Problem 71 - Chemistry The Science in Context

Question

How much glucose must be metabolized to completely evaporate \(1.00 \mathrm{g}\) of water at \(37^{\circ} \mathrm{C},\) given \(\Delta H_{\text {comb, }, \text { glucose }}=\) \(-2803 \mathrm{kJ} / \mathrm{mol} ?\)

Step-by-Step Solution

Answer

Answer: Approximately 0.145g of glucose.

Step 1: Calculate the heat required to evaporate the water

First, we need to determine the heat required to evaporate 1.00g of water at 37°C. The heat (Q) required for evaporation can be found using the formula: Q = m × ΔHvap where m is the mass of water and ΔHvap is the heat of vaporization of water. The heat of vaporization of water is approximately 40.7 kJ/mol (at 37°C), and the molar mass of water is 18.02 g/mol. First, let's find out the number of moles of water: moles of water = mass of water / molar mass of water moles of water = 1.00g / 18.02 g/mol = 0.0555 mol Next, we will use this value to calculate the heat required to evaporate the water: Q = 0.0555 mol × 40.7 kJ/mol = 2.26 kJ So, the heat required to evaporate 1.00g of water at 37°C is 2.26 kJ.

Step 2: Determine the moles of glucose needed to provide the required heat

Given that the heat of combustion of glucose is -2803 kJ/mol, we can determine the moles of glucose necessary to provide the heat required to evaporate the water using the formula: moles of glucose = heat required / heat of combustion of glucose moles of glucose = 2.26 kJ / (-2803 kJ/mol) = -0.000807 mol The negative sign indicates that the heat is released during the combustion of glucose.

Step 3: Calculate the amount of glucose that must be metabolized, in grams

Finally, we need to find the mass of glucose that must be metabolized. To do this, we will use the molar mass of glucose (which is 180.16 g/mol): mass of glucose = moles of glucose × molar mass of glucose mass of glucose = (-0.000807 mol) × (180.16 g/mol) = -0.145 g The negative sign indicates that the mass of glucose is decreased during metabolism. Thus, approximately 0.145g of glucose must be metabolized to completely evaporate 1.00g of water at 37°C.

Key Concepts

Heat of Vaporization

The heat of vaporization is a key concept in understanding how substances, like water, convert from liquid to vapor. This process requires a specific amount of heat, known as the heat of vaporization. For water, at a temperature of 37°C, this value is approximately 40.7 kJ/mol. It describes the energy needed to convert one mole of liquid water into vapor without changing its temperature.
Understanding this is crucial in calculations involving phase changes, like the evaporation of water. Given the molar mass of water is 18.02 g/mol, determining the moles of water allows us to use the heat of vaporization formula:

• \( Q = m \times \Delta H_{\text{vap}} \)

Here, \( Q \) is the heat required for evaporation, \( m \) is the moles of water, and \( \Delta H_{\text{vap}} \) is the heat of vaporization. Calculations like this are pivotal in thermodynamic studies, especially those that involve heating processes.

Heat of Combustion

The heat of combustion is another essential concept, particularly when discussing fuel and food energy, such as glucose metabolism. It is the amount of energy released as heat when a compound, like glucose, undergoes complete combustion with oxygen. This is an exothermic process, meaning it releases energy, and is typically measured in kJ/mol.
In the exercise, the heat of combustion for glucose is given as -2803 kJ/mol. This negative sign indicates the reaction releases energy. By knowing how much energy is required to perform a task, such as vaporizing water, we can calculate how much glucose needs to be metabolized:

• \( \text{Moles of glucose} = \frac{\text{Heat required}}{-2803 \text{ kJ/mol}} \)

Understanding the heat of combustion helps in determining the energy efficiency and requirements of metabolic processes, both in biological and chemical contexts.

Molar Mass Calculation

Calculating molar mass is fundamental for converting mass to moles and vice versa. The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It plays a crucial role in stoichiometry, allowing scientists and students to relate quantities in grams to quantities in moles.
For instance, water has a molar mass of 18.02 g/mol, while glucose has a molar mass of 180.16 g/mol. To find the number of moles from a given mass, one can use the formula:

• \( \text{Moles} = \frac{\text{Mass in grams}}{\text{Molar Mass}} \)

Conversely, to find the mass from a given number of moles, the formula rearranges to:

• \( \text{Mass in grams} = \text{Moles} \times \text{Molar Mass} \)

This concept is pivotal in chemical reactions where mass conservation is observed, and in practical applications such as dosing in pharmaceuticals and calculating fuel needs.