Question

Potentially, how much 25 ̊C water could you boil to gas using the energy released from the aerobic fermentation of 1 kg of glucose? Assume a Cp for water of 75 J/(mol K) and a heat of vaporization of water of 41 kJ / mol.

Answer 1

Solution :-

The balanced reaction equation is as follows

C6H12O6 + 6O2 ---- > 6CO2(g) + 6H2O(l)

Delta H rxn = sum of delta Hf product - sum of delta Hf reactant

                = [(CO2*6)+(H2O*6)] - [C6H12O6*1]

                =[(-393.5*6)+(-285.8*6)] -[-1272.3*1]

               = -2803 kJ

So 1 mole glucose gives -2803 kJ heat

now lets calculate the amount of energy give by the 1 kg glucose

(1 kg * 1000 g / 1 kg)*(-2803 kJ / 180.15 g) = -15560 kJ

-15560 kJ * 1000 J / 1 kJ = -15560000 J

now lets calculate the moles of water that can be vaporized from 25 C to gas using this amount of energy

41 kJ * 1000 J / 1 kJ= 41000 J

q = (n* Cp*delta T) + (delta H vap * n)

n= moles

15560000 J = (n * 75 J per mol C * 75 C) + (41000 J per mol * n)  

15560000 J = (n*5625)+(41000*n)

15560000 J = 46625 * n

15560000 J / 46625 = n

333.7 = n

so the moles of water that can be vaporized are 333.7 mol

now lets convert moles to mass

mass of water = moles *molar mass

                     = 333.7 mol * 18.0148 g per mol

                    = 6012 g water

So the amount of water that can vaporized from 25 C to gas is 6012 g water