Question

(15 points). The oxidation of glucose provides the principal energy source for animal cells. The reactants are glucose [C6H12O6(s)] and oxygen [O2(g)]. The products are carbon dioxide [CO2(g)] and water [H2O(l)]. a. (5 points). Write a balanced chemical reaction for glucose oxidation, and determine the standard heat of reaction at 298 K. Use the data tables in Smith and Van Ness Appendix C. b. (5 points). During a day, an average person consumes about 150 kJ energy per kg of body mass. Assuming glucose is the sole energy source, estimate the mass (grams) of glucose required daily to sustain a person of 57 kg. Ignore the effect of the effect of temperature on the heat of reaction. c. (5 points). For the U.S. population of 325 million persons, what mass of CO2 (a greenhouse gas) is produced daily by respiration? Ignore the effect of temperature on the heat of reaction.

Answer 1

Check the explanation

Step-by-step explanation:

The balanced reaction

C6H12O6(s) + 6O2(g) = 6CO2(g) + 6H2O(l)

Standard heat of reaction

Hrxn = 6*Hf(CO2) + 6*Hf(H2O) - 6*Hf(O2) - Hf(C6H12O6)

= 6*(-393.5) + 6*(-285.8) - 6*(0) - (-1274.4)

= - 2801.4 kJ/mol

Part b

Energy consumed by a person = 150 kJ/kg x 57 kg = 8550 kJ

Moles of glucose required = 8550 kJ / (2801.4 kJ/mol)

= 3.052 mol

Mass of glucose required = moles x molecular weight

= 3.052 mol x 180.156 g/mol

= 549.84 g

Part c

1 person requires = 3.052 mol

275 million person require = 275*10^6*3.052 = 8.39 x 10^8 mol

From the stoichiometry of the reaction

1 mol glucose produces = 6 mol CO2

8.39 x 10^8 mol glucose produces = 6*8.39*10^8

= 5.036 x 10^9 mol CO2

Mass of CO2 produced = moles x molecular weight

= 5.036 x 10^9 mol x 44 g/mol

= 2.22 x 10^11 g x 1kg/1000g

= 2.22 x 10^8 kg x 1million/10^6

= 222 million kg