multiply 3rd eqn by 4
... 2 N2 + 6 O2 + 2 H2 —> 4 HNO3 -696kJ
reverse 2nd eqn and multiply by 2
... 4 HNO3 —> 2 N2O5 + 2 H2O +154kJ
reverse 1st eqn and multiply by 2
... 2 H2O —> 2 H2 + O2 +572Kj
add the the three eqn, cancel identical items on both sides
-696 + 154 + 572 = ? ... check your math
This is not going to be formatted we’ll because there is no subscript here.
H2 + 1/2 O2 —> H2O -286kJ
N2O5 + H2O —> 2HNO3 -77kJ
1/2N2 + 3/2 O2 + 1/2 H2 —> HNO3 -174kJ
Calculate the *Delta* H for
2N2 + 5O2 —-> 2N2O5
I attempted this problem and got +20kJ. Was I right?
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