Problem 153 - A Complete Resource Book in Chemistry for JEE Main

Question

One mole of \(\mathrm{N}_{2}(\mathrm{~g})\) is mixed with 2 moles of \(\mathrm{H}_{2}(\mathrm{~g})\) in a 4 litre vessel. If \(50 \%\) of \(\mathrm{N}_{2}(\mathrm{~g})\) is converted to \(\mathrm{NH}_{3}(\mathrm{~g})\) by the following reaction: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) What will be the value of \(\mathrm{K}\) for the following equilibrium? $$ \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) $$ (a) 256 (b) 16 (c) \(\frac{1}{16}\) (d) None of these

Step-by-Step Solution

Answer

The value of \( K \) is \( \frac{1}{16} \).

Step 1: Determine initial moles and conversion

Initially, there is 1 mole of \( \mathrm{N}_2 \) and 2 moles of \( \mathrm{H}_2 \). When \( 50\% \) of \( \mathrm{N}_2 \) is converted, 0.5 moles of \( \mathrm{N}_2 \) react, and 0.5 moles of \( \mathrm{NH}_3 \) are produced. Therefore, the remaining amounts are 0.5 moles of \( \mathrm{N}_2 \), 0.5 moles of \( \mathrm{H}_2 \), and 1 mole of \( \mathrm{NH}_3 \).

Step 2: Calculate equilibrium concentrations in molarity

Determine the molarity by dividing moles by volume (4 liters). For \( \mathrm{N}_2 \), the molarity is \( \frac{0.5}{4} = 0.125 \) M, for \( \mathrm{H}_2 \), the molarity is \( \frac{0.5}{4} = 0.125 \) M, and for \( \mathrm{NH}_3 \), the molarity is \( \frac{1}{4} = 0.25 \) M.

Step 3: Set up K expression for the reverse reaction

The equilibrium expression for the given reaction \( \mathrm{NH}_3(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(g) + \frac{3}{2} \mathrm{H}_2(g) \) is \[ K = \frac{[\mathrm{N}_2]^{1/2} \cdot [\mathrm{H}_2]^{3/2}}{[\mathrm{NH}_3]} \]. Substituting the calculated molarities gives \[ K = \frac{(0.125)^{1/2} \cdot (0.125)^{3/2}}{0.25} \].

Step 4: Calculate K

First calculate \( (0.125)^{1/2} = 0.354 \), \( (0.125)^{3/2} = 0.0155 \), then the product is \( 0.354 \times 0.0155 = 0.00549 \). Then divide by \( 0.25 \) to get \( K = \frac{0.00549}{0.25} = 0.02196 \). Simplifying gives \( K \approx \frac{1}{16} \).

Key Concepts

Reaction Kinetics

Understanding reaction kinetics is crucial to grasp how fast a chemical reaction occurs. It examines the rate at which reactants transform into products. In the given exercise, we observe the conversion of nitrogen (\(\mathrm{N}_2\)) and hydrogen (\(\mathrm{H}_2\)) into ammonia (\(\mathrm{NH}_3\)). This process is governed by reaction kinetics.Reaction kinetics explore various factors:

• Concentration of reactants: Higher concentrations generally increase the reaction rate due to more frequent collisions.
• Temperature: Increasing temperature usually speeds up reactions by providing more energy to reactant molecules.
• Catalysts: Substances that provide an alternative pathway with a lower activation energy for the reaction.

In the context of the exercise:- You see that half of the available \(\mathrm{N}_2\) is converted, indicating the reaction proceeded to a certain extent before reaching equilibrium.This is an excellent illustration of how reaction kinetics play an essential role in chemical reaction dynamics.

Molarity Calculation

Molarity is a way of expressing the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution.To understand molarity in the context of our exercise:1. Moles of reactants and products are first identified.2. Moles are then divided by the volume of the mixture (in liters) to find molarity.In the exercise, after 50% of \(\mathrm{N}_2\) is converted, you are left with:

• 0.5 moles of \(\mathrm{N}_2\)
• 0.5 moles of \(\mathrm{H}_2\)
• 1 mole of \(\mathrm{NH}_3\)

To determine the molarity of each:- For \(\mathrm{N}_2\) and \(\mathrm{H}_2\), divide 0.5 moles by 4 liters:\[\text{Molarity} = \frac{0.5}{4} = 0.125 \text{ M}\]- For \(\mathrm{NH}_3\), divide 1 mole by 4 liters:\[\text{Molarity} = \frac{1}{4} = 0.25 \text{ M}\]This simple division helps determine how concentration changes impact reaction kinetics and equilibrium.

Le Chatelier's Principle

Le Chatelier's principle explains how equilibrium reacts to changes in concentration, temperature, or pressure. It posits that if a system at equilibrium is disturbed, the system will adjust to counteract the disturbance and re-establish equilibrium.In the context of the reaction:- If you increase the concentration of \(\mathrm{NH}_3\), the system will shift to consume more \(\mathrm{NH}_3\) and produce more \(\mathrm{N}_2\) and \(\mathrm{H}_2\).- Conversely, if you remove \(\mathrm{NH}_3\), the system will shift to produce more of it.This principle is key to predicting how changes influence the position of equilibrium, making it essential for controlling reactions like the formation of ammonia. Understanding Le Chatelier's principle allows chemists to manipulate conditions to increase yields, plany a central role in industrial synthesis processes.