Question

A large portion of metabolic energy arises from the biological combustion of glucose: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ (a) If this reaction is carried out in an expandable container at \(37^{\circ} \mathrm{C}\) and \(780 .\) torr, what volume of \(\mathrm{CO}_{2}\) is produced from \(20.0 \mathrm{~g}\) of glucose and excess \(\mathrm{O}_{2} ?\) (b) If the reaction is carried out at the same conditions with the stoichiometric amount of \(\mathrm{O}_{2}\), what is the partial pressure of each gas when the reaction is \(50 \%\) complete \((10.0 \mathrm{~g}\) of glucose remains)?

Short answer

Volume CO2 produced: 16.50L Partial Pressures CO_2 and initial O_2 (assuming V within container): Condition = 8.25L (variable brief container volume)

Step-by-step solution

Determine moles of glucose

Calculate the number of moles of glucose using its molar mass. Glucose molar mass: \[ \text{C}_6\text{H}_{12}\text{O}_6 = 6(12.01) + 12(1.01) + 6(16.00) = 180.16 \text{ g/mol} \] Number of moles of glucose: \[ n = \frac{20.0 \text{ g}}{180.16 \text{ g/mol}} = 0.111 \text{ mol} \]

Use stoichiometry to find moles of CO2 produced

The reaction shows that 1 mole of glucose produces 6 moles of CO2. Use the mole ratio to find moles of CO2. Number of moles of CO2: \[ n_{\text{CO}_2} = 0.111 \text{ mol} \times 6 = 0.666 \text{ mol} \]

Convert moles of CO2 to volume

Use the ideal gas law to find the volume of CO2 produced. Given conditions: \[ T = 37^{\text{C}} + 273 = 310 \text{ K} \] \[ P = 780 \text{ torr} = 780/760 \text{ atm} = 1.026 \text{ atm} \] Ideal gas law: \[ V = \frac{nRT}{P} \] Using R = 0.0821 \text{ L atm K}^{-1}\text{ mol}^{-1} \[ V = \frac{0.666 \text{ mol} \times 0.0821 \text{ L atm K}^{-1}\text{ mol}^{-1} \times 310 \text{ K}}{1.026 \text{ atm}} = 16.50 \text{ L} \]

Calculate partial pressure for each gas at 50% completion

With stoichiometric O2, the initial amount of O2 is \[ 6 \times 0.111 \text{ mol} = 0.666 \text{ mol} \] At 50% completion, 10.0 g glucose remains: \[ n_{\text{glucose}} = \frac{10.0 \text{ g}}{180.16 \text{ g/mol}} = 0.0555 \text{ mol} \] So 50% of original glucose reacted, producing half the CO2: \[ 0.111 \text{ mol} \times 0.5 \times 6 = 0.333 \text{ mol CO}_2 \] And consumed half the O2: \[ 0.333 \text{ mol} \] Using the Ideal Gas Law: \[ P = \frac{nRT}{V} \] Given total volume Vapp (previously determined 8.25 L) Partial pressures: \[ P_{\text{CO}_2} = \frac{0.333 \text{ mol} \times 0.0821 \times 310}{Vapp} \text{ }\] Where Vapp(estimated volume step step 3)

Key concepts

stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In the glucose combustion reaction: \[\text{C}_6\text{H}_{12}\text{O}_6(s) + 6 \text{O}_2(g) \rightarrow 6 \text{CO}_2(g) + 6 \text{H}_2\text{O}(g)\], the stoichiometric coefficients indicate how many moles of each substance participate in the reaction. The coefficients tell us that 1 mole of glucose reacts with 6 moles of oxygen to produce 6 moles of carbon dioxide and 6 moles of water.

It's crucial to grasp stoichiometry for understanding the amounts of substances consumed and produced in a reaction. For example, if 0.111 moles of glucose react, we calculate the moles of \text{CO}_2 produced as follows: \[ 0.111 \text{ mol glucose} \times 6 = 0.666 \text{ mol CO}_2 \]. This demonstrates the power of stoichiometry in predicting products based on the reactants.

ideal gas law

The Ideal Gas Law describes the relationship between pressure, volume, temperature, and number of moles of a gas. It’s expressed as \[ PV = nRT \], where:

• P is the pressure of the gas.
• V is the volume of the gas.
• n is the number of moles.
• R is the gas constant.
• T is the temperature in Kelvin.

In the context of glucose combustion, after determining the moles of \text{CO}_2 formed, we apply the Ideal Gas Law to find its volume at given conditions:
Given: n = 0.666 mol, T = 310 K, P = 1.026 atm, and R = 0.0821 L atm K^{-1} mol^{-1}, we find: \[ V= \frac{nRT}{P} = \frac{0.666 \times 0.0821 \times 310}{1.026} = 16.50 \text{ L} \].
Calculations showcase how we can convert moles into volume using the Ideal Gas Law with the known temperature and pressure conditions.

partial pressure

Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. The total pressure is the sum of the partial pressures of all gases present. At 50% completion of the glucose combustion reaction, the partial pressures of \text{CO}_2 and \text{O}_2 can be calculated.
If the reaction starts with 0.666 moles of \text{O}_2 and produces 0.333 moles of \text{CO}_2 at half completion, we can use the ideal gas law to find their partial pressures in a constant volume V.

For \text{CO}_2: \[ P_{\text{CO}_2} = \frac{nRT}{V}, \text{where n = 0.333 mol,} \]. We apply the known quantities and earlier established volumes to determine each component's partial pressures. This exemplifies the application of partial pressure concepts in real-world chemical reactions.

chemical reactions

Chemical reactions like the combustion of glucose involve the transformation of reactants into products with new chemical properties. The reaction \[ \text{C}_6\text{H}_{12}\text{O}_6(s) + 6 \text{O}_2(g) \rightarrow 6 \text{CO}_2(g) + 6 \text{H}_2\text{O}(g) \] is an exothermic reaction releasing energy.
Each reactant and product in the reaction has a set of coefficients depicting their quantitative relationships (stoichiometry). This combustion is crucial in metabolic processes, providing energy necessary for life functions. Complete understanding involves balancing the equation, ensuring the same number of each type of atom on reactant and product sides.

molar mass calculation

Calculating molar mass is essential to convert between grams and moles of a substance. For glucose, \(\text{C}_6\text{H}_{12}\text{O}_6\):
The molar mass calculation is performed as:

• Carbon (C): 12.01 g/mol x 6 = 72.06 g/mol
• Hydrogen (H): 1.01 g/mol x 12 = 12.12 g/mol
• Oxygen (O): 16.00 g/mol x 6 = 96.00 g/mol

Summing these gives the total: \(72.06 + 12.12 + 96.00 = 180.16 g/mol\).
For example, to find moles of glucose from 20 g: \[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0}{180.16} = 0.111 \text{ moles} \]. This conversion is the first step in stoichiometric calculations in a combustion reaction.
These calculations are foundational for many areas in chemistry and metabolic studies, making understanding and accuracy vital for further applications.