Problem 98 - A Complete Resource Book in Chemistry for JEE Main

Question

The enthalpy change involved in the oxidation of glucose is \(-2880 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Twenty five per cent of this energy is available for muscular work. If \(100 \mathrm{~kJ}\) of muscular work is needed to walk one kilometre, what is the maximum distance that a person will be able to walk after consuming \(120 \mathrm{~g}\) of glucose? (a) \(7.9 \mathrm{~km}\) (b) \(9.7 \mathrm{~km}\) (c) \(4.8 \mathrm{~km}\) (d) \(8.4 \mathrm{~km}\)

Step-by-Step Solution

Answer

The maximum distance is approximately 4.8 km, so the answer is (c) 4.8 km.

Step 1: Calculate Moles of Glucose

The molar mass of glucose (C₆H₁₂O₆) is calculated by adding the atomic masses of the elements: \[(6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.16 \text{ g/mol}\] Given that we have 120 g of glucose, the moles of glucose consumed is: \[\text{Moles of glucose} = \frac{120 \text{ g}}{180.16 \text{ g/mol}} = 0.666 \text{ mol}\]

Step 2: Calculate Total Enthalpy Change

The enthalpy change for oxidizing 1 mole of glucose is given as \(-2880 \text{ kJ/mol}\). Thus, for 0.666 moles, the total enthalpy change is: \[\text{Total enthalpy change} = -2880 \text{ kJ/mol} \times 0.666 \text{ mol} = -1917.6 \text{ kJ}\]

Step 3: Determine Energy Available for Work

Only 25% of the enthalpy change can be used for muscular work: \[\text{Energy for work} = 0.25 \times 1917.6 \text{ kJ} = 479.4 \text{ kJ}\]

Step 4: Calculate Maximum Distance

Given that walking one kilometer requires 100 kJ of energy, we determine the maximum distance the person can walk: \[\text{Maximum distance} = \frac{479.4 \text{ kJ}}{100 \text{ kJ/km}} = 4.794 \text{ km}\] Thus, after consuming 120 g of glucose, the person can walk a maximum of approximately 4.8 km.

Key Concepts

Enthalpy Change

In thermodynamics, enthalpy change is central to understanding how much energy is absorbed or released during a chemical reaction. This involves a crucial principle called Hess's Law, which states that the total enthalpy change in a reaction is the same, regardless of the route taken. When we talk about the oxidation of glucose, for example, we are interested in the energy it releases. The enthalpy change of \( -2880 \text{ kJ/mol} \) for glucose signifies the energy released when one mole of glucose undergoes complete oxidation.
The reason we pay attention to enthalpy change is that it helps us gauge whether the energy from the reaction can be harnessed for work, such as muscular activity. In everyday terms, this tells us how much of that energy contributes to practical activities, like walking or running.

Oxidation of Glucose

The oxidation of glucose is a metabolic process that provides energy for cellular functions. In simple terms, glucose, a sugar molecule many animals and plants consume, breaks down to release energy. This process happens in stages through cellular respiration.
The chemical reaction for glucose oxidation is generally represented as: \[ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}\]
Given the enthalpy change of \( -2880 \text{ kJ/mol} \), it indicates that this much energy is released when one mole of glucose is consumed. However, only a fraction of this energy, precisely 25% as mentioned, is actually convertible into muscular work. The rest is lost as heat, which is a normal part of metabolic processes.
Understanding this helps to illustrate how much energy intake is needed to perform physical tasks, such as walking.

Molar Mass Computation

Molar mass computation involves determining the mass of one mole of a compound by summing the atomic masses of its constituent elements. This is fundamental in chemistry because it allows for conversion between the mass of substances and the moles when performing calculations.
Let's break it down using the example of glucose, which is \(C_6H_{12}O_6\). To calculate its molar mass, we multiply the number of each type of atom by its atomic mass:

• Carbon (C): 6 atoms \( \times 12.01 \text{ g/mol} = 72.06 \text{ g/mol} \)
• Hydrogen (H): 12 atoms \( \times 1.008 \text{ g/mol} = 12.096 \text{ g/mol} \)
• Oxygen (O): 6 atoms \( \times 16.00 \text{ g/mol} = 96.00 \text{ g/mol} \)

Adding these values gives the molar mass of glucose: \( 180.16 \text{ g/mol} \). Knowing the molar mass, we can determine how many moles are present in a given mass of glucose, say 120 grams. This calculation is crucial for converting mass to moles, which is often necessary for further chemical calculations, such as finding how much energy can be derived from food during metabolism.