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

Question

A. 1 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 \(\mathrm{HCHO}\) 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

The concentration of \( \mathrm{HCHO} \) at equilibrium is \( 1.60 \times 10^{-4} \text{ M} \) (option d).

Step 1: Understanding the Reaction and Given Data

We have to find the concentration of \(\mathrm{HCHO}\) at equilibrium. The given reaction is: \[ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \rightleftharpoons 6 \mathrm{HCHO} \] with an equilibrium constant \( K = 0.167 \times 10^{-22} \). Initially, the concentration of glucose \( [\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}] = 1 \text{ M} \).

Step 2: Setting Up the Equilibrium Expression

The equilibrium constant expression for the reaction is: \[ K = \frac{[\mathrm{HCHO}]^6}{[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}]} \]. At equilibrium, let \( x \) be the concentration of \( \mathrm{HCHO} \) formed, then \([\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}] = 1 - x\).

Step 3: Simplifying the Equilibrium Expression

Assuming \( x \) is much smaller than 1 (since \( K \) is very small), \([\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}] \approx 1 \text{ M}\). Then, the equilibrium expression simplifies to \[ K = [\mathrm{HCHO}]^6 \]. Therefore, \[ x^6 = 0.167 \times 10^{-22} \].

Step 4: Solving for x

Find \( x \) by taking the sixth root of both sides: \[ x = \sqrt[6]{0.167 \times 10^{-22}} \]. Calculate this value to find \( x \).

Step 5: Calculating the Concentration of HCHO

Upon calculating, \( x \approx 1.60 \times 10^{-4} \text{ M} \). This is the concentration of \( \mathrm{HCHO} \) at equilibrium.

Key Concepts

Equilibrium Constant

The equilibrium constant, commonly denoted as \( K \), is a crucial part of understanding chemical reactions at equilibrium. It gives us an idea of how far a reaction proceeds before reaching a state where the rates of the forward and reverse reactions are equal. In other words, it marks the balance point of a reversible reaction.

For the dissociation of glucose reaction, the expression for the equilibrium constant is based on the concentrations of the products and reactants at equilibrium. The formula looks like this: \[K = \frac{[\mathrm{HCHO}]^6}{[\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6}]}\] Here, the concentration of \( \mathrm{HCHO} \) is raised to the power of 6 because 6 molecules of \( \mathrm{HCHO} \) are produced for every molecule of glucose that dissociates.

• If \( K \) is large, the reaction tends to favor the formation of products.
• If \( K \) is small, which is the case in our glucose dissociation, the reaction favors the reactants.

In this scenario, the equilibrium constant is very small \( (0.167 \times 10^{-22}) \), suggesting that glucose strongly favors staying in its original form over forming \( \mathrm{HCHO} \).

Concentration Calculation

Calculating the concentration of substances at equilibrium starts with understanding the change in concentration from initial to equilibrium stages. In this case, we began with a \( 1 \text{ M} \) solution of glucose. At equilibrium, some has dissociated into \( \mathrm{HCHO} \), which we'll denote with \( x \) representing its concentration.

Given the small equilibrium constant, \( x \) is expected to be negligible compared to the initial glucose concentration. Thus, \([\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6}] \approx 1 \text{ M}\). This approximation simplifies the calculations dramatically, allowing us to write:\[K = [\mathrm{HCHO}]^6 \]By substituting for \( K \), we get:\[0.167 \times 10^{-22} = x^6 \]Solving this by taking the sixth root helps us find the equilibrium concentration of \( \mathrm{HCHO} \). The solving yields:\[x \approx 1.60 \times 10^{-4} \text{ M}\]This value reflects the very low concentration of \( \mathrm{HCHO} \), consistent with the small equilibrium constant.

Dissociation of Glucose

Understanding the dissociation of glucose involves recognizing how glucose molecules break down into simpler substances, like \( \mathrm{HCHO} \), when at equilibrium. This type of reaction is reversible, indicated by the double arrows in the chemical equation:

\[ \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6} \rightleftharpoons 6 \mathrm{HCHO} \]

The reaction suggests that one molecule of glucose can produce up to six molecules of \( \mathrm{HCHO} \), depending on conditions of the reaction and the equilibrium constant.

• Because \( K \) is extremely small, glucose primarily remains as \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6} \).
• Only a minute amount dissociates to form \( \mathrm{HCHO} \).

The very low concentration of \( \mathrm{HCHO} \) at equilibrium illustrates how overwhelmingly glucose molecules prefer staying intact rather than dissociating under standard conditions. This knowledge reaffirms that such reactions may not yield significant products, challenging the prediction of reaction outcomes based purely on initial reactant availability.