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

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

Thank you! I do get that!

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

Glad to help.

To calculate the ΔHf (standard enthalpy of formation) of C6H12O6 (glucose), you need to use the given data and Hess's Law. Hess's Law states that the overall enthalpy change in a reaction is independent of the pathway taken, so you can use the enthalpies of combustion and formation to calculate the ΔHf.

The balanced chemical equation for the combustion of glucose is:

C6H12O6 (s) + 6O2 (g) -> 6CO2 (g) + 6H2O (l)

Using the enthalpies of formation of CO2 and H2O provided, you can calculate the ΔH of this combustion reaction.

First, calculate the enthalpy change for the combustion of one mole of glucose:

ΔH = 6*ΔHf(CO2) + 6*ΔHf(H2O) - ΔH(combustion)

Substituting the given values:

ΔH = 6*(-393.5 kJ/mol) + 6*(-285.9 kJ/mol) - (-2816 kJ/mol)
= -2361 kJ/mol + (-1715.4 kJ/mol) + 2816 kJ/mol
= -1260.4 kJ/mol

So, the ΔH of the combustion of glucose is -1260.4 kJ/mol.

Since the formation of glucose can be considered the reverse of its combustion, the ΔHf of C6H12O6 can be determined by reversing the sign of the ΔH value:

ΔHf(C6H12O6) = -(-1260.4 kJ/mol) = 1260.4 kJ/mol

Therefore, the ΔHf of C6H12O6 (glucose) is 1260.4 kJ/mol.

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.