why O2 ignored in hess's law

[boomboom2dolla]

Here is the problem and the solution. How come O2 is not accounted for when adding it up in the solution? Do we assume O2 is already included in the problem?

H​

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2O(g) → H2O(l) ΔH (kJ/mole) = –44
C(s) + O2(g) → CO2(g) –394
H2(g) + 1/2O2(g) → H2O(l) –286
C2H5OH(l)+ 3O2(g) → 2CO2(g) + 3H2O(l) –1367
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4. Calculate the enthalpy change for the reaction:
2C(s) + 2H2(g) + H2O(l) → C2H5OH(l)
A. –226 kJ/mole
B. +7 kJ/mole
C. +109 kJ/mole
D. +344 kJ/mole​

E. +687 kJ/mole

2 C(​

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s) + 2 O2(g) → 2 CO2(g) 2(–394)
H2(g) + O2(g) → 2 H2O(l) 2(–286)
2 CO2(g) + 3 H2O(l) → C2H5OH(l)+ 3 O2(g) –(–1367)
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2 C(s) + 2 H2(g) + H2O(l) → C2H5OH(l) ΔH= 2(–394) + 2(–286) – (–1367)​

Adding the values to arrive at the net ΔH, the result is +7 kJ, choice B.

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[sugarsting]

there are 3 O2's on each side of the arrow so they cancel out

 

[joonkimdds]

you are probably asking why O2 is ignored in hess's law when you sum up the net delta H and it's becuz heat of formation of O2 or anything in its standard form is ZERO.


by the way, can I ask where u got this problem from? it's really important to me.

 

[Contach]

you are probably asking why O2 is ignored in hess's law when you sum up the net delta H and it's becuz heat of formation of O2 or anything in its standard form is ZERO.


by the way, can I ask where u got this problem from? it's really important to me.

to elaborate... All halogens are diatomic in their standard form.. like Cl2, O2, Br2 etc.
Carbon's standard form is C(s) like graphite.