Saw this one already answered, but did not understand at all...

Question

Calculate the delta Hf of C6H12O6 (s) from the following data:

Delta H combustion= -2816 kj/mol
delta Hf of Co2 = -393.5 kj/mol
delta Hf= of H2 O= -285.9kj/mol

Answer 1

C6H12O6 + 6O2 ==> 6CO2 + 6H2O
dHrxn = (n*dHf products) - (n*dHf reactants)
-2816 = [(6*-393.5)+(6*-285.9)] - (dHfglucose)= -2816
Solve for dHf glucose

If you still don't understand, explain what you don't understand about it instead of just a blanket I don't get it.

Answer 2

Thank you! I do get that!

I was having trouble understanding this old post: /display.cgi?id=1330914025

Answer 3

Glad to help.