Question

The enthalpy change involved in oxidation of glucose is \(-2880 \mathrm{~kJ} \mathrm{~mol}^{-1}, 25 \%\) of this energy is available for muscular work. If \(100 \mathrm{~kJ}\) of muscular work is needed to walk one \(\mathrm{km}\), the maximum distance \((\mathrm{km})\) that a person will be able to walk after taking \(150 \mathrm{~g}\) of glucose is

Short answer

A person can walk approximately 6 km after consuming 150 g of glucose.

Step-by-step solution

Calculate Moles of Glucose

First, find the number of moles of glucose in 150 g. The molar mass of glucose (C6H12O6) is approximately 180 g/mol. Use the formula: \( ext{moles of glucose} = \frac{150}{180} \approx 0.833 \text{ mol} \).

Determine Total Energy from Glucose

Calculate the total energy available from 0.833 mol of glucose using the enthalpy change: \( -2880 \text{ kJ/mol} \times 0.833 \approx -2398.64 \text{ kJ} \).

Compute Energy Available for Work

Only 25% of the energy from glucose is available for muscular work. Thus, the energy available for work is: \( 0.25 \times 2398.64 \approx 599.66 \text{ kJ} \).

Find Maximum Distance Walkable

Since 100 kJ is required to walk 1 km, the maximum distance a person can walk with 599.66 kJ is: \( \frac{599.66}{100} \approx 5.9966 \text{ km} \).

Key concepts

Enthalpy Change

Enthalpy change is a fundamental concept in thermodynamics, especially useful in chemistry. It represents the total energy change in a system when a chemical reaction occurs under constant pressure. When glucose is oxidized, the enthalpy change is quite significant. Glucose oxidation releases energy, denoted by a negative enthalpy change of \(-2880 \, \text{kJ/mol}\). This indicates that the reaction is exothermic, meaning it releases energy in the form of heat.

• Enthalpy is often symbolized as \(\Delta H\), which stands for the heat absorbed or released during a process or reaction.
• In our exercise, only 25% of this energy becomes available for muscular work.

Understanding enthalpy change helps predict how much energy can be harnessed from a reaction, like the one that occurs when our bodies utilize glucose.

Oxidation of Glucose

The oxidation of glucose is a crucial process in biological systems, allowing organisms to get energy from food. Glucose, a simple sugar with the formula \(\text{C}_6\text{H}_{12}\text{O}_6\), undergoes oxidation as follows:

• Glucose reacts with oxygen to produce carbon dioxide, water, and energy.
• The energy released is partly stored by the body in the form of ATP, which powers cellular functions.

In our exercise, we calculated the energy released per mole. However, only a quarter of this energy is utilized for muscular activity. This fraction is critical because it aligns with the body's efficiency in converting chemical energy into kinetic energy. Understanding this process is crucial for fields such as biochemistry and nutrition.

Molar Mass Calculation

Molar mass is an essential calculation in chemistry that helps us determine the number of moles in a given sample. To calculate the molar mass of a compound like glucose, we sum the atomic masses of all the atoms present in its chemical formula.

• Glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) has: 6 Carbon atoms, 12 Hydrogen atoms, and 6 Oxygen atoms.
• The molar mass is calculated as: \((6 \times 12.01) + (12 \times 1.01) + (6 \times 16.00) = 180\, \text{g/mol}\).

In our scenario, understanding the molar mass of glucose allows us to compute how many moles are present in 150 g of glucose, resulting in approximately 0.833 mol. Recognizing the importance of these basic calculations enables more complex analyses and deeper insight into reaction energetics.