Problem 125 - A Complete Resource Book in Chemistry for JEE Main

Question

A \(1 \mathrm{M}\) solution of glucose reaches dissociation equilibrium given below. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \rightleftharpoons 6 \mathrm{HCHO} $$ If the equilibrium constant is \(0.167 \times 10^{-22}\), the concentration of HCHO in the equilibrium is (a) \(1.60 \times 10^{-8} \mathrm{M}\) (b) \(3.20 \times 10^{-6} \mathrm{M}\) (c) \(3.20 \times 10^{-4} \mathrm{M}\) (d) \(1.60 \times 10^{-4} \mathrm{M}\)

Step-by-Step Solution

Answer

(d) \(1.60 \times 10^{-4} \text{ M}\).

Step 1: Write the Equilibrium Expression

The reaction given is \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \rightleftharpoons 6\mathrm{HCHO}\). The equilibrium constant expression for this reaction is: \[K = \frac{[\text{HCHO}]^6}{[\text{C}_6\text{H}_{12}\text{O}_6]}\] Given, \( K = 0.167 \times 10^{-22} \). Initially, the concentration of \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\) is 1 M.

Step 2: Define Change and Equilibrium Concentrations

Assume some concentration \( x \) of \( \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \) dissociates to form \( 6x \) concentration of \( \mathrm{HCHO} \) at equilibrium. Hence, the equilibrium concentration of \( \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \) is \( (1 - x) \) and of \( \mathrm{HCHO} \) is \( 6x \).

Step 3: Substitute into Equilibrium Expression

Substitute these expressions into the equilibrium constant expression: \[0.167 \times 10^{-22} = \frac{(6x)^6}{1 - x}\] Due to the very small equilibrium constant value, assume \( x \) is very small, so \( 1 - x \approx 1 \).

Step 4: Simplify and Solve for x

The expression simplifies to: \[0.167 \times 10^{-22} = (6x)^6\] Solve for \( x \):\[(6x)^6 = 0.167 \times 10^{-22}\] \[6x = (0.167 \times 10^{-22})^{1/6}\]

Step 5: Calculate the Concentration of HCHO

Calculate \( 6x \): \[6x = (0.167 \times 10^{-22})^{1/6}\] This simplifies to: \[6x \approx 1.60 \times 10^{-4} \] Therefore, the concentration of \( \mathrm{HCHO} \) at equilibrium is \( 1.60 \times 10^{-4} \text{ M}\).

Key Concepts

Equilibrium Constant

In the study of chemical reactions, the equilibrium constant, often represented as \( K \), is a critical value that expresses the ratio of the concentrations of products to reactants at equilibrium. This ratio is specific for a given reaction at a constant temperature.

For the reaction

• \( ext{C}_{6} ext{H}_{12} ext{O}_{6} \rightleftharpoons 6 ext{HCHO} \)

the equilibrium constant expression can be written as follows:

• \( K = \frac{[ ext{HCHO}]^6}{[ ext{C}_6 ext{H}_{12} ext{O}_6]} \)

The reaction quotient \( Q \) mirrors the form of the equilibrium expression, but \( K \) is specific to when the reaction is at equilibrium.

In our example, the value of \( K \) is provided as \( 0.167 \times 10^{-22} \), indicating that the reaction favors the reactants, suggesting minimal dissociation of glucose.

Dissociation Equilibrium

Dissociation equilibrium occurs when a compound breaks down into its constituent parts and these parts exist in a dynamic balance. In our reaction, glucose (\( ext{C}_{6} ext{H}_{12} ext{O}_{6} \)) dissociates into formaldehyde (\( ext{HCHO} \)). At equilibrium, the rate of glucose dissociating into formaldehyde is equal to the rate of formaldehyde combining to form glucose.

When glucose dissociates, a part of it changes into many molecules of formaldehyde. In mathematical terms:

• If \( x \) is the amount of glucose that dissociates, then \( 6x \) is the resulting concentration of \( ext{HCHO} \).

This dissociation forms a new equilibrium where the concentration of glucose is \( (1 - x) \) and of \( ext{HCHO} \) is \( 6x \). Understanding this dynamic helps in calculating the concentrations of each at equilibrium.

Concentration Calculations

The challenge with equilibrium calculations is determining the concentration of each species at equilibrium, particularly when the dissociation is minimal, as shown by the small \( K \) value. Here’s a brief rundown of how it is done:

• Start with initial concentrations; for glucose, it is given as \( 1 ext{ M} \).
• Assuming \( x \) is the concentration of glucose that dissociates, the concentration of glucose at equilibrium becomes \( (1 - x) \).

The equilibrium expression:

• \( 0.167 \times 10^{-22} = \frac{(6x)^6}{1 - x} \)

Assume that \( 1 - x \approx 1 \) due to the minimal dissociation expected with such a small \( K \), it simplifies calculations:

• \( (6x)^6 \approx 0.167 \times 10^{-22} \)
• Solve for \( x \) to find \( 6x = 1.60 \times 10^{-4} \, ext{M} \), which is the concentration of \( ext{HCHO} \) at equilibrium.

Solution Chemistry

Solution chemistry involves the study of reactions in solutions where solutes mix to form a homogeneous mixture with a solvent. An important aspect in understanding these reactions is discerning how solutes behave in solutions and reach equilibrium.

In the context of our exercise:

• Initially, glucose is completely dissolved into the solution.
• Reaching equilibrium, the solution contains both dissolved glucose and the resulting formaldehyde pigments, in specified proportions as calculated.

With equilibrium reached, the low \( K \) value implies a strong tendency for glucose to remain in undissociated form in solution, versus forming formaldehyde. This insight can be used to predict the behavior of similar substances under analogous conditions in solution chemistry.