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

Question

If the equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g})\) at 750 is 49 , then the equilibrium constant for the reaction, \(\mathrm{NH}_{3}(\mathrm{~g})=\) \(1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})\) 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 for the reaction is \( \frac{1}{7} \), so the correct answer is (c).

Step 1: Understand the given reactions

The original reaction is \( \mathrm{N}_2(\mathrm{~g}) + 3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g}) \). The equilibrium constant for this reaction, \( K_c \), is given as 49.

Step 2: Identify the reverse reaction

The second reaction given is the reverse of the original but with different stoichiometry: \( \mathrm{NH}_3(\mathrm{~g}) \rightarrow \frac{1}{2} \mathrm{N}_2(\mathrm{~g}) + \frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \). This means we need the equilibrium constant for the reversed and modified stoichiometric reaction.

Step 3: Calculate the equilibrium constant for the reverse reaction

The relationship between the equilibrium constants of a reaction and its reverse is \( K_{c, \text{reverse}} = \frac{1}{K_c} \). Hence, the equilibrium constant for the reverse reaction without changes in stoichiometry is \( \frac{1}{49} \).

Step 4: Adjust for stoichiometry changes

The reverse reaction alters the stoichiometry: the reverse reaction's stoichiometry is halved. When the coefficients in a balanced chemical equation are multiplied by a factor, the equilibrium constant \( K' \) is adjusted by raising \( K_c \) to the power of this multiplication factor. Here, the coefficients are divided by 2, so:\[ K' = \left( \frac{1}{49} \right)^{1/2} = \frac{1}{\sqrt{49}} = \frac{1}{7} \].

Step 5: Choose the correct answer

According to the calculations, the equilibrium constant for the given reaction \( \mathrm{NH}_3(\mathrm{~g}) \rightarrow \frac{1}{2} \mathrm{N}_2(\mathrm{~g}) + \frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \) is \( \frac{1}{7} \). This matches option (c).

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a crucial concept in chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction under constant temperature.

• For a general reaction: \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression is \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \).
• Here, \([A]\), \([B]\), \([C]\), and \([D]\) represent the molar concentrations of the substances, and \(a\), \(b\), \(c\), and \(d\) are their stoichiometric coefficients in the balanced equation.

This constant remains unchanged unless the temperature changes.
It's important to understand that the value of the equilibrium constant tells us the extent to which a reaction proceeds; a large \( K_c \) indicates products are favored at equilibrium, whereas a small \( K_c \) suggests that reactants are predominant.
In the given exercise, the equilibrium constant for the reaction \( \mathrm{N}_2(\mathrm{~g}) + 3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g}) \) is 49, meaning the formation of ammonia is significantly favored over the reactants at this temperature.

Reversible Reactions

A reversible reaction is one where the reactants can form products, which can simultaneously decompose back into the reactants.
This means both forward and backward reactions occur at the same time. At equilibrium, the rates of the forward and backward reactions are equal, leading to a constant concentration of reactants and products.

• For example, in the reaction \(\mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\), the forward reaction forms ammonia, while the reverse reaction breaks it into nitrogen and hydrogen gases.
• The equilibrium constant for the reverse reaction is the inverse of that for the forward reaction, which is \( K_{c, \text{reverse}} = \frac{1}{K_c} \).

Reversible reactions are dynamic, with chemical equilibrium being a state of balance as opposed to complete conversion of reactants to products. This understanding is crucial for predicting how changes in conditions, such as pressure or temperature, can shift the equilibrium position.

Stoichiometry Changes

Stoichiometry changes refer to adjustments in the coefficients of a balanced chemical equation. These changes directly affect the calculation of the equilibrium constant.

• When the stoichiometric coefficients of a reaction are altered, the equilibrium constant must be adjusted accordingly. If the coefficients are multiplied by a factor \( n \), then the new equilibrium constant \( K' \) is given by \( K' = (K_c)^{1/n} \).
• In the given exercise, the reverse reaction \( \mathrm{NH}_3(\mathrm{~g}) \rightarrow \frac{1}{2} \mathrm{N}_2(\mathrm{~g}) + \frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \) has stoichiometric coefficients halved compared to its forward reaction counterpart.
• Thus, the calculated equilibrium constant \( K' \) becomes \( \left( \frac{1}{49} \right)^{1/2} \), simplifying to \( \frac{1}{7} \).

These adjustments are vital for accurately predicting the behavior of chemical systems under different stoichiometric constraints. Understanding stoichiometry changes helps in mastering chemical equilibrium concepts and solving related problems effectively.