Question

Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{N}_{2}, 15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2},\) and 6.2\(\%\) water vapor. (a) If the total pressure of the gases is 0.985 atm, calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is 455 \(\mathrm{mL}\) and its temperature is \(37^{\circ} \mathrm{C}\) , calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2} ?\) (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} .\) See Section 3.2 and Problem 10.57 )

Short answer

The partial pressures are \(P_{N2} = 0.737 \, \text{atm}\), \(P_{O2} = 0.151 \, \text{atm}\), \(P_{CO2} = 0.036 \, \text{atm}\), and \(P_{H2O} = 0.061 \, \text{atm}\). The number of moles of CO2 exhaled is \(0.00566 \, \text{mol}\). To produce this quantity of CO2, \(0.170 \, \text{g}\) of glucose would need to be metabolized.

Step-by-step solution

Calculate Partial Pressures

In this step, use the given percentages of each component and the total pressure to find the partial pressures of each gas. We will use the formula \(P_{gas} = P_{total} \cdot \%_{gas}\), where \(P_{gas}\) is the partial pressure of the gas, \(P_{total}\) is the total pressure, and \%_{gas} is the percentage of the gas in decimal form. a) For N2: - \(P_{N2} = (0.985 \, \text{atm}) \cdot 0.748 = 0.737 \, \text{atm}\) For O2: - \(P_{O2} = (0.985 \, \text{atm}) \cdot 0.153 = 0.151 \, \text{atm}\) For CO2: - \(P_{CO2} = (0.985 \, \text{atm}) \cdot 0.037 = 0.036 \, \text{atm}\) For water vapor: - \(P_{H2O} = (0.985 \, \text{atm}) \cdot 0.062 = 0.061 \, \text{atm}\) The partial pressures are: \(P_{N2} = 0.737 \, \text{atm}\), \(P_{O2} = 0.151 \, \text{atm}\), \(P_{CO2} = 0.036 \, \text{atm}\), and \(P_{H2O} = 0.061 \, \text{atm}\).

Calculate the Moles of CO2

In this step, use the Ideal Gas Law, \(PV = nRT\), to find the number of moles of CO2 by rearranging the formula: \(n = \frac{PV}{RT}\). We have the partial pressure of CO2 from step 1, the volume of the exhaled gas (455 mL), and the temperature (37°C). The gas constant R is \(0.0821\, \frac{\text{L atm}}{\text{mol K}}\). First, convert the volume to L and the temperature to K: - Volume: \(455 \, \text{mL} = 0.455 \, \text{L}\) - Temperature: \(37^{\circ}\text{C} + 273.15 = 310.15 \, \text{K}\) Now, use the Ideal Gas Law to find the number of moles of CO2: - \(n_{CO2} = \frac{(0.036 \, \text{atm})(0.455 \, \text{L})}{(0.0821 \, \frac{\text{L atm}}{\text{mol K}})(310.15 \, \text{K})} = 0.00566 \, \text{mol CO2}\)

Calculate the Grams of Glucose Metabolized

In this step, use the stoichiometry of the chemical reaction of glucose combustion to find the grams of glucose needed to produce the moles of CO2 calculated in step 2. For the combustion of glucose, the balanced chemical reaction is: \[C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O\] From the balanced chemical equation, 1 mol of glucose produces 6 mol of CO2. Calculate the moles of glucose needed to produce 0.00566 mol of CO2: - \(\text{mol glucose} = \frac{0.00566 \, \text{mol CO2}}{6} = 9.43 \times 10^{-4} \, \text{mol glucose}\) Now, convert moles of glucose to grams using the molar mass of glucose (\(180.16 \, \frac{\text{g}}{\text{mol}}\)): - \(9.43 \times 10^{-4} \, \text{mol glucose} \cdot \frac{180.16 \, \text{g}}{\text{mol}} = 0.170 \, \text{g glucose}\) Thus, to produce the given quantity of CO2, \(0.170 \, \text{g}\) of glucose would need to be metabolized.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry and physics that is used to relate the physical properties of gases. It combines several simpler gas laws, making it an essential tool for calculating various properties such as pressure, volume, and temperature. The formula is expressed as: \[ PV = nRT \] where:

• \( P \) is the pressure of the gas in atmospheres (atm).
• \( V \) is the volume of the gas in liters (L).
• \( n \) is the number of moles of the gas.
• \( R \) is the ideal gas constant, approximately \( 0.0821 \, \frac{L \, atm}{mol \, K} \) in the context of these units.
• \( T \) is the temperature in Kelvin (K).

Understanding this relationship allows us to solve for any unknown when the other variables are known. In our exercise, the Ideal Gas Law was used to calculate the number of moles of \( CO_2 \) in an exhaled breath by rearranging the formula to solve for \( n \):\[ n = \frac{PV}{RT} \]This step is crucial when dealing with gas mixtures and determining how much of a substance is involved in a reaction.

Stoichiometry

Stoichiometry is the area of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It is based on the conservation of mass, meaning that the total mass of reactants equals the total mass of products. This concept is represented through balanced chemical equations which reflect the fixed ratios in which substances react or form.For example, consider the combustion of glucose:\[C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O\]This equation demonstrates the stoichiometric relationships:

• One mole of \( C_6H_{12}O_6 \) produces six moles of \( CO_2 \).
• The same amount of moles of \( O_2 \) is consumed as \( CO_2 \) is produced.

This relationship was utilized to determine how much glucose needed to be metabolized to produce a given amount of \( CO_2 \). The exercise required the conversion from moles of \( CO_2 \) to moles of glucose by using the stoichiometric ratio from the balanced equation:\[\text{mol glucose} = \frac{\text{mol } CO_2}{6}\]

Chemical Reactions

Chemical reactions involve transformations of substances through re-arrangements of atoms. These reactions are represented by chemical equations essentially showing reactants converting into products. Each reaction has its unique characteristics, like releasing energy (exothermic) or absorbing energy (endothermic). For instance, the combustion of glucose is an exothermic reaction:\[C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O\]This reaction releases energy stored in the chemical bonds of glucose and oxygen.Moreover, understanding a chemical reaction requires knowing the reactants and products, which is essential for practical applications like energy production or biological metabolism. Chemical reactions follow the law of conservation of mass, meaning the mass of reactants must equal the mass of products - no atoms are lost or gained, only rearranged. In our exercise, understanding this balance helped determine the amount of glucose needed to match the \( CO_2 \) produced through metabolic processes.