Question

Calculate the DHrxn of
C6H12O6 (s) + O2 (g) --> CO2 (g) + H2O (l)
with DHf.
C6H12O6(s) -1260.04
O2(g) 0
CO2(g) -393.5
H2O(l) -285.840

I balanced the equation and my work:
DHrxn = [14mol(-393.5+(-285.840))] - [7mol(-1260.04+0)] = -690.48 KJ/mol

Is this right? If not, what did I do wrong?

Answer 1

You don't show the balanced equation but I think that is the problem. My balanced equation is 6CO2 and not 14
C6H12O6 + 6O2 ==> 6CO2 + 6H2O

Answer 2

668.2 KJ/mol? Endothermic