Question

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?

Answer 1

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