Chemistry Where am I wrong in calculating the heat of reaction for CH4 + 2O2?

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SUMMARY

The heat of reaction for the combustion of methane (CH4 + 2O2 -> CO2 + 2H2O) is calculated using the formula ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants). The heat of formation values are -393.5 kJ/mol for CO2 and -285.8 kJ/mol for H2O. To accurately compute the reaction's enthalpy change, the heat of formation for CH4 must also be included. The correct calculation is ΔHrxn = ΔHf(CO2) + 2ΔHf(H2O) - ΔHf(CH4), with ΔHf(O2) being zero.

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Homework Statement
given:
C + 2H2 -> CH4 What is the heat of reaction
Relevant Equations
C + O2 -> CO2 delta H = -393.5 kJ/mole
H2 + 1/2 O2 -> H2O delta H = -285.8 kJ/mole

CH4 + 2Os -> CO2 + 2H2O delta H = ?????

the book says delta H = -890.8 kL/mole
Since the heat of formtion of H2O is -285.8 and the heat of formation of CO2 is -393.5 I thought that the heat of reaction CH4 + 2O2
should be -393.5 + 2(-285.80 = -965.1

Where am I wrong?
 
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This is way too messy to follow, can you please relist given equations and the target one?
 
Lets put it this way...
What is the heat of reaction for this equaiton.

CH4 + 2O2 -> CO2 + 2H2O delta H = ?

The heat of formation of CO2 is -393.5
The heat of formation of H2O is -285.8
 
Not enough information, you need data related to CH4 to answer.
 
The heat of formation is the enthalpy change when you create a mole of substance from it's elements (atoms), or vice versa.
So first you need to compute the enthalpy change from CH4 and O2 to C,H and O. Then you need the enthalpy change from C,H,O to CO2 and H2O.
 
ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants)
So ΔHrxn = ΔHf(CO2) + 2ΔHf(H2O) - ΔHf(CH4)
(ΔHf(O2) is of course zero.)
(Oh, and by the way, heat of formation is from elements in their standard states - not from atoms when this is not the standard state. So from C(s), H2 and O2, not C(g), H and O.)
 
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I now understand, Thanks all.
 

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