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?
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
668.2 KJ/mol? Endothermic
Your approach is almost correct, but there is a small mistake in the stoichiometric coefficients of your equation. Let's go through the correct calculation step by step:
First, let's balance the equation:
C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)
Now, let's calculate the ΔHrxn using the given ΔHf values:
ΔHrxn = (6 mol × ΔHf[CO2]) + (6 mol × ΔHf[H2O]) - (1 mol × ΔHf[C6H12O6]) - (6 mol × ΔHf[O2])
Plug in the values:
ΔHrxn = (6 mol × (-393.5 kJ/mol)) + (6 mol × (-285.840 kJ/mol)) - (1 mol × (-1260.04 kJ/mol)) - (6 mol × 0 kJ/mol)
Simplify the equation:
ΔHrxn = (-2361 kJ) + (-1715.04 kJ) + (1260.04 kJ) + 0 kJ
Add the terms:
ΔHrxn = -2816 kJ
Therefore, the correct value for ΔHrxn is -2816 kJ/mol, not -690.48 kJ/mol.