Problem 39 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For a reaction \(\mathrm{X}+\mathrm{Y} \rightleftharpoons 2 \mathrm{Z}, 1.0 \mathrm{~mol}\) of \(\mathrm{X}, 1.5 \mathrm{~mol}\) of \(\mathrm{Y}\) and \(0.5 \mathrm{~mol}\) of \(Z\) were taken in a 1 L vessel and allowed to react. At equilibrium, the concentration of \(Z\) was \(1.0 \mathrm{~mol} \mathrm{~L}^{-1} .\) The equilibrium constant of the reaction is \(\frac{x}{15} .\) The value of \(x\) is _________.

Step-by-Step Solution

Answer

The value of \(x\) is 16.

Step 1: Initial concentration setup

Initially, we have 1.0 mol of X, 1.5 mol of Y, and 0.5 mol of Z in a 1 L container, so initial concentrations are \([X]_0 = 1.0\), \([Y]_0 = 1.5\), and \([Z]_0 = 0.5\) mol/L.

Step 2: Change in concentration

During the reaction, the concentration of Z increases by 0.5 mol up to 1.0 mol/L. Since the reaction is \(\text{X} + \text{Y} \rightleftharpoons 2 \text{Z}\), for every mole of Z formed, 0.5 moles of X and Y are consumed. Hence, let the change in concentration of Z be \(\Delta[Z] = +0.5\), and thus \(\Delta[X] = \Delta[Y] = -0.25\).

Step 3: Equilibrium concentrations

At equilibrium, the concentrations are \([X] = 1.0 - 0.25 = 0.75\), \([Y] = 1.5 - 0.25 = 1.25\), and \([Z] = 1.0\).

Step 4: Expression for the equilibrium constant

The equilibrium constant K is given by \(K = \frac{[Z]^2}{[X][Y]}\). Substituting the equilibrium concentrations, we have \(K = \frac{(1.0)^2}{(0.75)(1.25)}\).

Step 5: Calculation of equilibrium constant

Calculate \(K = \frac{1.0}{0.9375} = 1.0667\). The equilibrium constant is given as \(\frac{x}{15}\), therefore we equate \(1.0667 = \frac{x}{15}\).

Step 6: Solve for x

Solve \(x = 1.0667 \times 15\) to find \(x = 16.001\). Rounding, we have \(x = 16\).

Key Concepts

Equilibrium Constant

In chemical equilibrium, the equilibrium constant is a pivotal concept that helps understand the ratio of products to reactants in a balanced chemical reaction. For any given reaction, like the one in our problem \(\mathrm{X} + \mathrm{Y} \rightleftharpoons 2 \mathrm{Z}\), the equilibrium constant, denoted as \(K_c\), is calculated using the concentrations of the reactants and products. This concept is crucial because it gives insights into the position of equilibrium, indicating whether the reactants or products are favored in a chemical equilibrium.

To find the equilibrium constant, we first need to determine the concentrations at equilibrium. For our reaction, the equilibrium constant expression is \(K = \frac{[\mathrm{Z}]^2}{[\mathrm{X}][\mathrm{Y}]}\). This formula tells us how the concentration of products (Z in this case) squares due to its reaction coefficient and how it compares to the product of reactants' concentrations (X and Y).

Substituting the known equilibrium concentrations into the equilibrium expression helps us calculate \(K \). In our problem, solving this gives us a numerical value of \(1.0667\). This calculation and result are essential for making predictions about the direction of the reaction and understanding the composition of the equilibrium mixture.

Reaction Stoichiometry

Reaction stoichiometry is another key concept in understanding chemical reactions and equilibrium. This concept involves the quantitative relationship between the reactants and products in a chemical reaction. Using stoichiometry, we can predict the amount of product that will be formed from given amounts of reactants, and vice versa.

In our example reaction, \(\text{X} + \text{Y} \rightleftharpoons 2 \text{Z}\), the stoichiometry tells us that one mole of Z is produced for each mole of X and Y that react, but because the coefficient for Z is 2, two moles of Z actually form. Therefore, when the concentration of Z changed by 0.5 mol/L, X and Y each decreased by just 0.25 mol/L.

Understanding these stoichiometric relationships is crucial because they allow us to calculate exactly how much of each reactant is needed or how much of each product to expect. They form the backbone of balancing chemical equations and calculating the quantities of reactants and products at any stage in a reaction sequence.

Mole Concept

The mole concept is fundamental in chemistry, acting as a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and liters. It allows chemists to count particles by weighing them. One mole of a substance contains exactly \(6.022 \times 10^{23}\) particles, whether they be atoms, molecules, or ions.

In the study of chemical equilibrium, the mole concept aids in calculating concentrations and quantities of substances. In our given reaction involving X, Y, and Z, we started with initial amounts in moles and converted them into molarity because the volume of the vessel was 1 liter. This simplification makes the calculations straightforward since 1 mole in 1 liter solution directly equates to 1 mol/L.

Using the initial moles helps establish the starting point for any equilibrium calculation. As reactions progress towards equilibrium, the concept of moles links changes in concentration directly to changes in the stoichiometry of the reaction, predicting the shifts in amounts as equilibrium is approached. It's a powerful counting tool that helps make sense of the behavior of substances during chemical reactions.