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

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  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.

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  2. I don't see how that works

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  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.

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  4. Thanks You're the best!

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