4 FeO(s) + O2(g) �¨ 2 Fe2O3(s)

calculate value of delta H
i have done this
4(-272.0)+0->2(-825.5)
and the products minus reactants and i got -563 but it is not right?!

-556.7

Well, it seems like you've gotten yourself into a bit of a chemical comedy! Let me add some laughter to the equation and help you out.

First, let's calculate the change in enthalpy by multiplying the stoichiometric coefficients by their respective enthalpy values.

For reactant FeO(s), the enthalpy change is (-272.0) kJ/mol. Since we have 4 moles of FeO, the total enthalpy change for FeO is:

4 × (-272.0) kJ/mol = -1088.0 kJ

For product Fe2O3(s), the enthalpy change is (-825.5) kJ/mol. Since we have 2 moles of Fe2O3, the total enthalpy change for Fe2O3 is:

2 × (-825.5) kJ/mol = -1651.0 kJ

Now, let's sum up the enthalpy changes:

ΔH = (-1088.0 kJ) + (0 kJ) - (-1651.0 kJ)
= -1088.0 kJ + 1651.0 kJ
= 563.0 kJ

Well, well, well! Looks like you were just one step away from getting the right answer! The change in enthalpy (ΔH) for the given reaction is indeed 563.0 kJ. Keep up the good work, chemistry comedian!

To calculate the value of delta H, or the heat of reaction, you need to use the enthalpy values (in kilojoules/kmole) of the individual substances involved in the reaction.

In your reaction, you have 4 moles of FeO reacting with 1 mole of O2 to produce 2 moles of Fe2O3. You correctly identified the balanced equation:

4 FeO(s) + O2(g) → 2 Fe2O3(s)

Now, to calculate the value of delta H, you'll need to subtract the sum of the enthalpy of the reactants from the sum of the enthalpy of the products.

The enthalpy of FeO is -272.0 kJ/kmole, and the enthalpy of Fe2O3 is -825.5 kJ/kmole.

For the reactants, you have 4 moles of FeO, so the enthalpy of the reactants would be:

4 moles × (-272.0 kJ/kmole) = -1088.0 kJ

For the products, you have 2 moles of Fe2O3, so the enthalpy of the products would be:

2 moles × (-825.5 kJ/kmole) = -1651.0 kJ

Now, subtract the sum of the enthalpy of the reactants from the sum of the enthalpy of the products:

Delta H = Sum of products - Sum of reactants = -1651.0 kJ - (-1088.0 kJ) = -1651.0 kJ + 1088.0 kJ = -563.0 kJ

Therefore, the calculated value of delta H for this reaction should be -563.0 kJ, which matches your calculation. If you obtained a different result, please check your enthalpy values and make sure you are using the correct signs (+/-) for the reactants and products.

The answer is -560.8

My table says Fe2O3(s) = -824.2 kJ/mol which gives -560.4 kJ for the reaction.