Problem 204 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})\) if \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\mathrm{x}}\) where the symbols have usual meaning then the value of \(x\) is: (assuming ideality) [2014] (a) \(\frac{1}{2}\) (b) 1 (c) \(-1\) (d) \(-\frac{1}{2}\)

Step-by-Step Solution

Answer

The value of \(x\) is \(-\frac{1}{2}\).

Step 1: Understand the Reaction

The reaction given is \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2}\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})\). This is an equilibrium reaction where some \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) gases are converting to \(\mathrm{SO}_{3}\).

Step 2: Determine Change in Moles

To find \(x\), we need to determine the change in moles of gas for the reaction, which is given as \(\Delta n = n_{\text{products}} - n_{\text{reactants}}\). In this reaction:- Products: \(\mathrm{SO}_{3}(g)\): 1 mole- Reactants: \(\mathrm{SO}_{2}(g)\): 1 mole + \(\frac{1}{2}\mathrm{O}_{2}(g)\): \(\frac{1}{2}\) moleSo, \(\Delta n = 1 - (1 + \frac{1}{2}) = 1 - 1.5 = -0.5\) moles.

Step 3: Equilibrium Constant Relationship

The relationship between \(\mathrm{K}_{\mathrm{p}}\) and \(\mathrm{K}_{\mathrm{C}}\) is given by the equation \(\mathrm{K}_{\mathrm{p}} = \mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta n}\). Here, \(\Delta n\) we just calculated as \(-0.5\).

Step 4: Identify the Value of x

Since \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{x}\) must match \(\mathrm{K}_{\mathrm{p}} = \mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta n}\), we conclude that \(x\) must be equal to \(\Delta n = -0.5\).

Key Concepts

Equilibrium Constant

The equilibrium constant is essential in understanding chemical reactions at equilibrium. It helps us know how far a reaction proceeds before reaching a state where the reactants and products no longer change. There are two commonly used types:

• Kc: The Equilibrium Constant in Terms of Concentrations - It involves the concentrations of reactants and products. It tells us how much reactant is turned into product at equilibrium using their molar concentrations in the expression.
• Kp: The Equilibrium Constant in Terms of Partial Pressure - This is used when reactions involve gases. Instead of concentrations, it uses the partial pressures of the gases involved in the reaction. This is particularly useful when dealing with gases because pressure measurements are often easier in some situations.

Understanding these constants helps predict the position of equilibrium and the extent to which a reaction can proceed. Whether we use Kc or Kp depends on the states and the nature of the reacting substances.

Change in Moles

In equilibrium reactions involving gases, such as the conversion of SO\(\_2\) and O\(\_2\) to SO\(\_3\), it is crucial to consider the change in moles of the reactants and products. The change in moles (\( \Delta n \)) is calculated as the difference between the number of moles of products and the number of moles of reactants:

• For products: Count the moles of each product.
• For reactants: Count the moles of each reactant, including any stoichiometric coefficients.

In the example problem, this was expressed as \( \Delta n = n_{\text{products}} - n_{\text{reactants}} \). Here, for the reaction \( \text{SO}\_{2}(\text{g}) + \frac{1}{2} \text{O}\_{2}(\text{g}) \rightleftharpoons \text{SO}\_{3}(\text{g}) \), we calculated \( \Delta n = -0.5 \) moles, meaning the total amount of gas decreases as the reaction moves to equilibrium. Knowing this helps in understanding the reaction's behavior under pressure changes.

Relationship between Kp and Kc

The relationship between the equilibrium constants Kp and Kc is crucial when dealing with reactions in gaseous states.It is expressed by the equation:\[K_{p} = K_{c} (RT)^{\Delta n}\]where:

• \( R \) is the ideal gas constant.
• \( T \) is the temperature in Kelvin.
• \( \Delta n \) is the change in moles of gas during the reaction.

This relationship shows how pressure, temperature, and change in moles are interconnected. In the exercise, they wanted us to equate \( K_{p} \) and \( K_{c} (RT)^{x} \). When the exponent \( x \) matches \( \Delta n \), we can identify and solve for it using the given conditions. Here, since \( \Delta n = -0.5 \), \( x \) is also \(-0.5\). This tells us that changes in pressure and temperature can be compensated by the change in moles, adjusting the equilibrium constant accordingly.