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:
C6H12O6 (s) + 6O2 (g) --> 6CO2 (g) + 6H2O (l)
and my work:
DHrxn = [12mol(-393.5+(-285.840))] - [7mol(-1260.04+0)] = 668.2 KJ/mol
It is endothermic
Is this right?
Go back to the C57 compound and see how I did that. You are using TOTAL mols and that isn't the right way to do it.
-2816 Exothermic?
or -2551.916?
To calculate the ΔHrxn (the enthalpy change of a reaction) using the enthalpy of formation (∆Hf), you need to use the following equation:
ΔHrxn = ΣnΔHf(products) - ΣnΔHf(reactants)
First, let's ensure that the balanced equation is correct:
C6H12O6 (s) + 6O2 (g) --> 6CO2 (g) + 6H2O (l)
Now, substitute the ∆Hf values into the equation:
ΔHrxn = (6 mol × -393.5 kJ/mol) + (6 mol × -285.840 kJ/mol) - (1 mol × -1260.04 kJ/mol)
ΔHrxn = (-2361 kJ) + (-1715.040 kJ) + (1260.04 kJ)
ΔHrxn = -1815 kJ
So, the ΔHrxn for the given reaction is -1815 kJ. Since the value is negative, it indicates that the reaction is exothermic, not endothermic.
Please note that I used -2361 kJ as -393.5 kJ/mol × 6 mol, and -1715.040 kJ as -285.840 kJ/mol × 6 mol.