Problem 198 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant for the reaction \(\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})\) is \(K_{c}=4.9 \times 10^{-2}\). the value of \(K_{c}\) for the reaction \(2 \mathrm{SO}_{2}\) \((\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) will be (a) 416 (b) \(2.40 \times 10^{-3}\) (c) \(9.8 \times 10^{-2}\) (d) \(4.9 \times 10^{-2}\)

Step-by-Step Solution

Answer

The value of \(K_c\) is 416, corresponding to option (a).

Step 1: Understand the Given Reaction and its Equilibrium Constant

The problem provides the equilibrium constant, \(K_c = 4.9 \times 10^{-2}\), for the reaction \( \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2}\mathrm{O}_{2}(\mathrm{~g}) \). This means the forward reaction forms \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) from \(\mathrm{SO}_{3}\).

Step 2: Write the Expression for Given Reaction's Equilibrium Constant

For the given reaction:\[ K_c = \frac{[\mathrm{SO}_2][\mathrm{O}_2]^{1/2}}{[\mathrm{SO}_3]} = 4.9 \times 10^{-2} \]

Step 3: Identify the Required Reaction

We are looking for the equilibrium constant for the reverse reaction: \( 2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) \). This is essentially the reverse of the given reaction multiplied by 2.

Step 4: Relate the Equilibrium Constants with the Required Reaction

The equilibrium constant of a reverse reaction is the reciprocal of the constant for the forward reaction. Additionally, when reactions are multiplied by a factor, exponents are applied to the equilibrium constant.Since the reaction is reversed and doubled, the new equilibrium constant \(K_c\) can be calculated as follows:\[ K_{c,\text{new}} = \left(\frac{1}{K_c}\right)^2 = \left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \]

Step 5: Calculate New Equilibrium Constant

Compute the reciprocal and apply the exponential based on the stoichiometry change:\[ K_{c,\text{new}} = \left( 20.408 \right)^2 = 416 \]

Step 6: Select the Correct Option

Based on the calculation, the correct equilibrium constant \(K_c\) for the reaction \(2\mathrm{SO}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{SO}_3(\mathrm{g})\) is 416, which corresponds to option (a).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time. In a given reversible reaction, the rate at which the reactants convert to products is equal to the rate at which the products convert back to reactants.
This balance does not imply that the reactants and products are present in equal amounts, but rather that their concentrations have stabilized. There are some key features to understand when discussing chemical equilibrium:

• Dynamic Nature: While concentrations remain constant, the reaction continues to occur in both directions at equal rates.
• Equilibrium Position: This is the ratio of the concentrations of products to reactants at equilibrium, described by the equilibrium constant, K.
• Le Chatelier's Principle: If a change is applied to a system at equilibrium, the system shifts to counteract that change, re-establishing equilibrium.

Understanding these concepts helps in interpreting how different conditions affect the state of equilibrium in chemical systems.

Reversible Reactions

Reversible reactions are processes where the reactants transform into products and vice versa, potentially reaching a point of equilibrium. The fundamental nature of reversible reactions can be represented by a double arrow (↔️), indicating the forward and reverse paths of the reaction.
Here are the essentials of reversible reactions:

• Bidirectionality: Both forward (reactants to products) and reverse (products back to reactants) processes can occur.
• Equilibrium Attainment: Over time, the rates of these two processes equalize, achieving equilibrium.
• Temperature and Pressure Influence: Changes in temperature and pressure can shift equilibrium positions and impact the extent of these reversible reactions.

Being familiar with reversible reactions allows you to predict how different conditions affect the direction and extent of a reaction.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to assess the proportions of substances involved, central to understanding both equilibrium and the conversion of reactants to products.
Key points to consider in stoichiometry include:

• Balanced Chemical Equations: Essential for correctly representing the quantities of reactants and products.
• Mole Ratios: Coefficients from balanced equations provide ratios for calculating the amount of substances consumed or produced.
• Complex Reactions: For reactions like the one discussed in the exercise, changes in stoichiometry impact the calculation of equilibrium constants.

Grasping the principles of stoichiometry is fundamental in determining how changes to a reaction's conditions affect chemical equilibrium.

Reaction Kinetics

Reaction kinetics is the study of the speed, or rate, of chemical reactions and the factors affecting this rate. Understanding kinetics is crucial for predicting how fast a system reaches equilibrium.

• Reaction Rate: Defined as the change in concentration of reactants or products per unit time.
• Catalysts: Substances that increase the reaction rate without being consumed, often achieving equilibrium more quickly.
• Factors Affecting Rate: Temperature, concentration, and the presence of catalysts all influence reaction speed.

These concepts help elucidate how rapid changes in reaction conditions can affect the approach to equilibrium, altering the dynamics of reversible reactions. A deep understanding of reaction kinetics provides valuable insights into chemical processes across various fields.