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A typical fat in the body is glyceryl trioleate, \(\mathrm{C}_{57} \mathrm{H}_{104} \mathrm{O}_{6}\). When it is metabolized in the body, it combines with oxygen to produce carbon dioxide, water, and \(3.022 \times 10^{4} \mathrm{~kJ}\) of heat per mole of fat. (a) Write a balanced thermochemical equation for the metabolism of fat. (b) How many kilojoules of energy must be evolved in the form of heat if you want to get rid of five pounds of this fat by combustion? (c) How many nutritional calories is this? (1 nutritional calorie = \(1 \times 10^{3}\) calories)

Short Answer

Expert verified
Question: Write a balanced thermochemical equation for the metabolism of glyceryl trioleate and calculate the heat evolved when burning 5 pounds of this fat. Convert the heat energy to nutritional calories. Answer: The balanced thermochemical equation for the metabolism of glyceryl trioleate is C57H104O6 + 80O2 -> 57CO2 + 52H2O + 3.022 * 10^4 kJ/mol. When burning 5 pounds of glyceryl trioleate, the heat evolved is 7.72 * 10^4 kJ, which is equivalent to 1.84 * 10^4 nutritional calories.

Step by step solution

01

(a) Write a balanced thermochemical equation for the metabolism of glyceryl trioleate

First we have to write the general equation for the metabolism of glyceryl trioleate. Glyceryl trioleate (C57H104O6) combines with oxygen (O2) to produce carbon dioxide (CO2), water (H2O), and heat. C57H104O6 + O2 -> CO2 + H2O + heat Next, balance the equation: C57H104O6 + 80O2 -> 57CO2 + 52H2O + 3.022 * 10^4 kJ (a) Here is the balanced thermochemical equation for the metabolism of glyceryl trioleate: C57H104O6 + 80O2 -> 57CO2 + 52H2O + 3.022 * 10^4 kJ/mol
02

(b) Calculate the heat evolved for 5 pounds of glyceryl trioleate

We are given the weight of glyceryl trioleate in pounds, and we need to convert it into moles and calculate the heat evolved during combustion. First, we have to convert the weight to grams: 5 pounds * 454 g/pound ≈ 2270 g Now, we need to find the molar mass of glyceryl trioleate: Molar mass = (57 * C) + (104 * H) + (6 * O) Molar mass ≈ (57 * 12.01 g/mol) + (104 * 1.01 g/mol) + (6 * 16 g/mol) Molar mass ≈ 885.71 g/mol Next, we convert the weight of glyceryl trioleate to moles: Moles = weight/molar mass = 2270 g / 885.71 g/mol ≈ 2.56 moles Now, we can calculate the heat evolved: Heat evolved = moles * heat per mole = 2.56 moles * 3.022 * 10^4 kJ/mol ≈ 7.72 * 10^4 kJ (b) The heat evolved in the form of heat when burning 5 pounds of glyceryl trioleate is 7.72 * 10^4 kJ.
03

(c) Converting heat energy to nutritional calories

We have to convert the heat energy in kJ to nutritional calories. 1 nutritional calorie = 1 kcal = 10^3 cal 1 kJ = 0.239 kcal Heat in nutritional calories = heat in kJ * (0.239 kcal/kJ) = (7.72 * 10^4 kJ) * (0.239 kcal/kJ) ≈ 1.84 * 10^4 kcal (c) The heat evolved in the form of heat when burning 5 pounds of glyceryl trioleate is equivalent to 1.84 * 10^4 nutritional calories.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Metabolism of Fats
When we talk about the metabolism of fats, we're referring to how our bodies break down fat molecules to release energy. This is crucial for providing our bodies with the necessary fuel to perform daily functions. Fats are rich in energy, often stored in the body as a backup fuel source. The equation to represent the metabolism of a typical fat, such as glyceryl trioleate, looks something like this:

Glyceryl trioleate combines with oxygen, resulting in the production of carbon dioxide, water, and a significant amount of heat.

  • This process can be simplified into a formula: \[ \text{C}_{57}\text{H}_{104}\text{O}_{6} + 80\text{O}_{2} \rightarrow 57\text{CO}_{2} + 52\text{H}_{2}\text{O} + 3.022 \times 10^4 \text{ kJ/mol} \]
This formula shows us how the components combine and transform, releasing a substantial amount of energy. This energy is measured in kilojoules per mole of fat. Understanding this process helps us see why fats are such a rich energy source.
Balanced Chemical Equations
Balancing chemical equations ensures that we depict accurate representations of chemical reactions. The Law of Conservation of Mass dictates that matter can neither be created nor destroyed. Therefore, the number of atoms of each element must be the same on both sides of the equation.

In the metabolism of glyceryl trioleate, the equation starts as:\[ \text{C}_{57}\text{H}_{104}\text{O}_{6} + \text{O}_{2} \rightarrow \text{CO}_{2} + \text{H}_{2}\text{O} + \text{heat} \]

We then balance this by ensuring equal carbon, hydrogen, and oxygen atoms on both sides.
  • Carbon: 57 atoms from glyceryl trioleate translate to 57 \(\text{CO}_{2}\).
  • Hydrogen: 104 atoms require 52 \(\text{H}_{2}\text{O}\) molecules.
  • Oxygen: Starts with 80 \(\text{O}_{2}\) to balance the equation while producing the noted heat energy.
Balancing chemical equations is essential in accurately describing the metabolic process and understanding energy transformation.
Nutritional Calories
Nutritional calories are units of energy used to describe the energy content of food and how much energy the body can obtain from it. In scientific terms, one nutritional calorie is equivalent to 1 kcal, or 1000 small 'calories.'

When we talk about burning fat (like 5 pounds of glyceryl trioleate for energy), it's important to convert that energy to nutritional calories since that's how we measure food energy.
  • The total energy released when metabolizing 5 pounds of fat is 77200 kJ.
  • We convert this to nutritional calories using the conversion factor: \(1\text{ kJ} = 0.239\text{ kcal}\).
  • Thus, \(77200 \text{ kJ} \times 0.239 \text{ kcal/kJ} \approx 18400 \text{ kcal (nutritional calories)}\).
This calorie conversion helps us understand the potential energy gained or lost through diet and exercise, especially concerning fats.

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Most popular questions from this chapter

When one mole of caffeine \(\left(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\right)\) is burned in air, \(4.96 \times 10^{3} \mathrm{~kJ}\) of heat is evolved. Five grams of caffeine is burned in a bomb calorimeter. The temperature is observed to increase by \(11.37^{\circ} \mathrm{C}\). What is the heat capacity of the calorimeter in \(\mathrm{J} /{ }^{\circ} \mathrm{C} ?\)

Butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is widely used as a fuel for disposable lighters. When one mole of butane is burned in oxygen, carbon dioxide and steam are formed and \(2658.3 \mathrm{~kJ}\) of heat is evolved. (a) Write a thermochemical equation for the reaction. (b) Using Table \(8.3\), calculate the standard heat of formation of butane.

Which statement(s) is/are true about bond enthalpy? (a) Energy is required to break a bond. (b) \(\Delta H\) for the formation of a bond is always a negative number. (c) Bond enthalpy is defined only for bonds broken or formed in the gaseous state. (d) Because the presence of \(\pi\) bonds does not influence the geometry of a molecule, the presence of \(\pi\) bonds does not affect the value of the bond enthalpy between two atoms either. (e) The bond enthalpy for a double bond between atoms \(A\) and \(B\) is twice that for a single bond between atoms \(\mathrm{A}\) and \(\mathrm{B}\).

In photosynthesis, the following reaction takes place: \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 6 \mathrm{O}_{2}(g)+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \quad \Delta H=2801 \mathrm{~kJ}\) (a) Calculate \(\Delta H\) when one mole of \(\mathrm{CO}_{2}\) reacts. (b) How many kilojoules of energy are liberated when \(15.00 \mathrm{~g}\) of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), is burned in oxygen?

When \(35.0 \mathrm{~mL}\) of \(1.43 \mathrm{M} \mathrm{NaOH}\) at \(22.0^{\circ} \mathrm{C}\) is neutralized by \(35.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) also at \(22.0^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, the temperature of the final solution rises to \(31.29^{\circ} \mathrm{C}\). Assume that the specific heat of all solutions is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), that the density of all solutions is \(1.00 \mathrm{~g} / \mathrm{mL}\), and that volumes are additive. (a) Calculate \(q\) for the reaction. (b) Calculate \(q\) for the neutralization of one mole of \(\mathrm{NaOH}\).

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