What Is the Approximate Heat of Reaction for 2H2 + O2→2H2O?

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SUMMARY

The approximate heat of reaction for the equation 2H2 + O2 → 2H2O is calculated using bond energies, resulting in a value of -488 KJ/mol. The bond energies provided are 436 KJ/mol for H-H, 464 KJ/mol for H-O, and 496 KJ/mol for O=O. The initial calculation of -440 KJ/mol was incorrect due to an oversight in counting the hydrogen-oxygen bonds in the water molecules produced. The correct calculation involves breaking down the bonds and applying the formula Hproducts - Hreactants accurately.

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Homework Statement


Given that the bond energy of hydrogen-hydrogen bonds is 436 KJ/mol, that of hydrogen-oxygen bonds is 464 KJ/mol, and those in oxygen molecules 496 KJ/mol, what is the approximate heat of reaction for 2H2 + O2→2H2O

Homework Equations


Hproducts-Hreactants

The Attempt at a Solution


So, I understand that heats of reaction involving hydrogen gas and oxygen gas to liquid water is exothermic, so my answer must be negative. My book gives and answer of -488 KJ/mol but I keep ending up with -440 KJ/mol:

2(464) - [2(436) + 496] = -440 KJ/mol.

Calculation error? I went over the numbers twice. Thanks!
 
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How many molecules of water does the reaction yield, and how many hydrogen-oxygen bonds are in each molecule? I hope this helps.
 
Actually, I looked up a video to see where my reasoning went astray. With the original equation, we have:

2H2+ O2→ 2H2O

Bonds broken:
2H-H = 436 KJ/mol
02 = 496 KJ/mol

Bonds formed:
2H2O

Using Hp-Hr, we get:
4(464)-[2(436) + 496)] = 488 KJ/mol but must be negative because g-->l is exothermic.

Since this is the right answer, I feel better now, but I prefer the experts here to make sure my work is valid. Thanks in advance!
 
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Likes   Reactions: Calaver
Your answer is correct, and your original answer had the right concept, it just looked like to me that you forgot to count one of the water molecules produced (or one of the hydrogen-oxygen bonds within each molecule).
 
You are right. I had to draw out the bonds in the products to see what I did wrong. Thanks again!
 

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