (reactant) you need to get the desired amount of products. In stoichiometry, coefficients and ratios are crucial.
For example, in the reaction \( A + B \rightarrow 2C \), the coefficients indicate that one molecule of \( A \) combines with one molecule of \( B \) to produce two molecules of \( C \). When the reaction equation is like \( 2A + 2B \rightarrow 4C \), it essentially doubles everything.
This doubling affects the equilibrium constant, as more of each reactant and product are involved. The doubling causes the equilibrium constant to be squared, transforming \( K_1 \) into \( K_1^2 \), leading to \( K_2 \).

• The coefficients dictate how quantities relate in reactions.
• Changes in stoichiometry affect the equilibrium expressions as the relationships between quantities shift.
• Understanding these ratios helps predict outcomes in chemical reactions based on given conditions.

By mastering stoichiometry, you can adjust equations and predict changes in chemical reactions, just as shown when transforming \( K_1 \) to \( K_2 \).

Equilibrium Expression

An equilibrium expression mathematically expresses the ratio of the concentrations of products to reactants at equilibrium. It's essential for calculating the equilibrium constant \( K \), which differs for every reaction. Consider the reaction: \( A + B \rightleftharpoons 2C \). The equilibrium constant \( K \) for this is calculated using the expression: \( K = \frac{[\text{products}]}{[\text{reactants}]} \). For this reaction, it translates to: \( K_1 = \frac{[C]^2}{[A][B]} \).
In another reaction \( 2A + 2B \rightleftharpoons 4C \), the products and reactants are used differently, resulting in \( K_2 = \frac{[C]^4}{[A]^2[B]^2} \).
The key here is that the exponents in the expression match the coefficients in the balanced equation, profoundly affecting how \( K \) is related to different variations of the balanced equation.

• The equilibrium constant tells the direction and extent to which a reaction proceeds.
• Changes in coefficients of the chemical equation show up as exponents in \( K \)'s expression.
• A different equation requires calculating a unique \( K \) based on its stoichiometry.

By understanding the equilibrium expression, you can predict how changes in reactants or products at equilibrium affect the equilibrium constant, as shown between \( K_1 \) and \( K_2 \).