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

Question

One mole of \(\mathrm{N}_{2}(\mathrm{~g})\) is mixed with \(2 \mathrm{moles}\) 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 \(K_{c}\) 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 \(K_c\) is \(\frac{1}{16}\).

Step 1: Understand the Reaction and Changes

Initially, \[ \mathrm{N}_2(g): 1 \text{ mole}\quad \mathrm{H}_2(g): 2 \text{ moles}\quad \mathrm{NH}_3(g): 0 \text{ moles} \]The balanced reaction is:\[ \mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftharpoons 2\mathrm{NH}_3(g) \]If 50% of \(\mathrm{N}_2\) is converted, 0.5 mole of \(\mathrm{N}_2\) reacts to form ammonia.

Step 2: Calculate Changes in Moles

Based on the stoichiometry of the reaction, when 0.5 moles of \(\mathrm{N}_2\) react, 1.5 moles of \(\mathrm{H}_2\) are used producing 1 mole of \(\mathrm{NH}_3\). The final moles are:\[ \mathrm{N}_2(g): 0.5 \text{ moles}\quad \mathrm{H}_2(g): 0.5 \text{ moles}\quad \mathrm{NH}_3(g): 1 \text{ mole} \]

Step 3: Find Equilibrium Concentrations

The volume of the vessel is 4 liters. Thus, the concentrations at equilibrium are:\[ [\mathrm{NH}_3] = \frac{1}{4}\, M \quad [\mathrm{N}_2] = \frac{0.5}{4}\, M = \frac{1}{8}\, M \quad [\mathrm{H}_2] = \frac{0.5}{4}\, M = \frac{1}{8}\, M \]

Step 4: Write Expression for Kc of Given Reaction

The given equilibrium is:\[ \mathrm{NH}_3(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_2(g) + \frac{3}{2} \mathrm{H}_2(g) \]The expression for \(K_c\) is:\[ K_c = \frac{[\mathrm{N}_2]^{1/2}[\mathrm{H}_2]^{3/2}}{[\mathrm{NH}_3]} \]

Step 5: Plug in Equilibrium Concentrations

Substitute the concentrations:\[ K_c = \frac{\left(\frac{1}{8}\right)^{1/2}\left(\frac{1}{8}\right)^{3/2}}{\frac{1}{4}} \]This simplifies to:\[ K_c = \frac{\sqrt{\frac{1}{8}} \left(\frac{1}{8}\right)^{3/2} }{\frac{1}{4}} = \frac{\left(\frac{1}{8}\right)^2}{\frac{1}{4}} \]

Step 6: Simplify to Find Kc

Further simplifying:\[ K_c = \frac{\frac{1}{64}}{\frac{1}{4}} = \frac{1}{64} \times 4 = \frac{1}{16} \]

Key Concepts

Reaction Stoichiometry

Understanding reaction stoichiometry is like following a recipe in the chemical world. Here, we're dealing with the balanced chemical equation: \( \mathrm{N}_{2}(\mathrm{~g}) + 3\mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \). This equation tells us the exact proportion of reactants (nitrogen and hydrogen) needed to form the products (ammonia). Think of it as a guide that shows how many molecules or moles of one substance react with another to create something new. In our exercise, if 1 mole of nitrogen reacts according to the equation, it pairs with 3 moles of hydrogen to produce 2 moles of ammonia. Stoichiometry helps us calculate how much product can be formed or how much of a reactant is needed. It's the quantitative heart of reactions, revealing how 0.5 moles of nitrogen result in 1 mole of ammonia with 1.5 moles of hydrogen being consumed. The stoichiometry is crucial because it helps us predict and control the amounts of substances consumed or produced, ensuring the reaction occurs as expected.

Equilibrium Constant Calculation

The equilibrium constant, \(K_c\), serves as a snapshot of the balance point within a reaction. For our reaction, it is important to determine the concentrations of each species when the system is at equilibrium. This point is where the rate of the forward reaction equals the rate of the reverse reaction, and concentrations of reactants and products remain unchanged. The expression for \(K_c\) in our exercise is derived from the balanced reaction equation: \( \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \). Here, \(K_c\) is defined by the formula:

• \(K_c = \frac{[\mathrm{N}_2]^{1/2}[\mathrm{H}_2]^{3/2}}{[\mathrm{NH}_3]}\)

Substituting values from the equilibrium concentrations:

• \( [\mathrm{N}_2] = \frac{1}{8} \text{ M}\)
• \( [\mathrm{H}_2] = \frac{1}{8} \text{ M}\)
• \( [\mathrm{NH}_3] = \frac{1}{4} \text{ M}\)

We arrive at \(K_c = \frac{1}{16}\). This value demonstrates the position of equilibrium and helps in predicting how changes in conditions may shift this balance.

Mole Concept

The mole concept is a cornerstone of chemistry, allowing us to count atoms and molecules by weighing them. It's like using dozen as a measure for eggs. The word "mole" refers to Avogadro's number, \(6.022 \times 10^{23}\), showcasing how many atoms or molecules are present in a substance. In this exercise, the problem starts with 1 mole of nitrogen and 2 moles of hydrogen. The mole concept helps us to simplify understanding and working with chemical equations. It connects the mass of a substance to the number of particles it contains, using molar mass as a bridge. For instance, if 50% of \( \mathrm{N}_2 \) reacts, 0.5 moles participate in the reaction, translating to a precise amount of material conversion as per the equation. Recognizing moles in this context enables accurate and meaningful calculations in determining how much of each substance is present before and after the reaction.
In essence, moles determine initiative conversion into reality, grounding abstract numbers into tangible chemical reactions.