Help answering chemical engineering thermodynamics question

Help answering chemical engineering thermodynamics question. Equation 3.56 mentioned in part A: C6H1206(aq) + 602(g) rightarrow 6C02(g) + 6H20(L) This problem considers oxidation of glucose as a model carbohydrate. In the human body glucose is typically metabolized in aqueous solution Ignoring the enthalpy change due to dissolving glucose in the aqueous solution, calculate the standard heat of reaction for Eqn. 3.56. Using the standard lie at of reaction calculated in part together with the steady-state energy expenditure for sitting at rest, determine the rate that glucose is consumed (g/day). You may ignore sensible heats because the respiration streams are very close to 298.15 K. Compare the energy consumption in part (b) with a light bulb by putting the energy consumption in terms of W. which is usually used to characterize light bulbs. How many 70 kg humans collectively produce energy equivalent to a 1500 W hair dryer? On a mass basis, all carbohydrates have about the same heat of reaction. A soft drink has approximately 73 g carbohydrates (sugar) per 20 fluid oz. bottle and the manufacturer\'s label reports tilt the contents provide 275 kcal. (A dietary calorie is a thermodynamic kcal, so the conversion lies already been made, aim the bottle is labelled 275 calories.) Convert the answer from part (a) to kcal/g and compare with the manufacturer\'s ratio. What is the annual output of CO_2 (kg) for a 70 kg human using the energy expenditure for sitting as an average value?

Solution

1. C₆H₁₂O₆ + 6O₂ ------------> 6CO₂ + 6H₂O

Enthalpy change (dH)= dHf products - dHf reactants

dH= [dHf 6CO₂+ 6H₂O] - [dHf C₆H₁₂O₆ + 6 O₂)

dHf CO₂ =-393.5

dHf H₂O= -285.5

dHfC₆H₁₂O₆₌ -1273.3

dHf O_2=Zero

Putting all values in the above equation,

dH= [ 6 (-393.5) + 6 (-285.5) ] - [ (-1273.3) + 6 (0)]

= [-2361+ (-1713)] - [-1273.3]

= -4074 + 1273.3

= -2800 KJ