Consider the combustion of a mole of fructose, given by the reaction C6H12O6 + 6
ID: 575181 • Letter: C
Question
Consider the combustion of a mole of fructose, given by the reaction
C6H12O6 + 6O2 6CO2 + 6H2O.
This is the net reaction that generates the energy one would get from drinking a sugary drink.
(a) Calculate the enthalpy of combustion H.
(b)The atomic weight of fructose is about 180 u. A typical can of soda contains about 40 grams of sugar, which we’ll assume is fructose since most things use high fructose corn syrup. Use this information and the result of the previous part to calculate the amount of energy released in the combustion of the sugar in one soda can (i.e., 40 grams of fructose). Compare this to the energy needed to lift 60 kilograms a distance of 10 meters (representing, perhaps, a set on the bench press at the gym...).
Explanation / Answer
Ans. #a. Balanced reaction: C6H12O6(s) + 6O2(s) ----> 6CO2(g) + 6H2O(g)
Using Hess’s law, the standard free energy change for oxidation of fructose is –
dG0 = (6 x dGf0 of CO2) + (6 x dGf0 of H2O)
= 6 x (-394.39 kJ/mol) + 6 x (- 237.14 kJ/mol) = -3789.18 kJ/mol
Note: The negative sign indicates the release of energy during oxidation of fructose.
Therefore, dH for oxidation of fructose = -3789.18 kJ/mol
#b. Moles of fructose in 40.0 g sample = Mas / Molar mass
= 40.0 g / (180.0 g/ mol)
= 0.2222 mol
Energy released from combustion of 40 g fructose = Moles x dH,molar of combustion
= 0.2222 mol x (-3789.18 kJ/mol)
= -841.96 kJ
# The amount of energy spent to move a body through a vertical distance is given by-
Potential energy, E = mgh ,
where, m= mass of the body,
g= gravitational force,
h = height (vertical displacement)
Or, E = 60.0 kg x 9.8 ms-2 x 10.0 m [g = 9.8 ms-2]
Or, E= 5880 kg m2 s-2 ; [1 kg m2 s-2 = 1 J]
Hence, E = 5880.0 J
Therefore, energy required to left the mass = 5880.0 J = 5.88 kJ