(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

From the enthalpies of reaction for equations A) and B), calculate the heat of reaction for equation C).

A) 2H2 (g) + O2 (g) ----> 2H2O2 (g) DH= -483.6 kJ

B) 3O2 (g) ----> 2O3 (g) DH= +284.6 kJ

c) 3 H2 (g) + O3 (g) -----> 3H2O (g) DH= ??? kJ

2. Relevant equations

NONe

3. The attempt at a solution

My attempt is as follows:

I multiplied equation A) by 3 first including the DH as well and got and flipped equation be so the 2O3 can cancel out.

A) 6H2 (g) + 3O2 (g) ----> 6H2O (g) DH= -1450.8 kJ

B) 2O3 (g) ------> 3O2 DH= -284.6 kJ

adding A) + B) I got

6H2 (g) + 2O3 (g) -------> 6H2O (g) DH= -1699.4 kJ

since i need to get to c) 3 H2 (g) + O3 (g) -----> 3H2O (g) all I did is divide everything by 2 including the DH = -1699.4 kJ

the DH came out to for equation C) as -849.7 kJ and this is not correct because the enthalpy for H2O (g) is -241.82 and since there are 3 H2O it should be -725.46 which I can't seem to get.

Can someone please help where I have made a mistake? Thanks

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# Homework Help: Adding and subtracting thermalchemcial equations help needed

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