Problem 80 - A Complete Resource Book in Chemistry for JEE Main

Question

The ratio of \(\mathrm{Kp} / \mathrm{Kc}\) for the reaction \(\mathrm{SO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})\) is (a) \((\mathrm{RT})^{-1 / 2}\) (b) \((\mathrm{RT})^{1 / 2}\) (c) RT (d) 1

Step-by-Step Solution

Answer

The ratio \(\frac{\mathrm{Kp}}{\mathrm{Kc}}\) is \((\mathrm{RT})^{-1 / 2}\), so the correct option is (a).

Step 1: Identify the Chemical Reaction and Expressions for Kp and Kc

The balanced chemical equation given is \(\mathrm{SO}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})\). Here, \(\mathrm{Kp}\) is the equilibrium constant in terms of partial pressures, and \(\mathrm{Kc}\) is the equilibrium constant in terms of concentration.

Step 2: Apply the Relation between Kp and Kc

The relationship between \(\mathrm{Kp}\) and \(\mathrm{Kc}\) is given by: \(\mathrm{Kp} = \mathrm{Kc} \cdot (\mathrm{RT})^{\Delta n}\). Here, \(\Delta n\) is the change in moles of gas between products and reactants.

Step 3: Determine Δn for the Reaction

In the given reaction, \(\Delta n\) is calculated by subtracting the sum of moles of gaseous reactants from the moles of gaseous products. It is \(1 - (1 + 0.5) = 1 - 1.5 = -0.5\).

Step 4: Calculate the Ratio Kp/Kc

Substitute \(\Delta n = -0.5\) into the relation: \(\mathrm{Kp} = \mathrm{Kc} \cdot (\mathrm{RT})^{-0.5}\). Rearranging, we find \(\frac{\mathrm{Kp}}{\mathrm{Kc}} = (\mathrm{RT})^{-0.5}\).

Step 5: Identify the Correct Option

Compare the result \((\mathrm{RT})^{-0.5}\) to the given options. The correct option that matches \((\mathrm{RT})^{-0.5}\) is (a) \((\mathrm{RT})^{-1 / 2}\).

Key Concepts

Kp and Kc relationship

In chemical equilibrium, understanding the relationship between the equilibrium constants in terms of pressure and concentration is crucial. These are denoted as \( K_p \) and \( K_c \), respectively. To relate the two constants, scientists use the following relation: \[ K_p = K_c \cdot (RT)^{\Delta n} \] Here, \( R \) represents the gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gases in the reaction. This equation allows the conversion between concentration-based (\( K_c \)) and pressure-based (\( K_p \)) equilibrium constants, considering how changes in pressure and volume relate to temperature and the number of gas particles.

Equilibrium constant

The equilibrium constant is a value that describes the ratio of the concentrations or partial pressures of products to reactants at equilibrium. It tells us how far a reaction will proceed.

• For concentration-based reactions, this is referred to as \( K_c \). It is derived from the concentrations of the reactants and products.
• For pressure-based gas reactions, we use \( K_p \), which uses partial pressures instead of concentrations.

At equilibrium, these constants indicate whether the balance of the reaction lies more towards the products or reactants. Large values of \( K \) suggest that the reaction greatly favors the products, while small values indicate a preference towards reactants. It is essential to note that \( K \) varies with temperature, ensuring that each condition at equilibrium is unique.

Change in moles (Δn)

The change in moles, represented by \( \Delta n \), is crucial for understanding how a chemical equilibrium responds to changes in conditions, like pressure and temperature. Specifically, \( \Delta n \) refers to the difference in the number of moles of gaseous products and gaseous reactants: \[ \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} \] Knowing \( \Delta n \) helps in determining the effect of pressure changes on the equilibrium position. For reactions where \( \Delta n = 0 \), pressure changes have no effect, since the number of moles of gases don’t change. If \( \Delta n > 0 \), the forward reaction will be favored by an increase in volume (or decrease in pressure), and if \( \Delta n < 0 \), reducing the volume (or increasing the pressure) will favor the forward reaction. Thus, \( \Delta n \) is fundamental in predicting how changes in conditions can shift the position of equilibrium.