Problem 48 - A Complete Resource Book in Chemistry for JEE Main

Question

In which of the following case, the value of \(K_{p}\) is less than \(\mathrm{K}_{\mathrm{c}}\) ? (a) \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\) (b) \(\mathrm{H}_{2}+\mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{HCl}\) (c) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) (d) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\)

Step-by-Step Solution

Answer

Option (c): \(2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) is where \(K_p < K_c\).

Step 1: Review the relationship between Kp and Kc

The relationship between the equilibrium constants in terms of partial pressures \(K_p\) and concentrations \(K_c\) for any reaction at a given temperature is given by: \[ K_p = K_c (RT)^{\Delta n} \] where \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants).

Step 2: Analyze option (a)

For the reaction \( \mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} \), the \( \Delta n = (2 - (1 + 1)) = 0 \). Therefore, \[ K_p = K_c (RT)^0 = K_c \]. So, \(K_p = K_c\) in this case.

Step 3: Analyze option (b)

For the reaction \( \mathrm{H}_{2} + \mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{HCl} \), the \( \Delta n = (2 - (1 + 1)) = 0 \). Therefore, \[ K_p = K_c (RT)^0 = K_c \]. So, \(K_p = K_c\) in this case as well.

Step 4: Analyze option (c)

For the reaction \( 2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3} \), the \( \Delta n = (2 - (2 + 1)) = -1 \). Therefore, \[ K_p = K_c (RT)^{-1} \]. Since \((RT)^{-1} < 1\), it implies \(K_p < K_c\) in this case.

Step 5: Analyze option (d)

For the reaction \( \mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3} + \mathrm{Cl}_{2} \), the \( \Delta n = (1 + 1 - 1) = 1 \). Therefore, \[ K_p = K_c (RT)^{1} \]. Since \((RT)^{1} > 1\), it implies \(K_p > K_c\) in this case.

Step 6: Determine the correct option

From the analysis in steps 2 to 5, the reaction \(2 \mathrm{SO}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3} \) in option (c) is the only case where \(K_p < K_c\).

Key Concepts

Kp and Kc relationship

The equilibrium constants, represented as \( K_p \) and \( K_c \), are essential in understanding how chemical reactions behave under different conditions.
\( K_p \) is expressed in terms of partial pressures of gases, while \( K_c \) is based on molar concentrations. The relationship between them is crucial for solving many equilibrium problems: \[ K_p = K_c (RT)^{\Delta n} \]

• \(R\) is the universal gas constant, which is \(0.0821 \, \text{L atm/mol K}\).
• \(T\) represents the temperature in Kelvin.
• \(\Delta n\) denotes the change in the number of moles of gas.

The expression \((RT)^{\Delta n}\) accounts for changes in the reaction under different temperature and pressure conditions. Therefore, understanding \(\Delta n\) is essential in predicting the relation between \(K_p\) and \(K_c\). If \(\Delta n\) is zero, both equilibrium constants are equal, simplifying the evaluation.

Le Chatelier's Principle

This principle is a fundamental concept that predicts how a change in conditions can affect chemical equilibria. It asserts that if an external change is applied to a system at equilibrium, the system adjusts to minimize that change. Here's how it generally works:

• An increase in pressure by decreasing volume shifts equilibrium towards the side with fewer moles of gas.
• A decrease in pressure by increasing volume favors the side with more moles of gas.
• Adding a reactant or product shifts the equilibrium to the opposite side to counteract the change.
• Temperature changes depend on the reaction being exothermic or endothermic.

Le Chatelier’s Principle provides insights into how changes affect \(K_p\) and \(K_c\). Since these constants reflect concentrations or pressures at equilibrium, understanding how equilibrium shifts aids in predicting changes to these values.

Chemical Equilibrium

Chemical equilibrium occurs when a reaction and its reverse process happen at the same rate. In such a state, the macroscopic properties like concentration, pressure, and color remain constant over time.
Key features of equilibrium include:

• The reaction continues, but because the forward and reverse rates are equal, reactant and product concentrations remain stable, giving the impression of a "static" state.
• Equilibrium can be altered by changing conditions such as temperature, pressure, or concentration, leading the system to readjust according to Le Chatelier's Principle.

It is crucial to understand that equilibrium constants \(K_c\) and \(K_p\) do not change with concentration mixtures but are temperature-dependent. Recognizing this helps anticipate how a system will respond to environmental changes and predicts shifts in equilibrium.

Delta n in equilibrium reactions

\(\Delta n\) plays a significant role in determining the relationship between \(K_p\) and \(K_c\). It is defined as the difference in the number of moles of gaseous products and reactants in a balanced chemical equation. Here's the breakdown:

• If \(\Delta n = 0\), \(K_p\) equals \(K_c\) because \((RT)^0 = 1\).
• If \(\Delta n > 0\), \(K_p\) becomes greater than \(K_c\) because \((RT)^{\Delta n}\) will be greater than 1.
• If \(\Delta n < 0\), \(K_p\) will be less than \(K_c\) as \((RT)^{\Delta n}\) will be less than 1.

Predicting whether \(K_p\) is larger or smaller than \(K_c\) helps in understanding the behavior of gases in an equilibrium system, including compression, expansion, and variations influenced by temperature.