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 4.8 km, which is option (c).

Step 1: Convert Mass to Moles

Given that the molar mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol, first calculate how many moles of glucose are in 120 g. Use the formula: \[ \text{Moles of glucose} = \frac{\text{Mass of glucose (g)}}{\text{Molar mass of glucose (g/mol)}}.\] Thus, \[ \text{Moles of glucose} = \frac{120\, \text{g}}{180\, \text{g/mol}} = 0.6667\, \text{mol}.\]

Step 2: Calculate Total Energy Available From Glucose

The enthalpy change for the oxidation of glucose is \(-2880\, \text{kJ/mol}\). To find the total energy released by 0.6667 mol of glucose, multiply the moles by the enthalpy change: \[ \text{Total energy} = 0.6667\, \text{mol} \times 2880\, \text{kJ/mol} = 1920\, \text{kJ}.\]

Step 3: Determine Usable Energy for Work

Only 25% of the energy from glucose is available for muscular work. Therefore, calculate 25% of the total energy: \[ \text{Usable energy} = 0.25 \times 1920\, \text{kJ} = 480\, \text{kJ}.\]

Step 4: Calculate Maximum Distance Walked

Given that 100 kJ of energy are needed to walk 1 km, determine the total distance that can be covered using the available energy: \[ \text{Distance} = \frac{\text{Usable energy (kJ)}}{\text{Energy per km (kJ/km)}} = \frac{480\, \text{kJ}}{100\, \text{kJ/km}} = 4.8\, \text{km}.\]

Step 5: Select Correct Answer

After finishing the calculations, the maximum distance a person can walk is 4.8 km. Therefore, the correct option is (c) 4.8 km.

Key Concepts

Oxidation of Glucose

The oxidation of glucose is a crucial biochemical reaction that provides energy for living organisms. This process involves the breakdown of glucose into carbon dioxide and water while releasing energy. The chemical equation for the oxidation of glucose is: \[C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + ext{energy}\] In this reaction, glucose (C_6H_{12}O_6) combines with oxygen to produce carbon dioxide, water, and energy. The energy released during this process is in the form of heat, with an enthalpy change of -2880 ext{ kJ/mol}. This means that each mole of glucose releases 2880 kilojoules of energy.Understanding the enthalpy change in this context is essential as it represents the potential of glucose as an energy source in biological systems. This energy supports various bodily functions like maintaining body temperature, muscle contractions, and daily activities.

Molar Mass of Glucose

The molar mass of a substance tells us how much one mole of that substance weighs, and it is usually measured in grams per mole (g/mol). The molar mass of glucose, whose chemical formula is C_6H_{12}O_6, is approximately 180 g/mol. To calculate this, we sum up the atomic masses of all the atoms in a glucose molecule:

• Carbon (C): 12 atomic mass units (amu) × 6 = 72 amu
• Hydrogen (H): 1 amu × 12 = 12 amu
• Oxygen (O): 16 amu × 6 = 96 amu

Adding these together gives us 180 amu, which translates to a molar mass of 180 g/mol. Knowing the molar mass allows us to convert between grams of glucose and moles of glucose using the formula: \[\text{Moles of glucose} = \frac{\text{Mass of glucose (g)}}{\text{Molar mass of glucose (g/mol)}}\] This conversion is crucial for understanding how much energy can be extracted from a given mass of glucose.

Energy Conversion

Energy conversion refers to the transformation of one form of energy into another, and it is a key concept in both physics and chemistry. In the context of glucose oxidation, chemical energy stored in the glucose molecule is converted into other forms of energy, mainly heat and usable biochemical energy. When glucose oxidizes, we capture only a portion of the energy released and convert it into usable energy for the body known as ATP (adenosine triphosphate). Not all energy is efficiently captured; instead, only about 25% is available for muscular work as per the exercise provided. This means:

• Of the -2880 ext{ kJ/mol} released, only 0.25 imes 2880 = 720 ext{ kJ/mol} is harnessed for activities like physical work.

Understanding this partial energy conversion is important for assessing how dietary glucose contributes to physical performance and energy needs in the human body.

Muscular Work

Muscular work requires energy, which is derived from the biochemical processes occurring within our body. The energy needed for movement and exercise comes from the oxidation of nutrients, primarily glucose.In the exercise, it is noted that 100 kJ of energy are required to walk 1 kilometer. This means that to determine how far someone can walk after consuming glucose, we must first know how much usable energy results from its oxidation.Using the information given, after consuming 120 g of glucose:

• We find that 0.6667 ext{ mol} of glucose can release a total of 1920 kJ of energy.
• Only 25%, or 480 kJ, is available for muscular work.
• Dividing the usable energy by energy required per kilometer gives us the distance: \[\frac{480 ext{ kJ}}{100 ext{ kJ/km}} = 4.8 ext{ km}\]

This calculation illustrates the direct link between the caloric intake from glucose and physical activity. It shows how our body metabolizes consumed glucose to support daily physical activities like walking.