Problem 71 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

At a certain temperature equilibrium constant \((K)\) is 16 for the reaction. $$ \begin{gathered} {[1987-5 \text { Marks }]} \\ \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) \end{gathered} $$ If we take one mole each of all the four gases in a one litre container, what would be the equilibrium concentrations of \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{NO}_{2}(\mathrm{~g})\) ?

Step-by-Step Solution

Answer

The equilibrium concentrations are approximately \([NO] = 2\, \text{M}\) and \([NO_2] = 0\, \text{M}\).

Step 1: Understanding the Reaction and Initial Conditions

We have the reaction: \( \mathrm{SO}_{2} + \mathrm{NO}_{2} \rightleftharpoons \mathrm{SO}_{3} + \mathrm{NO} \). The initial concentrations are 1 M for each species, given that the container has a volume of 1 litre. We denote the change in concentration at equilibrium by \(x\).

Step 2: Write the Equilibrium Expression

For the reaction, the equilibrium constant \(K\) is given by \( K = \frac{[\mathrm{SO}_3][\mathrm{NO}]}{[\mathrm{SO}_2][\mathrm{NO}_2]} \). Substituting \(K = 16\), we get: \( 16 = \frac{(1+x)^2}{(1-x)^2} \).

Step 3: Solving for x

Set up the equation: \( 16 = \frac{(1+x)^2}{(1-x)^2} \). By cross-multiplying, we get: \(16(1-x)^2 = (1+x)^2\). Simplifying, we obtain: \(16(1-2x+x^2) = 1 + 2x + x^2\). That simplifies further to \(16 - 32x + 16x^2 = 1 + 2x + x^2\).

Step 4: Solving the Quadratic Equation

Re-arrange terms to form a quadratic equation: \(15x^2 - 34x + 15 = 0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 15\), \(b = -34\), and \(c = 15\), compute \(x\).

Step 5: Calculating Equilibrium Concentrations

We find two potential solutions for \(x\), but only the one that keeps concentrations positive is valid. Using the positive valid \(x\), the equilibrium concentrations are \([\mathrm{NO}_2] = 1-x\) and \([\mathrm{NO}] = 1+x\). Calculate these values.

Key Concepts

Equilibrium Concentration

Equilibrium concentration refers to the concentrations of reactants and products in a chemical reaction that have reached a state where their amounts remain constant over time. This occurs when the forward and reverse reactions happen at the same rate. In our example, we start with 1 mole of each gas in a 1-litre container. At equilibrium, the concentrations change depending on the extent of the reaction, which is denoted by \( x \).

The equilibrium concentrations are calculated using this change \( x \). For every increase in product concentration due to the forward reaction, there is an equivalent decrease in reactant concentration because the reaction reaches a balance. Hence, if \( x \) moles of \( \mathrm{SO}_3 \) and \( \mathrm{NO} \) are formed, the concentration of these products is \( 1 + x \). Similarly, the concentrations of \( \mathrm{SO}_2 \) and \( \mathrm{NO}_2 \) are \( 1 - x \) moles. Understanding these changes helps predict the behavior of the components in equilibrium, which is crucial for steering reactions in practical applications.

Quadratic Equation in Chemistry

Quadratic equations often arise in chemistry when dealing with equilibrium problems, especially when equilibrium constants are involved. In this exercise, we encounter a quadratic equation while calculating the concentration changes in the equilibrium state.

Once we express the equilibrium constant \( K \) as a function of \( x \), solving the resulting equation usually leads to a quadratic form. Specifically, by substituting \( K = 16 \) into the equilibrium expression \( 16 = \frac{(1+x)^2}{(1-x)^2} \), we derive O the equation as \( 15x^2 - 34x + 15 = 0 \). This is a standard quadratic equation that can be solved by using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

where \( a = 15 \), \( b = -34 \), and \( c = 15 \). The quadratic equation solutions guide us in finding feasible values for \( x \) that maintain physically meaningful (positive) concentrations for all substances involved.

This process is critical because only positive concentration values are valid, ensuring they represent physically possible solutions in a chemical system.

Chemical Equilibrium Calculation

Calculating chemical equilibrium involves determining the concentrations of reactants and products at equilibrium, using the reaction's equilibrium constant \( K \). For the given reaction at equilibrium:

• \( \mathrm{SO}_3 \) and \( \mathrm{NO} \) are formed as products
• \( \mathrm{SO}_2 \) and \( \mathrm{NO}_2 \) are reactants
• Each participates in the forward and reverse reactions

In our case, starting from given initial conditions where all gas present has a concentration of 1 M, we use the relationship between \( x \) and \( K \) to find the equilibrium concentrations by solving the quadratic equation. After calculating \( x \) from the quadratic equation, it's essential to verify if the values make chemical sense, i.e., concentrations must remain positive.

Such calculations form the basis of predicting the direction of the reaction, optimizing conditions for maximum yield, and separating useful products from reaction mixtures. By mastering chemical equilibrium calculations, students can better understand how systems reach stability and how one might control these reactions in industrial or laboratory chemistry settings.