[greenfloss]

calculate the C=O bond energy given the following equations.

C (g ) + O2 (s) --> CO2 (g ) (-393 KJ/mol)

C (s ) --> C (g ) --> (+715 KJ/mol)

1/2O2 (g ) --> O (g ) ( +248 KJ/mol)


Please please please explain in the duh-est way possible. Imagine you're talking to an idiot

Thanks

[Hello12]

I dont think this is Hess law. All you need to do is bonds broken, which is the C-C bond and the O=O bond minus the bond that is formed. Seeing as you are trying to find the bond that is formed, you simply rearrange the Bonds broken-bonds formed= delta H. To Bonds broken + Delta H = Bonds formed. so it should look something like this, +715+248+-393= C=O bond energy.

[Blitzkrieg]

I'm not sure about how to find the bond energy, but I can help with the Hess's Law part.

Okay, so Hess's Law is all about canceling things out and rearranging equations that make another equation. So let's look at the C(g) part of that equation first. The only other equation out of the three that uses C(g) is the second one: C (s ) --> C (g ) --> (+715 KJ/mol)

You have to rearrange that equation and thus manipulate the energy next to it (+715 KJ/mol) so that the C(g) in C (g) + O₂ (s) --> CO₂ (g ) will cancel out with the C(g) in C (s)--> C (g). In order for anything to cancel out, they must have the same amount of moles, be in the same state of matter, and one must be on the products side and the other on the reactants side. So comparing them:

C (g ) + O₂ (s) --> CO₂ (g ) (-393 KJ/mol)
C (s ) --> C (g ) --> (+715 KJ/mol)

Both C(g)'s have the same amount of moles, they are in the same state of matter, and one is on the products side and another is on the reactants side. You can leave this alone (and if you have this written on a piece of paper, I would cross out the C(g)'s in both equations).

Now for the O₂(s) part. First, I think this is supposed to be O₂(g)? (I'm going to continue pretending that it is. )

The only other equation where there's an O₂(g) is 1/2O₂ (g) --> O (g). So:

C (g ) + O₂ (g) --> CO₂ (g) (-393 KJ/mol)
1/2O₂ (g) --> O (g) ( +248 KJ/mol)

Since both O₂'s are on the products side, you must switch one of them, because in order for them to cancel out one must be on the products side and another on the reactants side. We don't want to flip the first equation, because we've already dealt with the C(g), and if we switch them then the C(g) would be on the products side (and we would have to revise our work earlier). So we can flip the second equation:

O (g) --> 1/2O₂ (g)  ( -248 KJ/mol)

Notice how I switched the delta H to -248 KJ/mol. When you flip an equation like that, you negate the delta H. (Remember, we're manipulating these equations all so we can manipulate the delta H's, and find the energy in a C=O bond.)

To cancel out, they must also have the same amount of moles. Since it's 1/2O₂ (g), we can double it to make it O₂(g).

2O (g) --> O₂ (g)  ( -496 KJ/mol)
The delta H is now -496 KJ/mol, because I doubled everything.

C (g ) + O₂ (g) --> CO₂ (g) (-393 KJ/mol)
2O (g) --> O₂ (g)  ( -496 KJ/mol)

They now have the same amount of moles, are in the same state of matter, and are on opposite sides of the equation. We can cross out the O₂ (g).

This leaves us with:

C (g ) + O₂ (g) --> CO₂ (g ) (-393 KJ/mol)
C (s ) --> C (g ) --> (+715 KJ/mol)
2O (g) --> O₂ (g)  ( -496 KJ/mol)

So to find bond energy, it's simply the [bonds breaking] - [bonds forming] = delta H. So the bond breaking of O2(g) and C(g) (?) minus the bond forming of two C=O bonds should equal -393 kJ/mol.... unfortunately I don't know how to use equations for the heats of formation of C(g) and O2(g) to find that out.

I think these pages might be helpful, they're all about bond energy:
http://chemistry.about.com/od/workedchemistryproblems/a/bondenergyexmpl.htm
http://www.saskschools.ca/curr_content/chem30/modules/module3/lesson5/hessbonde.htm