Question

For the hypothetical reactions, the equilibrium constant \((\mathrm{K})\) values are given: \(\mathrm{A} \rightleftharpoons \mathrm{B} \mathrm{K}_{1}=2\) \(\mathrm{B} \rightleftharpoons \mathrm{C} \mathrm{K}_{2}^{1}=4\) \(\mathrm{C} \rightleftharpoons \mathrm{D} \mathrm{K}_{3}^{2}=3\) The equilibrium constant \((\mathrm{K})\) for the reaction: \(\mathrm{A} \rightleftharpoons \mathrm{D}\) is (a) 3 (b) 6 (c) 12 (d) 24

Short answer

The equilibrium constant \( \mathrm{K} \) for \( \mathrm{A} \rightleftharpoons \mathrm{D} \) is 24, so the answer is (d) 24.

Step-by-step solution

Understand the Reaction Path

To find the equilibrium constant for \( \mathrm{A} \rightleftharpoons \mathrm{D} \), we need to follow the sequence of reactions: \( \mathrm{A} \rightleftharpoons \mathrm{B} \rightleftharpoons \mathrm{C} \rightleftharpoons \mathrm{D} \). Each individual reaction has its own equilibrium constant.

Identify the Equilibrium Constants

The given equilibrium constants are: \( \mathrm{K}_{1} = 2 \) for \( \mathrm{A} \rightleftharpoons \mathrm{B} \), \( \mathrm{K}_{2}^{1} = 4 \) for \( \mathrm{B} \rightleftharpoons \mathrm{C} \), and \( \mathrm{K}_{3}^{2} = 3 \) for \( \mathrm{C} \rightleftharpoons \mathrm{D} \). These constants will help us determine the overall equilibrium constant.

Calculate Overall Equilibrium Constant

To find the equilibrium constant for the overall reaction \( \mathrm{A} \rightleftharpoons \mathrm{D} \), multiply all individual equilibrium constants together: \( \mathrm{K} = \mathrm{K}_{1} \times \mathrm{K}_{2}^{1} \times \mathrm{K}_{3}^{2} \). Substitute the known values: \( \mathrm{K} = 2 \times 4 \times 3 \).

Perform the Calculation

Multiply the constants together: \( 2 \times 4 = 8 \). Then, \( 8 \times 3 = 24 \). Therefore, the equilibrium constant \( \mathrm{K} \) for the reaction \( \mathrm{A} \rightleftharpoons \mathrm{D} \) is 24.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that refers to a state where the concentrations of reactants and products do not change with time. This happens when the forward and reverse reactions occur at the same rate.
At equilibrium, the system's composition remains constant, although the reactions continue to occur. The equilibrium state is dynamic, not static.
For any reversible reaction, such as \[ ext{A} ightleftharpoons ext{B}, \]the equilibrium constant \( K \) gives the ratio of the concentrations of products to reactants:\[ K = \frac{[ ext{B}]}{[ ext{A}]} \] where \([\text{B}]\) and \([\text{A}]\) are the equilibrium concentrations of B and A, respectively.

• If \( K \) is large, the reaction favors the formation of products.
• If \( K \) is small, the reaction favors the formation of reactants.

Understanding chemical equilibrium helps predict the concentration of materials present at equilibrium, which is crucial in various chemical processes.

Reaction Pathways

Reaction pathways describe the step-by-step sequence of elementary reactions by which an overall chemical change occurs. In the given problem, the pathway starts from A and ends at D, moving through intermediate steps B and C.
Each step or elementary process has its own equilibrium constant.
These reactions are:

• \( \text{A} \rightleftharpoons \text{B} \) (Equilibrium constant \( K_1 \))
• \( \text{B} \rightleftharpoons \text{C} \) (Equilibrium constant \( K_2 \))
• \( \text{C} \rightleftharpoons \text{D} \) (Equilibrium constant \( K_3 \))

By understanding each individual step, we can follow the overall reaction pathway effectively. The pathway provides valuable insights into the stepwise transformations of reactants into products, essential for calculating the overall equilibrium constant from the individual constants.

Multiplication of Constants

When dealing with a sequence of reactions, it is often necessary to find the overall equilibrium constant for the total transformation of starting materials to final products.
This is achieved by multiplying the equilibrium constants of each step along the pathway.
For example, given the reactions:

• \( \text{A} \rightleftharpoons \text{B} \) with \( K_1 = 2 \)
• \( \text{B} \rightleftharpoons \text{C} \) with \( K_2 = 4 \)
• \( \text{C} \rightleftharpoons \text{D} \) with \( K_3 = 3 \)

We calculate the overall equilibrium constant for \( \text{A} \rightleftharpoons \text{D} \) as:\[ K_{overall} = K_1 \times K_2 \times K_3 = 2 \times 4 \times 3 = 24. \]
Multiplying the constants provides the equilibrium constant for the complete transformation from A to D, reflecting the overall tendency of all reactions combined.

Hypothetical Reactions

Hypothetical reactions are used to model and analyze chemical processes without lab experimentation. They help us understand complex chemical systems and facilitate calculations.
In the given example, three hypothetical reactions are examined:

• \( \text{A} \rightleftharpoons \text{B} \)
• \( \text{B} \rightleftharpoons \text{C} \)
• \( \text{C} \rightleftharpoons \text{D} \)

Each individual reaction is assigned an equilibrium constant. This allows for the calculation of the equilibrium constant for an overarching reaction, such as \( \text{A} \rightleftharpoons \text{D} \), without needing direct experimental data.
Hypothetical reactions enable a straightforward exploration of theories and calculations, such as understanding the combined effects of multiple reaction pathways and equilibrium constants. They are invaluable in predicting the behavior of chemical systems under different conditions.