- #1
Intrusionv2
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Hello, I have a question regarding enthalpy.
I know that when calculating ΔS(o), you just do all of the products minus all of the reactants.
However, when calculating ΔHf(o), I am confused as to what you include in the products and reactants? My book is sometimes disregarding some of the products/reactants, example below.
N2(g) + 2O2(g) --> 2NO2(g)
ΔH(o) = 2ΔHf(NO2)
ΔH(o) = 2mol*33.2kj/mol = 66.4 kJ
Why wouldn't it be ΔH(o) = 2ΔHf(NO2) - [ΔHf(N2) + 2ΔHf(O2)] ?
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Another example:
2KClO3 (s) --> 2KCl(s) + 3O2(g)
ΔH = 2ΔHf(KCl) - 2ΔHf(KClO3)
Why not ΔH = [2ΔHf(KCl) + 3ΔHf(O2)] - 2ΔHf(KClO3) ?
For ΔS everything IS included, so that's why I am confused.
Thanks!
I know that when calculating ΔS(o), you just do all of the products minus all of the reactants.
However, when calculating ΔHf(o), I am confused as to what you include in the products and reactants? My book is sometimes disregarding some of the products/reactants, example below.
N2(g) + 2O2(g) --> 2NO2(g)
ΔH(o) = 2ΔHf(NO2)
ΔH(o) = 2mol*33.2kj/mol = 66.4 kJ
Why wouldn't it be ΔH(o) = 2ΔHf(NO2) - [ΔHf(N2) + 2ΔHf(O2)] ?
------------
Another example:
2KClO3 (s) --> 2KCl(s) + 3O2(g)
ΔH = 2ΔHf(KCl) - 2ΔHf(KClO3)
Why not ΔH = [2ΔHf(KCl) + 3ΔHf(O2)] - 2ΔHf(KClO3) ?
For ΔS everything IS included, so that's why I am confused.
Thanks!