Problem 6 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

If the equilibrium constant for \(\mathrm{A} \rightleftharpoons \mathrm{B}+\mathrm{C}\) is \(\mathrm{K}_{\mathrm{eq}}^{(1)}\) and that of \(\mathrm{B}+\mathrm{C} \rightleftharpoons \mathrm{P}\) is \(\mathrm{K}_{\mathrm{eq}}^{(2)}\), the equilibrium constant for \(\mathrm{A} \rightleftharpoons \mathrm{P}\) is : (a) \(\mathrm{K}_{\mathrm{eq}}^{(1)} / \mathrm{K}_{\mathrm{eq}}^{(2)}\) (b) \(\mathrm{K}_{\mathrm{eq}}^{(2)}-\mathrm{K}_{\mathrm{eq}}^{(1)}\) (c) \(\mathrm{K}_{\mathrm{eq}}^{(1)}+\mathrm{K}_{\mathrm{eq}}^{(2)}\) (d) \(\mathrm{K}_{\mathrm{eq}}^{(\mathrm{I})} \mathrm{K}_{\mathrm{eq}}^{(2)}\)

Step-by-Step Solution

Answer

Option (d) \( \mathrm{K}_{\mathrm{eq}}^{(1)} \mathrm{K}_{\mathrm{eq}}^{(2)} \).

Step 1: Understand the Reactions

First, identify the reactions involved: \( \text{A} \rightleftharpoons \text{B} + \text{C} \) and \( \text{B} + \text{C} \rightleftharpoons \text{P} \). You need to find the equilibrium constant for the overall reaction \( \text{A} \rightleftharpoons \text{P} \).

Step 2: Use the Property of Equilibrium Constants

The overall reaction \( \text{A} \rightleftharpoons \text{P} \) is the sum of the two given reactions. In chemical equilibrium, when reactions are added, their equilibrium constants are multiplied. Thus, \( K_{eq}^{(overall)} = K_{eq}^{(1)} \times K_{eq}^{(2)} \).

Step 3: Identify the Correct Option

From the calculated product of equilibrium constants, the correct option that reflects \( K_{eq}^{(1)} \times K_{eq}^{(2)} \) is (d) \( \mathrm{K}_{\mathrm{eq}}^{(1)} \mathrm{K}_{\mathrm{eq}}^{(2)} \).

Key Concepts

Chemical Equilibrium

Understanding chemical equilibrium is central to solving problems like the one presented in the exercise. At chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This balance means that the concentrations of reactants and products remain constant over time, even though reactions are still occurring at the molecular level.
The equilibrium constant, denoted as \( K_{eq} \), quantifies the ratio of products to reactants at equilibrium for a given reaction at a specific temperature. In our exercise, this concept helps us find how maneuvers made in the reaction impact the overall equilibrium.
Several factors affect equilibrium:

• Concentration of reactants/products
• Temperature
• Pressure (for gases)

An important principle to remember is Le Chatelier's Principle, which states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust itself to counteract partially the effect of the change and a new equilibrium will be established.
Thus, understanding how these factors work together will allow you to predict and calculate changes in equilibrium efficiently.

Reaction Kinetics

While chemical equilibrium tells us about concentrations at rest, reaction kinetics gives us the rate at which the reactions proceed. This knowledge is essential to understand how quickly a system can reach equilibrium and what influences these speeds.
Kinetics looks at the activation energy and the transition state through which reactions must pass. The higher the activation energy, the slower the reaction, because fewer molecules have the sufficient energy to surpass the energy barrier.
Factors influencing reaction rates include:

• Temperature: increased temperatures generally increase the rate of reaction.
• Catalysts: substances that lower the activation energy of a reaction, increasing the reaction rate without being consumed.
• Concentration: higher concentrations of reactants can increase the likelihood of collisions, thereby increasing the reaction rate.

Understanding reaction kinetics allows us to manipulate conditions to achieve an equilibrium state more quickly or stabilize certain products over time. In the given exercise, it ensures we see how \( K_{eq} \) from one reaction interplays with the kinetics to reach equilibrium in another combined reaction.

Equilibrium Reactions

Equilibrium reactions are those that proceed in both forward and reverse directions. In these, the formation of products and the conversion back into reactants occur simultaneously.
In our original exercise, we dealt with equilibrium reactions, specifically combining them to determine a new equilibrium constant. When equilibrium reactions are considered, one reaction's equilibrium constant provides a snapshot of the balance between reactants and products. When two such reactions are combined, their equilibrium constants help us understand the broader system's equilibrium.
When reactions are added as in the problem's scenario, the equilibrium constants multiply. This result is a direct consequence of the relationship between the concentrations at equilibrium in the connected reactions.
For instance, if reaction 1 has equilibrium constant \( K_{eq}^{(1)} \) and reaction 2 has \( K_{eq}^{(2)} \), the combined reaction equilibrium is given by \( K_{eq}^{(1)} \times K_{eq}^{(2)} \). This multiplication tells us about the proportional presence of reactants and products when these reactions are considered together, as seen in part (d) of the exercise. Understanding how these reactions interlink helps in problem-solving and predicting reaction behavior.