Question

ExampleThe production of 5 tons of beer requires the formation of 250 kg of ethanol MW 46, how much heat need to be removed from the fermentation chamber in order to keep the temperature constant at 37 C?
C6H12O6 → 2C2H5OH + 2CO2
enthalpies of formation: glucose(-1274kJ/mol), ethanol (-597 kJ/mol) and CO2(-394 kj/mol) .
heat capacites (J/mol/K) : are 218 (glucose),111(ethanol) and 37 (carbondioxide)

Answer 1

C6H12O6 → 2C2H5OH + 2CO2

HRxn = Hproducts - Hreactants

HRxn= (2*-597 + 2*-394) - (1*1274)

HRxn = -3256 kJ/2 mol ofethanol

HRxn = 3256/2 = -1628 kJ per mol of ethanol

at 37°C

Hheat = 2*(111) + (2*37) - (218 ) = 78*(37-25) = 936 J / 2

we can ignore amount of sensible heat, mostly it is heat of reaction

now..

we need

m = 250 kg = 250/46 = 5.4347 kmol of ethanol

heat required:

-1628 kJ / mol of ethanol * 5.4347 kmol of ethanol

-1628 kJ / mol of ethanol * 5.4347*10^3 mol of ethanol

8847691.6 kJ must be removed