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 \(-\frac{1}{2}\), option (d).

Step 1: Write the General Equation Relating Kp and Kc

The relationship between the equilibrium constants \( K_{p} \) and \( K_{C} \) is given by the formula: \[ K_{p} = K_{C} (RT)^{\Delta n} \] where \( \Delta n \) is the change in the number of moles of gas.

Step 2: Identify the Change in Moles (Delta n)

For the reaction \( \mathrm{SO}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) \), compute \( \Delta n \): - Moles of gaseous products = 1 (for \( \mathrm{SO}_{3} \)) - Moles of gaseous reactants = 1.5 (\( \mathrm{SO}_{2} + 0.5 \mathrm{O}_{2} \)) \[ \Delta n = \text{Moles of Products} - \text{Moles of Reactants} = 1 - 1.5 = -0.5 \]

Step 3: Equate Delta n to x

Since \( K_{p} = K_{C} (RT)^{x} \) and \( \Delta n = -0.5 \) from our equation, we equate \( x = \Delta n \). Hence, \[ x = -0.5 \]

Key Concepts

Equilibrium Constants

Equilibrium constants are vital in chemistry to describe the balance point of reversible reactions. They help determine the concentration of reactants and products at equilibrium. There are two main types of equilibrium constants you should know about:

• \(K_C\): This is the equilibrium constant for a reaction in terms of concentrations. It applies to reactions that occur in the liquid or gas phase. For gaseous reactions, concentrations are measured in moles per liter.
• \(K_p\): This constant relates to partial pressures for reactions involving gases. Partial pressure is used because it's easier to measure gases in this way, using units of atmospheres.

To understand how these constants interrelate, consider the formula shown in our exercise: \[ K_{p} = K_{C} (RT)^{\Delta n} \]Here, \(R\) is the universal gas constant, \(T\) is temperature in Kelvin, and \(\Delta n\) is the change in moles of gas during the reaction. This equation helps convert between \(K_C\) and \(K_p\) values when dealing with gases.

Reaction Stoichiometry

Reaction stoichiometry is about understanding and balancing the quantitative relationships between reactants and products in a chemical reaction. It involves coefficients in a balanced chemical equation that tell you how much of each substance is involved in the reaction.For example, in our reaction:\[ \mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2}\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g}) \]The stoichiometry tells us that:

• One mole of \(\mathrm{SO}_2\) reacts with half a mole of \(\mathrm{O}_2\) to produce one mole of \(\mathrm{SO}_3\).

Using stoichiometry helps in determining \(\Delta n\), which is crucial for using the \(K_p = K_C (RT)^{\Delta n}\) formula. It includes calculating the moles of reactants and products and determining their difference:

• \(\Delta n = \text{Moles of Products} - \text{Moles of Reactants}\)
• \(\Delta n = 1 - 1.5 = -0.5\)

This value of \(\Delta n\) plays a role in altering the equilibrium constant from \(K_C\) to \(K_p\).

Ideal Gas Law

The Ideal Gas Law is a simplified equation used to understand gases' behavior under different conditions. It's an essential tool in thermodynamics and helps to connect various gas properties. You'll often see it expressed as:\[PV = nRT\]where:

• \(P\) is the pressure of the gas.
• \(V\) is the gas' volume.
• \(n\) is the number of moles of gas.
• \(R\) is the universal gas constant.
• \(T\) is the temperature in Kelvin.

In our exercise, the Ideal Gas Law helps justify the use of pressures (\(K_p\)) rather than concentrations (\(K_C\)) for the equilibrium constant involving gases. In systems where gases are involved, assuming ideal behavior allows this relationship.Remember, the assumption of an ideal gas simplifies calculations and allows us to use the equilibrium formula \(K_{p} = K_{C} (RT)^{\Delta n}\). Understanding this link is essential to solve problems involving reactions and changes in gas systems.