Problem 108 - General Chemistry: Principles and Modern Applications

Question

The heat required to sustain animals while they hibernate comes from the biochemical combustion of fatty acids, such as arachidonic acid, \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\) What volume of air, measured at \(298 \mathrm{K}\) and \(1.00 \mathrm{atm}\) is required to burn \(2.00 \mathrm{kg} \mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2} ?\) Air is approximately \(78.1 \% \mathrm{N}_{2}\) and \(20.9 \% \mathrm{O}_{2},\) by volume. Other gases make up the remaining \(1.0 \%\)

Step-by-Step Solution

Answer

In conclusion, roughly 23950 litres of air are required to burn 2.00 kg of arachidonic acid under the given conditions

Step 1: Write down the combustion reaction equation

To solve the problem, one needs to first write down the combustion reaction equation to see how arachidonic acid reacts with oxygen to produce carbon dioxide and water: \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2} + 31\mathrm{O}_{2} \rightarrow 20\mathrm{CO}_{2} + 16\mathrm{H}_{2}\mathrm{O} \) Note that the coefficients in front of reactants tell us the stoichiometric amounts needed (or produced) per each molecule (or mole) of arachidonic acid.

Step 2: Calculate the amount of \(\mathrm{O}_{2}\) needed

First, convert the mass of \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\) to moles by dividing it with its molecular weight. The molecular weight of \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\) is \(304 \mathrm{g/mole}\). Therefore, for \(2.00 \mathrm{kg} = 2000 \mathrm{g}\) of \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\), we have \(2000 \mathrm{g} / 304 \mathrm{g/mole} = 6.58 \mathrm{moles}\) of arachidonic acid. According to the stoichiometry of the combustion reaction, for each mole of \(\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{O}_{2}\) we need \(31 \mathrm{moles}\) of \(\mathrm{O}_{2}\). Therefore, we need \(6.58 \mathrm{moles} * 31 = 204 \mathrm{moles}\) of \(\mathrm{O}_{2}\) to burn \(2.00 \mathrm{kg}\) of arachidonic acid.

Step 3: Calculate the volume of air required

Knowing that air consists of \(20.9 \%\ \mathrm{O}_{2}\) by volume, we can use this percentage to calculate the total volume of air needed to provide \204 \mathrm{moles}\ \mathrm{O}_{2}\. First, convert the moles of \(\mathrm{O}_{2}\) to volume at the given conditions using the Ideal Gas Law: \(V = nRT/P\), where \(n\) is the number of moles, \(R\) is the ideal gas constant (0.0821 \(\bm{L·atm/mol·K}\)), \(T\) is the temperature in Kelvin, and \(P\) is the pressure in atm. Substituting these values, we get \(V = 204 \mathrm{moles} * 0.0821 \bm{L·atm/mol·K} * 298 \bm{K} / 1.00 \bm{atm} = 5004 \bm{L}\). As this represents \(20.9 \%\ of the total air volume, if we let \(x\) be the total volume of air, we get: \(0.209x = 5004 \bm{L}\), solving for \(x\) we get an approximation of \(x = 23950 \bm{L}\) of air.

Key Concepts

Stoichiometry

Stoichiometry involves calculating the right amounts of reactants and products in a chemical reaction based on the balanced equation. In combustion reactions, like that of arachidonic acid, stoichiometry lets us understand the precise quantities of oxygen required to completely combust a given mass of a substance.

Key points about stoichiometry include:

• The coefficients in a balanced chemical equation represent the molar ratio of the reactants and products.
• For example, in the combustion of arachidonic acid, the coefficients show that 31 moles of oxygen are needed to burn 1 mole of the acid.
• This ratio helps convert the mass of reactants to moles and determine how much of a reactant is needed or how much product will be formed.

The stoichiometric calculation from the exercise illustrated that 204 moles of \(\mathrm{O}_{2}\) are needed for complete combustion of 6.58 moles of arachidonic acid, derived from its mass. This understanding is crucial for scaling processes in industrial and biochemical applications.

Ideal Gas Law

The Ideal Gas Law relates the macroscopic properties of gases: pressure \( (P) \), volume \( (V) \), temperature \( (T) \), and number of moles \( (n) \), through the equation \( PV = nRT \). Here, \( R \) is the gas constant which fits consistent units for calculations. This law is fundamental for calculating gas volumes under certain conditions in reactions.

For instance, in the calculation of the volume of air needed to supply oxygen for combustion, the Ideal Gas Law is utilized to find the oxygen volume from its mole count:

• First, the number of moles of \( \mathrm{O}_{2} \) was determined using stoichiometry.
• Then, the volume is derived using \( V = nRT/P \), substituting the values for \( n \), \( R \), \( T \), and \( P \).
• The outcome was 5004 L for 204 moles of \( \mathrm{O}_{2} \) under the conditions given (298 K and 1 atm).

The versatility of the Ideal Gas Law shows in how it bridges the gap between chemical equations and real-world applications, by converting moles to measurable volumes of gases.

Biochemical Combustion

Biochemical combustion refers to the breakdown of organic molecules by living organisms to release energy. In hibernating animals, like bears, stored fats undergo biochemical combustion to produce heat, water, and carbon dioxide. Arachidonic acid serves as an example fatty acid in these processes.

Some important aspects of biochemical combustion include:

• It is an efficient process as it stores large amounts of energy in small quantities of fat.
• In the exercise, combustion of arachidonic acid generates heat that is vital for sustaining biological functions during hibernation.
• The stoichiometric balance of reactants such as oxygen and fatty acids ensures the production of energy and metabolic byproducts.

This context shows how chemical reactions are not just theoretical exercises but crucial for survival, highlighting the intersection of chemistry and biology. Understanding these reactions aids in comprehending how organisms, especially during energy-intensive processes like hibernation, manage their metabolic needs.