Problem 97 - A Complete Resource Book in Chemistry for JEE Main

Question

If the equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at 750 is 49 , then the equilibrium constant for the reaction, \(\mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons\) \(1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})\) at the same temperature will be (a) \(1 / 49\) (b) 49 (c) \(1 / 7\) (d) \(49^{2}\)

Step-by-Step Solution

Answer

The equilibrium constant is \( \frac{1}{7} \). The correct option is (c).

Step 1: Identify the Original Reaction

The given original reaction is \( \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \). The equilibrium constant for this reaction is given as \( K = 49 \).

Step 2: Write and Understand the Reverse Reaction

The reverse reaction we need the equilibrium constant for is \( 2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \). For the given problem, a new reaction is \( \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g}) + \frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \).

Step 3: Determine the Equilibrium Constant for the Reverse Reaction

For the reaction \( 2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \), the equilibrium constant \( K' \) is the inverse of the original reaction's equilibrium constant because it is the reverse reaction. So, \( K' = \frac{1}{49} \).

Step 4: Adjust the Equilibrium Constant for the Given Reaction

The new reaction is \( \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g}) + \frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \), which is half of the reverse reaction \( 2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \). For a reaction scaled by a factor, the equilibrium constant is the original constant raised to the power of the inverse of that factor. Thus, \( K_{new} = (K')^{1/2} = \left( \frac{1}{49} \right)^{1/2} = \frac{1}{7} \).

Step 5: Choose the Correct Option

Based on the calculations, the equilibrium constant for the given reaction at the same temperature is \( \frac{1}{7} \). Therefore, the correct answer is (c) \( \frac{1}{7} \).

Key Concepts

Reversible Reactions

In chemistry, a reversible reaction is one that can proceed in both the forward and reverse directions. This means the products can recombine to form the reactants again, unlike irreversible reactions that go to completion in one direction only. An example discussed here is the reaction of nitrogen and hydrogen gas to form ammonia: \\[ \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \] \
At any point during a reversible reaction, both the forward and reverse reactions occur simultaneously. However, the rates of these can differ. Over time, the system may reach a state where the forward and reverse reaction rates are equal, leading us to chemical equilibrium. This dynamic nature is crucial to understanding why the concentrations of reactants and products remain constant even though reactions continue to take place.

• Reversible reactions are marked with the symbol "\(\rightleftharpoons\)" to indicate their two-way nature.
• These reactions do not run out of reactants or products as they can continuously convert back and forth.

Chemical Equilibrium

Chemical equilibrium occurs when a reversible chemical reaction's forward and backward processes happen at equal rates. This balance means that the concentrations of reactants and products stay constant over time. The equilibrium state is dynamic; although molecules continuously react with each other, there's no overall concentration change.
In our example with ammonia, \\[ \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g}) + \frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \] \
equilibrium is reached when the rate of formation of ammonia equals the rate at which it decomposes back to nitrogen and hydrogen.

• At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.
• The system is in a state of dynamic balance, not static.
• The equilibrium position can be shifted by changing conditions like temperature, pressure, or concentration.

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a measure that helps determine the direction in which a reaction will proceed to reach equilibrium. It's calculated in the same way as the equilibrium constant \( K \), but using concentrations or pressures at any point in time, not just equilibrium. For the reaction \\[ \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g}) + \frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \] \\( Q \) is determined by the current concentrations of the products and reactants:
\[ Q = \frac{[\mathrm{N}_2]^{1/2} \, [\mathrm{H}_2]^{3/2}}{[\mathrm{NH}_3]} \]
Comparing \( Q \) with \( K \), the equilibrium constant, helps predict whether a reaction needs to proceed in the forward or reverse direction to achieve equilibrium:

• If \( Q < K \), the reaction will proceed forward to form more products.
• If \( Q > K \), the reaction will proceed in reverse to form more reactants.
• If \( Q = K \), the system is already at equilibrium.

Understanding \( Q \) is crucial in adjusting the conditions to drive reactions toward desired outcomes.