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

Question

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

Step-by-Step Solution

Answer

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

Step 1: Write the Reaction and Identify Δn

The given reaction is \( \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})=\mathrm{SO}_{3}(\mathrm{~g}) \). Calculate the change in the number of moles of gas, \( \Delta n \), which is the difference between the moles of gaseous products and reactants. Here, \( \Delta n = (1) - (1 + \frac{1}{2}) = -\frac{1}{2} \).

Step 2: Apply the Formula for Kₚ and Kc Relationship

We know the relationship between \( \mathrm{K}_{\mathrm{p}} \) and \( \mathrm{K}_C \) for gaseous reactions: \( \mathrm{K}_{\mathrm{p}} = \mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta n} \). Here, \( \Delta n = -\frac{1}{2} \).

Step 3: Determine the Value of x

From the equation \( \mathrm{K}_{\mathrm{p}} = \mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{x} \), we equate \( x = \Delta n \). Therefore, \( x = -\frac{1}{2} \).

Step 4: Choose the Correct Option

We found that \( x = -\frac{1}{2} \). Therefore, the correct answer is option (d) \( -\frac{1}{2} \).

Key Concepts

Kp and Kc Relationship

In the realm of chemical equilibrium, specific constants are used to describe the state of equilibrium in reactions involving gases. Two such constants are the equilibrium constant in terms of pressure, \( K_p \), and the equilibrium constant in terms of concentration, \( K_c \). For gaseous reactions, these constants are related through the formula:

• \( K_p = K_c(RT)^{\Delta n} \)

Here, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) represents the change in moles of gas. This relationship is crucial because it allows chemists to convert between pressure-based and concentration-based constants, depending on the conditions of the reaction. Keep in mind that this equation is used assuming ideal behavior of gases.

Gaseous Reaction

When dealing with equilibrium constants, it's important to consider whether a reaction involves gases, as this will determine how we express and measure these constants. Gaseous reactions are those that involve gaseous reactants and/or products. The given example: \[ \mathrm{SO}_{2} (\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{~g}) \rightarrow \mathrm{SO}_{3} (\mathrm{~g}) \]represents a gaseous reaction.In this context, understanding how to express equilibrium conditions for reactions with gases involves recognizing that the concentrations can often be replaced by partial pressures. That's why \( K_p \), which uses these partial pressures, is valuable in providing insights about the status of the reaction under various conditions.

Change in Moles (Δn)

The change in the number of moles of gas, \( \Delta n \), is a key element in understanding the relationship between \( K_p \) and \( K_c \). It is calculated as the difference between the moles of gaseous products and gaseous reactants:

• \( \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} \)

For the given reaction \( \mathrm{SO}_{2} (\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{~g}) \rightarrow \mathrm{SO}_{3} (\mathrm{~g}) \),we determine \( \Delta n \) by subtracting the sum of the moles on the reactant side from the moles on the product side. Here:

• Products: 1 mole of \( \mathrm{SO}_{3} \)
• Reactants: 1 mole of \( \mathrm{SO}_{2} \) and 0.5 mole of \( \mathrm{O}_{2} \)

Thus, \( \Delta n = 1 - (1 + 0.5) = -\frac{1}{2} \). This value represents the net change in moles of gas and plays a pivotal role in using the \( K_p \) and \( K_c \) relationship formula.