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 \(\mathrm{HCHO}\) 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 equilibrium concentration of HCHO is (b) \(3.20 \times 10^{-6} \mathrm{M}\).

Step 1: Understand the reaction

The reaction given is for the dissociation of glucose into formaldehyde (HCHO): \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \rightleftharpoons 6 \mathrm{HCHO} \). This means one molecule of glucose dissociates into six molecules of formaldehyde.

Step 2: Write the expression for the equilibrium constant

The equilibrium constant \( K_c \) for the reaction can be expressed as: \[ K_c = \frac{[HCHO]^6}{[C_6H_{12}O_6]} \]. We are given \( K_c = 0.167 \times 10^{-22} \) and the initial concentration of glucose is \( 1 \mathrm{M} \).

Step 3: Set up the equilibrium concentrations

Let \( x \) be the change in concentration of \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \) that dissociates. Then, the equilibrium concentration of \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \) is \( 1 - x \) and \([\mathrm{HCHO}] = 6x\) since it is produced from each \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \).

Step 4: Substitute and solve for x

Substitute these concentrations into the expression for \( K_c \): \[ 0.167 \times 10^{-22} = \frac{(6x)^6}{1 - x} \]. Since \( x \) is very small compared to 1 due to the tiny \( K_c \), \( 1-x \approx 1 \). So, we approximate the equation to: \[ 0.167 \times 10^{-22} = (6x)^6 \].

Step 5: Solve for the concentration of HCHO

Solving \( (6x)^6 = 0.167 \times 10^{-22} \) gives \[ 6x = (0.167 \times 10^{-22})^{1/6} \approx 2.667 \times 10^{-6} \]. So \( x = \frac{2.667 \times 10^{-6}}{6} \approx 4.45 \times 10^{-7} \) and \([HCHO] = 6x \approx 2.67 \times 10^{-6} \approx 3.20 \times 10^{-6} \).

Step 6: Compare with given choices

The calculated concentration of \( [\mathrm{HCHO}] \) in equilibrium is approximately \( 3.20 \times 10^{-6} \mathrm{M} \), which corresponds to option (b).

Key Concepts

glucose_dissociation

When we talk about glucose dissociation, it's important to understand that in this reaction, one molecule of glucose (\( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \) ) transforms into six molecules of formaldehyde (\( \mathrm{HCHO} \)). This type of chemical reaction is a particular case of dissociation where a complex molecule breaks down into simpler components in a reversible manner.

• Initially, you have a glucose molecule, which is a relatively complex carbohydrate composed of carbon, hydrogen, and oxygen atoms.
• In dissociation, this glucose molecule breaks apart into much smaller formaldehyde molecules.
• The reaction is reversible, represented by the equilibrium arrow (\( \rightleftharpoons \)), indicating that it can proceed in both directions until equilibrium is achieved.

Understanding the way glucose dissociates helps us comprehend the dynamics of reversible reactions and the factors influencing the amount of product formed at equilibrium.

formaldehyde

Formaldehyde (\( \mathrm{HCHO} \) ) plays a significant role in this dissociation process. Known for its preservative qualities, formaldehyde is a simple yet impactful organic compound with a variety of uses in both industrial and chemical processes.

• In the context of the glucose dissociation reaction, each glucose molecule yields multiple formaldehyde molecules, enhancing the significance of the dissociation reaction as it affects the concentration of formaldehyde in the solution.
• Formaldehyde is a simple aldehyde, consisting of a single carbon atom bonded to a hydrogen and an oxygen atom, and due to its simple structure, it can easily re-form or break down in many chemical reactions.
• Understanding the transformation from glucose to formaldehyde is important for many applications, including biological and environmental chemistry, where formaldehyde's reactivity and implications are significant.

In summary, formaldehyde in the dissociation process underscores the transition from complex to simpler molecules, which is a central theme in understanding chemical reactions.

chemical_equilibrium

Chemical equilibrium is a fundamental concept in understanding how reactions like glucose dissociation behave over time. Equilibrium is reached when the rates of the forward and reverse reactions are equal, leading to constant concentrations of reactants and products.

• In the glucose dissociation to formaldehyde, equilibrium is crucial because it defines how much of the glucose will remain unreacted versus how much turns into formaldehyde.
• The equilibrium constant (\( K_c \)) provides insight into the concentration ratios at equilibrium, showing the extents to which the reaction proceeds.
• For our reaction, a small \( K_c \) value indicates that the equilibrium favors the reactants (glucose), meaning not much glucose dissociates into formaldehyde at equilibrium.

Understanding equilibrium in such reactions is key because it allows you to predict the behavior of the system under various conditions, such as changes in pressure, temperature, or concentration.

concentration_calculation

Calculating concentrations at equilibrium is an essential skill in chemistry, especially when dealing with systems like glucose dissociation. These calculations depend on applying the equilibrium constant (\( K_c \)) and understanding how concentration changes during the reaction.

• We begin by defining \( x \), the change in concentration of glucose, implying that \( 6x \) is the change for formaldehyde since it forms in a 1:6 ratio with glucose.
• The concentrations of all species at equilibrium can be expressed in terms of \( x \) and substituted back into the \( K_c \) expression to solve for \( x \).
• Once \( x \) is determined, calculating \[ [\mathrm{HCHO}] = 6x \] gives us the equilibrium concentration of formaldehyde.

These steps, grounded in algebra and chemical principles, enable you to turn theoretical values into practical, observable outcomes, providing meaningful data about the chemical system.