Problem 1 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the reaction \(\mathrm{Fe}_{2} \mathrm{~N}(\mathrm{~s})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Fe}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{~g})\) (a) \(K_{\mathrm{c}}=K_{\mathrm{p}}(\mathrm{RT})\) (b) \(K_{\mathrm{c}}=K_{\mathrm{p}}(\mathrm{RT})^{\frac{-1}{2}}\) (c) \(K_{\mathrm{c}}=K_{\mathrm{p}}(\mathrm{RT})^{\frac{1}{2}}\) (d) \(K_{\mathrm{c}}=K_{\mathrm{p}}(\mathrm{RT})^{\frac{3}{2}}\)

Step-by-Step Solution

Answer

The correct answer is (b) \(K_c = K_p (RT)^{-\frac{1}{2}}\).

Step 1: Determine the Reaction's Delta n (Δn)

In chemical equilibrium, we need to calculate Δn, where Δn is the change in the number of moles of gas. For the given reaction,\[ \text{Fe}_2 \text{N} (s) + \frac{3}{2} \text{H}_2 (g) \rightleftharpoons 2\text{Fe} (s) + \text{NH}_3 (g) \]Δn is calculated by subtracting the moles of gaseous reactants from the moles of gaseous products. Thus,\[ Δn = 1 - \frac{3}{2} = -\frac{1}{2} \].

Step 2: Use the Kc and Kp Relationship Formula

The relationship between Kc and Kp is given by the formula:\[ K_c = K_p (RT)^{Δn} \]This formula allows us to relate the equilibrium constant in terms of concentration (Kc) and the equilibrium constant in terms of pressure (Kp), using the calculated Δn.

Step 3: Calculate the Expression for Kc

Substitute the value of Δn into the formula:\[ K_c = K_p (RT)^{-\frac{1}{2}} \]Which implies that the correct choice matches option (b) from the given options.

Key Concepts

Reaction Equilibrium

In chemistry, reaction equilibrium refers to the state in which the concentrations of reactants and products remain constant over time. This occurs because the rate of the forward reaction equals the rate of the reverse reaction. When a reaction reaches this balance, it is said to be at equilibrium. It's important to understand that while the overall concentrations remain constant, both the forward and reverse reactions are still occurring.
This dynamic balance allows us to express the equilibrium state with an equilibrium constant. The equilibrium constant ( K ) is a numerical value that relates the concentrations of reactants and products at equilibrium. For gases, it can be expressed in terms of concentration ( K_c ) or in terms of pressure ( K_p ). Both constants provide insights into the equilibrium position of a reaction and tell us whether the reactants or the products are favored at equilibrium.
Knowing whether a reaction has reached equilibrium can help predict the behavior of the reaction given changes in conditions, ensuring the optimal outcomes whether in industrial or laboratory processes.

Kc and Kp Relationship

The relationship between the equilibrium constant in terms of concentration (K_c) and the equilibrium constant in terms of pressure (K_p) is vital in understanding how gaseous systems behave. These constants are related through the formula:\[ K_c = K_p (RT)^{Δn} \]where R is the universal gas constant, T is the temperature in Kelvin, and Δn is the change in the number of moles of gas in the reaction. This equation is particularly useful when the reaction involves gases, as their concentrations are frequently more convenient to measure in terms of pressure rather than molarity.
Knowing this relationship allows chemists to convert between K_c and K_p based on the conditions and the specific requirements of their experiment or industrial application. It also highlights the influence of temperature and the difference in gaseous moles on the equilibrium position. Understanding these conversions provides a comprehensive insight into equilibrium systems and aids in predicting reaction behavior under varying conditions.

Change in Moles (Δn)

Change in moles, denoted by Δn, plays a crucial role in the relationship between K_c and K_p. It represents the difference in the number of moles of gaseous products and reactants. Specifically, Δn is defined as:- Moles of gaseous products - Moles of gaseous reactants
For example, in the reaction \(\mathrm{Fe}_{2} \mathrm{~N}(\mathrm{~s})+\frac{3}{2}\mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2\mathrm{Fe}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{~g})\), the only gases are \(\mathrm{H}_{2}\) and \(\mathrm{NH}_{3}\). Thus, Δn will be\[ Δn = 1 - \frac{3}{2} = -\frac{1}{2} \]This calculation is integral to determining how changes in pressure and temperature will affect the equilibrium constant that expresses the state of a gaseous reaction. Correctly identifying Δn is essential to applying the K_c and K_p relationship formula accurately, enabling you to predict how the equilibrium will shift with changes in reaction conditions.