RXN #1: H2 (g) + O2 (g) --> H2O2 (l) ; delta-H =?

Question

Calculate the enthalpy change (delta-H1) for the reaction above (RXN #1) using the information below and Hess' Law.

RXN A: H2O2 (l) --> H2O (l) + 1/2 O2 (g); delta-H=-98.0 kJ/mol

RXN B: 2 H2 (g) + O2 --> 2 H2O (l); delta-H= -571.8 kJ/mol

Answer 1

Reverse eqn A and add to 1/2 eqn B. Reversing an eqn changes sign of H. Taking 1/2 eqn changes H by 1/2.

Answer 2

I don't see how that works

Answer 3

1/2 eqn b.
2H2 + O2 ==> 2H2O becomes
H2 + 1/2 O2 ==> H2O

reverse eqn A.
H2O2 ==> H2O + 1/2 O2 becomes
H2O + 1/2 O2 ==> H2O2

.....H2 + 1/2 O2 ==>H20
+....H2O + 1/2O2 ==> H2O2
------------------------------
H2 + 1/2 O2 + H2O + 1/2 O2 >H2O + H2O2
1/2 O2 + 1/2 O2 = O2
H2O on left cancels with H2O on the right and you are left with
H2 + O2 ==> H2O2.
Now MY question. IF you actually tried this, and I suspect you didn't, what did you not understand? You've got to get your hands a little dirty if you want to work through these things.

Answer 4

Thanks You're the best!