Problem 124 - A Complete Resource Book in Chemistry for JEE Main

Question

Determine the value of equilibrium constant \(\left(\mathrm{K}_{\mathrm{C}}\right.\) ) for the reaction $$ \mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g}) $$ If 10 moles of \(A_{2} ; 15\) moles of \(B_{2}\) and 5 moles of \(A B\) are placed in a 2 litre vessel and allowed to come to equilibrium. The final concentration of \(\mathrm{AB}\) is \(7.5 \mathrm{M}\) : (a) \(4.5\) (b) \(1.5\) (c) \(0.6\) (d) None of these

Step-by-Step Solution

Answer

The value of the equilibrium constant \(K_C\) is 4.5.

Step 1: Determine Initial Concentrations

First, let's calculate the initial concentrations of each species in the reaction vessel, which has a volume of 2 litres.- Initial concentration of \(A_2\) is \( \frac{10 \, \text{moles}}{2 \, \text{litres}} = 5 \, \text{M} \).- Initial concentration of \(B_2\) is \( \frac{15 \, \text{moles}}{2 \, \text{litres}} = 7.5 \, \text{M} \).- Initial concentration of \(AB\) is \( \frac{5 \, \text{moles}}{2 \, \text{litres}} = 2.5 \, \text{M} \).

Step 2: Calculate Change in Concentration

We know the concentration of \(AB\) at equilibrium is 7.5 M. Therefore, there has been a change in concentration of \(\Delta[AB] = 7.5 - 2.5 = 5 \, \text{M}\).Since the stoichiometry of \(AB\) is 2:1 in the reaction, this change in \(AB\) implies a change of \(2.5 \, \text{M}\) in both \(A_2\) and \(B_2\) (i.e., \(-2.5 \, \text{M}\) for each).

Step 3: Determine Equilibrium Concentrations

Subtract the change from the initial concentrations to find the equilibrium concentrations of \(A_2\) and \(B_2\):- Equilibrium concentration of \(A_2\) is \(5 - 2.5 = 2.5 \, \text{M} \).- Equilibrium concentration of \(B_2\) is \(7.5 - 2.5 = 5 \, \text{M} \).- Equilibrium concentration of \(AB\) is provided as 7.5 M.

Step 4: Calculate Equilibrium Constant

The equilibrium constant \(K_C\) is expressed as:\[ K_C = \frac{[AB]^2}{[A_2][B_2]} \]Substitute the equilibrium concentrations into the equation:\[ K_C = \frac{(7.5)^2}{(2.5)(5)} \]Calculate the value:\[ K_C = \frac{56.25}{12.5} = 4.5 \]

Key Concepts

Understanding Initial Concentrations

To determine the equilibrium constant for a chemical reaction, we first need to analyze the initial concentrations of the reactants and products. This involves calculating how many moles of each substance exist in a given volume before any reaction takes place. In the exercise you're working with, you're given 10 moles of \(A_2\), 15 moles of \(B_2\), and 5 moles of \(AB\) in a 2-litre container. These figures allow us to determine their respective initial concentrations by dividing the number of moles by the volume of the vessel.

Here's how you would calculate the initial concentrations:

• The initial concentration of \(A_2\) is \( \frac{10\, \text{moles}}{2\, \text{litres}} = 5\, \text{M} \).
• The initial concentration of \(B_2\) is \( \frac{15\, \text{moles}}{2\, \text{litres}} = 7.5\, \text{M} \).
• The initial concentration of \(AB\) is \( \frac{5\, \text{moles}}{2\, \text{litres}} = 2.5\, \text{M} \).

Establishing these concentrations is the first step in understanding how the reaction will progress towards equilibrium.

Change in Concentration: Tracking the Reaction Progress

During a chemical reaction, the concentrations of reactants and products change until equilibrium is reached. For the given reaction, we know that the concentration of \(AB\) at equilibrium is \(7.5 \, \text{M}\). This means that there has been an observable change from its initial concentration of \(2.5 \, \text{M}\).

The change in concentration for \(AB\) is calculated as follows:
\[ \Delta[AB] = 7.5 - 2.5 = 5 \, \text{M} \]
Since for every mole of \(AB\) produced, half a mole of \(A_2\) and \(B_2\) are consumed (based on reaction stoichiometry), we find:

• \( \Delta[A_2] = -2.5 \, \text{M} \)
• \( \Delta[B_2] = -2.5 \, \text{M} \)

Understanding these changes helps us calculate the equilibrium concentrations of all substances involved.

Equilibrium Concentrations: Where Balance is Achieved

After establishing how concentrations change, we can determine the equilibrium concentrations of all reactants and products. The reaction reaches a balance where the rates of the forward and reverse reactions are equal. Based on the initial concentrations and the changes calculated previously, we now subtract these changes from the starting values to find the equilibrium concentrations.

Here are the calculations for equilibrium concentrations:

• The equilibrium concentration of \(A_2\) is \(5 - 2.5 = 2.5 \, \text{M} \).
• The equilibrium concentration of \(B_2\) is \(7.5 - 2.5 = 5 \, \text{M} \).
• The equilibrium concentration of \(AB\) is given as \(7.5 \, \text{M} \).

These concentrations are critical for determining the equilibrium constant \(K_C\). They illustrate the balance between reactants and products at equilibrium.

Reaction Stoichiometry: The Guiding Principle

Reaction stoichiometry gives us the coefficients from the balanced equation and tells us how moles of reactants are converted to moles of products. In this reaction:

\[ A_2(g) + B_2(g) \rightleftharpoons 2 AB(g) \]
The stoichiometry provides a ratio that is essential for understanding how changes in concentration affect each species when the reaction moves towards equilibrium. It implies that one mole of \(A_2\) and one mole of \(B_2\) yield two moles of \(AB\).

When we see a 5 M increase in \(\text{AB}\), the reaction stoichiometry tells us that there is a corresponding \(-2.5\,\text{M}\) change in both \(A_2\) and \(B_2\) since they are reactants consumed in forming \(AB\).

This stoichiometric understanding paves the way for determining the equilibrium constant \(K_C\), using the formula:
\[ K_C = \frac{[AB]^2}{[A_2][B_2]} \] Substituting our equilibrium concentrations here, the calculated value of \(K_C\) reflects the equilibrium state of the system.