Question

Compare the heat evolved at constant pressure per mole of oxygen in the combustion of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) and palmitic acid \(\left(\mathrm{C}_{16} \mathrm{H}_{32} \mathrm{O}_{2}\right)\) with the combustion of a typical protein, for which the empirical formula is \(\mathrm{C}_{4.3} \mathrm{H}_{6.6} \mathrm{NO}\) Assume for the protein that the combustion yields \(\mathrm{N}_{2}(g)\) \(\mathrm{CO}_{2}(g),\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Assume that the enthalpies for combustion of sucrose, palmitic acid, and a typical protein are \(5647 \mathrm{kJ} \mathrm{mol}^{-1}, 10,035 \mathrm{kJ} \mathrm{mol}^{-1},\) and \(22.0 \mathrm{kJ} \mathrm{g}^{-1},\) respectively. Based on these calculations, determine the average heat evolved per mole of oxygen consumed, assuming combustion of equal moles of sucrose, palmitic acid, and protein.

Short answer

The average heat evolved per mole of oxygen consumed for the combustion of equal moles of sucrose, palmitic acid, and protein is \(462.60\, kJ/mol\, O_{2}\).

Step-by-step solution

Find the moles of oxygen consumed in each combustion reaction

For each compound, set up a balanced combustion equation, and find the ratio of the moles of oxygen to the number of moles of the compound. For sucrose: \(C_{12}H_{22}O_{11}(s) + 12O_{2}(g) \rightarrow 12CO_{2}(g) + 11H_{2}O(l)\) 12 moles of oxygen are consumed per mole of sucrose. For palmitic acid: \(C_{16}H_{32}O_{2}(s) + 23O_{2}(g) \rightarrow 16CO_{2}(g) + 16H_{2}O(l)\) 23 moles of oxygen are consumed per mole of palmitic acid. For a typical protein: \(C_{4.3}H_{6.6}NO + 4.3O_{2}(g) \rightarrow 1.3N_{2}(g) + 4.3CO_{2}(g) + 3.3H_{2}O(l)\) 4.3 moles of oxygen are consumed per mole of the protein.

Compute the heat evolved per mole of oxygen for each combustion reaction

Divide the given enthalpies of combustion by the number of moles of oxygen consumed in each reaction. For sucrose: \(\frac{5647\, kJ/mol}{12\, mol\, O_{2}} = 470.58\, kJ/mol\, O_{2}\) For palmitic acid: \(\frac{10,035\, kJ/mol}{23\, mol\, O_{2}} = 436.30\, kJ/mol\, O_{2}\) For a typical protein, first, convert the enthalpy from kJ/g to kJ/mol using the protein's molar mass, which is approximately 94 g/mol: \(22.0 \frac{kJ}{g} \times 94 \frac{g}{mol}= 2068\, kJ/mol\). Then, \(\frac{2068\, kJ/mol}{4.3\, mol\, O_{2}} = 480.93\, kJ/mol\, O_{2}\)

Calculate the average heat evolved per mole of oxygen consumed

Compute the mean of the heat evolved per mole of oxygen for sucrose, palmitic acid, and typical protein. \(\frac{470.58 + 436.30 + 480.93}{3} = 462.60\, kJ/mol\, O_{2}\) The average heat evolved per mole of oxygen consumed for the combustion of equal moles of sucrose, palmitic acid, and protein is 462.60 kJ/mol O₂.

Key concepts

Enthalpy of Combustion

Enthalpy of combustion is a fundamental concept in chemistry that measures the heat released when a substance completely reacts with oxygen, typically under constant pressure. In simpler terms, it's the amount of energy you can obtain from burning a fuel. When we say that the enthalpy of combustion for sucrose is 5647 kJ/mol, for palmitic acid is 10,035 kJ/mol, and for a typical protein is 22 kJ/g, we're talking about the energy released when one mole of each of these substances burns entirely in oxygen to form carbon dioxide and water.
To understand this concept better, imagine lighting a sugar cube on fire. The heat you feel and the light you see are part of the energy released, which can be quantified as the enthalpy of combustion for sucrose. Similarly, for palmitic acid and the protein, the enthalpies represent the heat energy that could keep you warm or cook your food. This concept is crucial when comparing different fuels or food energy content.

Moles of Oxygen Consumed

In a combustion reaction, oxygen from the air reacts with the substance being burned. The 'moles of oxygen consumed' is exactly what it sounds like: the amount of oxygen, measured in moles, that reacts with a specific amount of fuel.
For our examples, 12 moles of oxygen react with a mole of sucrose, 23 react with a mole of palmitic acid, and 4.3 react with a mole of the typical protein. This number is crucial because it reflects the stoichiometry of the reaction, which is the proportional relationship between the reactants and products in a chemical equation. Knowing the moles of oxygen consumed helps us understand how much oxygen we need to burn a fuel completely, which is particularly important in industrial applications where efficiency and cost-effectiveness are key.

Heat Evolution Per Mole

The 'heat evolution per mole' refers to the heat released per mole of the substance being reacted or per mole of oxygen consumed. For the latter perspective, it provides insight into the efficiency of the oxidation process. By dividing the enthalpy of combustion by the moles of oxygen consumed, we arrive at the heat evolved per mole of oxygen.
It's like comparing the fuel efficiency of different cars—knowing the miles per gallon gives you an idea of how far you can travel on a gallon of gasoline. Similarly, the heat evolution per mole of oxygen tells us how much energy we can expect to get for each mole of oxygen that participates in the combustion. It's an essential concept for environmental and economic considerations. For instance, the calculation from the exercise indicates sucrose releases 470.58 kJ/mol O₂, palmitic acid releases 436.30 kJ/mol O₂, and the protein releases 480.93 kJ/mol O₂. This information helps us understand which substance provides the most energy per unit of oxygen, which can be used to maximize energy output while minimizing waste.