Problem 57 - Chemistry: The Central Science in SI Units, Global Edition

Question

The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in our bodies produces \(\mathrm{CO}_{2}\), which is expelled from our lungs as a gas: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at normal body temperature, \(37^{\circ} \mathrm{C},\) and \(101.33 \mathrm{kPa}\) when \(10.0 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at \(100 \mathrm{kPa}\) and \(298 \mathrm{~K},\) to completely oxidize \(15.0 \mathrm{~g}\) of glucose.

Step-by-Step Solution

Answer

(a) The volume of dry CO2 produced at normal body temperature (37°C) and pressure (101.33 kPa) when 10.0g of glucose is consumed is 8.36 L. (b) The volume of oxygen required to completely oxidize 15.0g of glucose at 100 kPa and 298 K is 12.39 L.

Step 1: Convert the temperature to Kelvin

The temperature given for the body is 37°C. We need to convert this to Kelvin by adding 273.15: Temperature in Kelvin = 37°C + 273.15 = 310.15 K

Step 2: Find the moles of glucose used

The mass of glucose (C6H12O6) used is 10.0g. First, let's determine the molar mass of glucose: Molar mass of C6H12O6 = (6 × 12.01) + (12 × 1.01) + (6 × 16.00) = 180.18 g/mol Now, let's find the moles of glucose used: Moles of glucose = (10.0 g) / (180.18 g/mol) = 0.05553 mol

Step 3: Find the moles of CO2 produced

The stoichiometric ratio of glucose to CO2 in the reaction is 1:6. Therefore, for every mole of glucose, 6 moles of CO2 are produced. Moles of CO2 = 0.05553 mol × 6 = 0.3332 mol

Step 4: Apply the ideal gas law for CO2

Now, we can apply the ideal gas law to find the volume of CO2 produced: \(PV = nRT\) where P is the pressure, V is the volume, n is the moles of CO2, R is the ideal gas constant, and T is the temperature in Kelvin. We will use the ideal gas constant value R = 8.314 J/(mol·K), as our pressure is given in kPa. Rearrange the equation to solve for volume: V = nRT / P Substitute the pressure, moles of CO2, ideal gas constant, and temperature into the equation: V = (0.3332 mol × 8.314 J/(mol·K) × 310.15 K) / (101.33 kPa) = 8.36 L Now, following the same approach, we will find the volume of O2 needed for part (b).

Step 5: Find the moles of glucose for part (b)

The mass of glucose used in this part is 15.0g. Moles of glucose = (15.0g) / (180.18g/mol) = 0.0833 mol

Step 6: Find the moles of O2 needed

The stoichiometric ratio of glucose to O2 in the reaction is 1:6. Therefore, for every mole of glucose, 6 moles of O2 are required. Moles of O2 = 0.0833 mol × 6 = 0.4998 mol

Step 7: Apply the ideal gas law for O2

We are given the pressure (100 kPa) and temperature (298 K) for part (b). Apply the ideal gas law to find the volume of O2 needed: V = nRT / P Substitute the pressure, moles of O2, ideal gas constant, and temperature into the equation: V = (0.4998 mol × 8.314 J/(mol·K) × 298 K) / (100 kPa) = 12.39 L So, the volume of dry CO2 produced at normal body temperature and pressure when 10.0g of glucose is consumed is 8.36 L, and the volume of oxygen required to completely oxidize 15.0g of glucose at 100 kPa and 298 K is 12.39 L.

Key Concepts

Stoichiometry

Stoichiometry is like the recipe for chemical reactions. It helps us understand the precise amounts of each substance needed or produced. In our exercise, stoichiometry tells us how much student as input glucose is oxidized. You work out from the balanced equation:

• For every mole of glucose \((C_6H_{12}O_6)\), you get 6 moles of carbon dioxide \((CO_2)\).
• Also, for every mole of glucose, you need 6 moles of oxygen \((O_2)\).

This ratio is crucial because it shows how reactants convert into products in fixed proportions.
Whenever you use stoichiometry, start with the balanced equation of the reaction. This balanced equation provides the ratios (coefficients) that make all calculations possible.
Keep in mind, stoichiometry is valid in all states of matter: gases, liquids, and solids. It ensures the law of conservation of mass by balancing the number of atoms of each element on both sides of the reaction. In our example, calculating the moles of glucose and using stoichiometry allow us to figure out the necessary oxygen and produced carbon dioxide.

Ideal Gas Law

The Ideal Gas Law is a formula that connects four key properties of a gas: Pressure (P), Volume (V), Temperature (T), and the amount in moles (n). It is represented as \(PV = nRT\). Here, \(R\) is the ideal gas constant, typically \(8.314 \, J/(mol \cdot K)\).
In our exercise, this law helps calculate the volume of gases involved at specific conditions:

• The law shows that by knowing any three out of the four variables, you can calculate the fourth one.
• We rearrange the formula to find the volume, \(V = \frac{nRT}{P}\), using conditions like pressure and temperature known from the problem.
For the oxidation of glucose in the body at \(37^\circ C\) and atmospheric pressure \(101.33 kPa\), the Ideal Gas Law predicts how much \(CO_2\) is released, showing how gases behave under different conditions.
The law assumes gas molecules don’t interact and occupy no volume, which isn't always the case, but provides a good approximation for most gases under moderate pressures and temperatures.

Glucose Metabolism

Glucose metabolism is a crucial process for our bodies, transforming glucose into energy and releasing by-products like carbon dioxide \((CO_2)\) and water. In a chemical reaction, glucose \((C_6H_{12}O_6)\) reacts with oxygen \((O_2)\) to produce these by-products.
Understanding the pathway of glucose in our metabolism not only reveals how energy is extracted but also how waste products are formed:

• The chemical equation \( C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \) is central in cellular respiration.
• During metabolism, the energy stored in glucose is released in a biologically useful form known as ATP (Adenosine Triphosphate).

This reaction is a combustion reaction at a biochemical level. It shows the oxygen we inhale oxidizing glucose, leading to the production of energy and the elimination of carbon dioxide—the latter expelled from our bodies.
This entire process is vital for maintaining the energy needs of cells but also involves a sophisticated set of biochemical pathways beyond this simple reaction, including glycolysis and the citric acid cycle.