Chemistry - Thermochemistry Question with Solution | TestHub
An athlete takes 20 breaths per minute at room temperature. The air inhaled in each breath is 164.2 ml which contains 20% oxygen by volume, while exhaled air contains 10% oxygen by volume. Assuming that all the oxygen consumed is used for converting glucose into carbon dioxide and water, how much heat is produced (in kJ) in the body in one hour? Body temperature is 310 K and enthalpy of combustion of glucose is -2820 kJ/mol at 310 K.
Answer:
Solution:
O₂ consumed per breath = (20% - 10%) × 164.2 = 16.42 ml at 298 K (room temperature). For glucose, the Respiratory Quotient (RQ) is 1, so the total gas moles remain unchanged.
Per hour, the O₂ consumed is 16.42 × 1200 = 19704 ml = 19.704 l.
Thus, the moles of O₂ (nO₂) can be calculated using the ideal gas law:
nO₂ = 19.704/(0.0821 × 298) = 0.8054 mol
The balanced chemical equation for the combustion of glucose is:
C₆H₁₂O₆ + 6O₂ tarrow 6CO₂ + 6H₂O
From the stoichiometry, 1 mole of glucose reacts with 6 moles of O₂.
Therefore, the moles of glucose (nglucose) consumed are:
nglucose = 0.8054/6 = 0.1342 mol
Given that the heat of combustion for glucose is 2820 kJ/mol, the total heat produced is:
Heat = 0.1342 × 2820 = 378.5 ≈ 379 kJ
The answer is 0379.
