Problem 77 - A Complete Resource Book in Chemistry for JEE Main

Question

In which of the following gaseous reaction, the value of \(K_{p}\) is less than \(K_{c}\) ? (a) \(\mathrm{PCl}_{3} \rightleftharpoons \mathrm{PCI}_{3}+\mathrm{Cl}_{2}\) (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) (c) \(2 \mathrm{HI} \rightleftharpoons{ } \mathrm{H}_{2}+\mathrm{I}_{2}\) (d) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\)

Step-by-Step Solution

Answer

Options (b) and (d) have \( K_p < K_c \).

Step 1: Understand the Relationship between Kp and Kc

The relationship between the equilibrium constants of gases, with respect to pressure \(K_p\) and concentration \(K_c\), is given by the equation: \[K_p = K_c(RT)^{\Delta n}\] where \(R\) is the gas constant in units that match the units of pressure and concentration, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas (moles of products - moles of reactants). To find when \(K_p < K_c\), \(\Delta n\) must be negative, making \((RT)^{\Delta n} < 1\).

Step 2: Analyze Option (a) PCl3 ⇌ PCl3 + Cl2

For option (a), the reaction is \(\mathrm{PCl}_3 \rightleftharpoons \mathrm{PCl}_3+\mathrm{Cl}_2\). The number of moles of gaseous products is 2 (1 mole of PCl3 and 1 mole of Cl2) and the number of moles of reactants is 1, hence \(\Delta n = 2 - 1 = 1\). This makes \((RT)^{\Delta n} > 1\), therefore \(K_p > K_c\).

Step 3: Analyze Option (b) 2 SO2 + O2 ⇌ 2 SO3

For option (b), the reaction is \(2 \mathrm{SO}_2 + \mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3\). The total moles of products are 2, and moles of reactants are 3, hence \(\Delta n = 2 - 3 = -1\). This results in \((RT)^{-1} < 1\), so \(K_p < K_c\).

Step 4: Analyze Option (c) 2 HI ⇌ H2 + I2

For option (c), the reaction is \(2 \mathrm{HI} \rightleftharpoons \mathrm{H}_2+\mathrm{I}_2\). Both sides have a total of 2 moles (1 mole of H2 and 1 mole of I2 as products, 2 moles of HI as reactants), thus \(\Delta n = 2 - 2 = 0\). Therefore, \((RT)^{\Delta n} = 1\), so \(K_p = K_c\).

Step 5: Analyze Option (d) N2 + 3 H2 ⇌ 2 NH3

For option (d), the reaction is \(\mathrm{N}_2 + 3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3\). The moles of products are 2, and the moles of reactants are 4, giving \(\Delta n = 2 - 4 = -2\). This means \((RT)^{-2} < 1\), which implies \(K_p < K_c\).

Step 6: Conclude from the Analysis

From the analysis, option (b) and (d) satisfy the condition where \(\Delta n\) is negative, thus making \(K_p < K_c\).

Key Concepts

Equilibrium Constant

The equilibrium constant is a fundamental concept in chemistry that helps us understand the extent to which a chemical reaction proceeds under a set of conditions. It is temperature-dependent and can be expressed in terms of concentrations or pressures. For gaseous reactions, we often talk about two forms of equilibrium constants:

• \(K_c\) - The equilibrium constant in terms of the concentration of the substances involved in the reaction.
• \(K_p\) - The equilibrium constant in terms of the partial pressures of the gases involved in the reaction.

The essence of the equilibrium constant is to describe the ratio of the concentrations or partial pressures of products to reactants, raised to the power of their coefficients in the balanced chemical equation, at equilibrium. This ratio gives us a number that provides valuable insight into the position of equilibrium of a given reaction.

Gaseous Reactions

Gaseous reactions are reactions where the reactants and products are in the gaseous state. These reactions are often discussed in terms of their equilibrium position and shifts under different conditions. They are characterized by their ability to be affected by changes in pressure and temperature. Some properties of gaseous reactions include:

• They often exhibit significant volume changes because gases expand to fill their containers.
• The equilibrium may shift with pressure changes due to Le Chatelier's Principle, which states that a system at equilibrium will respond to a change in a way that tends to counteract that change.
• Temperature changes can significantly alter the position of equilibrium, and thus the equilibrium constant for gaseous reactions, because they only occur under specific temperature conditions.

For example, in the reaction between \(2 \text{SO}_2 + \text{O}_2 \rightleftharpoons 2 \text{SO}_3\), changing the total pressure or temperature could affect the amounts of reactants and products present at equilibrium.

Kp and Kc Relationship

The relationship between \(K_p\) and \(K_c\) for a gaseous reaction is crucial in understanding how pressure and concentration are interconnected in chemical equilibrium. The formula linking them is:\[K_p = K_c(RT)^{\Delta n}\]where:

• \(R\) is the ideal gas constant.
• \(T\) is the temperature in Kelvin.
• \(\Delta n\) is the change in moles of gas, calculated as the difference between moles of gaseous products and reactants.

This relationship highlights that the two constants are equal (\(K_p = K_c\)) if \(\Delta n = 0\), meaning there is no net change in moles of gas. If \(\Delta n\) is negative, as seen in the reaction \(2 \text{SO}_2 + \text{O}_2 \rightleftharpoons 2 \text{SO}_3\), \(K_p\) will be less than \(K_c\) because \((RT)^{\Delta n} \) would be smaller than one, contributing to the overall reduction in \(K_p\) value. Conversely, if \(\Delta n\) is positive, \(K_p\) would be greater. This relationship provides a deeper understanding of how gases' behavior in chemical systems can fluctuate based on conditions such as pressure and concentration changes.