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

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.

I don't see how that works

Thanks You're the best!

To calculate the enthalpy change (delta-H1) for RXN #1 using Hess' Law, we need to apply the Law of Conservation of Energy.

Hess' Law states that if a reaction can be expressed as the sum of multiple reactions, then the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the individual reactions.

First, let's align the reactions so that the reactants and products match:

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

Now, we need to manipulate RXN A so that the number of moles of the desired species matches RXN #1. Multiply RXN A by 2 to get:

2 H2O2 (l) --> 2 H2O (l) + O2 (g) ; delta-H = (-98.0 kJ/mol) * 2 = -196.0 kJ/mol

Next, we need to manipulate RXN B so that the number of moles of the desired species matches RXN #1. Divide RXN B by 2 to get:

H2 (g) + 1/2 O2 --> H2O (l) ; delta-H = (-571.8 kJ/mol) / 2 = -285.9 kJ/mol

Lastly, rearrange the manipulated reactions so that their reactants and products match RXN #1:

RXN C: 2 H2O2 (l) --> 2 H2O (l) + O2 (g) ; delta-H = -196.0 kJ/mol

RXN D: 2 H2 (g) + O2 (g) --> 2 H2O (l) ; delta-H = -285.9 kJ/mol

Now, we can sum up RXN C and RXN D to obtain RXN #1:

RXN C + RXN D: 2 H2O2 (l) + 2 H2 (g) + O2 (g) --> 2 H2O (l) + 2 H2O (l) + O2 (g)

Since H2O and O2 appear on both sides of the equation, they cancel out:

2 H2O2 (l) + 2 H2 (g) --> 2 H2O (l) + 2 H2O (l) + O2 (g)

Simplifying further, we get:

2 H2O2 (l) + 2 H2 (g) --> 4 H2O (l)

The sum of the delta-H values for RXN C and RXN D gives us the delta-H1 for RXN #1:

delta-H1 = -196.0 kJ/mol + (-285.9 kJ/mol) = -481.9 kJ/mol

Therefore, the enthalpy change (delta-H) for RXN #1 is -481.9 kJ/mol.

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.