[Jabus]

Hi there,

Just a few questions that have got me stumped and hopefully someone can either answer or lead me on the right direction so I can figure them out.

- Identify two attractive and repulsive forces that influences the formation of chemical bonds between two atoms. - I wrote down that ionic and covalent forces were two attractive forces, but I'm not sure what would repel atoms from each other. I wasn't aware that atoms even repeled each other I simply thought they did not attract each other.

- The next question is "how would you designate the hybrid orbitals formed by 'mixing' one d, one s, and two p orbitals? - I have this written down http://www.jabussucks.com/hybrid.jpg but to be honest that doesn't look right nor do I believe it to be right. I don't think the text explained hybridization that well so if anyone knows how to explain it easily or if you have links to some useful sites that would be awesome.

- Finally I took a look at Hess' Law in my text and on wikipedia but I'm still not understanding when to do what. When do reactions have t be reversed, can I reverse them at any time? Essentially I'm suppose to calculate the heat of combustion for C2H6 given this:

equation #1 C2H4  + 3O2 --> 2CO2 + 2H2O  heat of formation = -1409.5 kJ
equation #2 C2H4 + H2 --> C2H6  heat of formation = -136.7 kJ
equation #3 H2 + 1/2O2 --> H2O  heat of formation = -285.5 kJ

But I'm not quite sure where to start. I decided to get the combustion equation which was C2H6 + 7/2O2 --> 2CO2 + 3H2O but I'm not sure really where to go next. If someone can help out with any of these and put me on the right track I'd like to learn where I went wrong. Thanks!

[jennielynn_1980]

I am not sure if I totally understand the first question but my guess would be that you are to look at attractive and repulsive forces within the actual atom.

For your second question, it is probably alot easier than you think.  If you mix one s and one p you get an sp orbital so just apply that to your question.

For your third question, you really need to focus on how you can get to the combustion equation from the three equations given.  Think about about ADDITIVITY of reaction heats.  Also have a look at this
http://www.ilc.org/cfmx/AAT/BrainBank/QuestionFinder/aat_problem.cfm?topic_id=31

I hope that helps.

[Yggdrasil]

- Identify two attractive and repulsive forces that influences the formation of chemical bonds between two atoms. - I wrote down that ionic and covalent forces were two attractive forces, but I'm not sure what would repel atoms from each other. I wasn't aware that atoms even repeled each other I simply thought they did not attract each other.

Well consider what would happen if there were no repulsive forces.  If only attractive forces existed, instead of bonding, atoms would just collide with each other and fuse together.  However, there are repulsive forces, and these repulsive forces keep atoms from colliding.  In fact, the equilibrium bond distance is the interatomic distance where the attractive and repulsive forces balance.  The following Wikipedia article, while a little advanced, may give you some insight into the repulsive and attractive forces in chemical bonds:

http://en.wikipedia.org/wiki/Lennard-Jones_potential

If you still have questions afterward, we'd be glad to clarify any confusions you have.

[Jabus]

Ok so I'll start with how far I've gotten with the messy one (number 3). btw, thanks jennie for the help I had actually gone to the ILC site looking for help on this and managed not to find it until you posted the direct link to it, yay.

So for heat of combustion of C2H6 with info:
equation #1 C2H4  + 3O2 --> 2CO2 + 2H2O  heat of formation = -1409.5 kJ
equation #2 C2H4 + H2 --> C2H6  heat of formation = -136.7 kJ
equation #3 H2 + 1/2O2 --> H2O  heat of formation = -285.5 kJ

The combustion reaction for C2H6 is C2H6 + 7/2O2 -> 2CO2 +3H2O

From what I understand I can manipulate the equation as long as I don't change any of the base values so in that case I did this:

C2H4 + 3O2 + H2 + 1/2 O2 -> 2CO2 + 2H2O + H2O
(-1409.5 kJ) + (-285.5kJ)  -> (-1409.5kJ) + (-285.5kJ)

Which gets me -1695kJ on either side (does it matter?)

That is what I've come up with in regards to that, if I've gone wrong somewhere feel free to mention where. The ILC site shows crossing out certain formulas so suddenly I'm not so sure this is as right as I wished it to be.

As for point 1, You're probably right that the wiki article may be advance as the chapter the question comes from has not even delt with Van Der Walls forces yet. Would it be fair to say that a repulsive force is the amount of protons located in the nucleus? This is sort of a guess but since they are positively charged wouldn't they repel the negatively charged electroncs? An attractive force, after reading a bit more, make me think I'm supposed to answer that the smaller the electron cloud the more attractive it is. And there is high electronegativity but those are the same in a sense.

As for #2: I am still fairly confused. You mentioned mixing a s and a p gives you sp but then with a p left over and a d would that mean sppd? Now I think I'm just losing it.

How'd I do 0 for 3?

[jennielynn_1980]

For number three, you need to use the three equations that you have to come up with your combustion equation.  If you look at step four on the ILC site under "how do I start" it shows how you would sort of add the equations together - hence the ADDITIVITY of heats of reaction.  I realize that you took a shot at doing this but instead of adding the heats of reaction,  try actually adding all the equations together as well to see what you get.  So if you have equation 1 as A + B --> C  and equation 2 as E + F --> G + H
if you add those together you would have A+ B + E + F --> C + G + H.  Then your next step would be to see if you have any common molecules or atoms on either side that you can cross out as it shows in the example on the ILC site. 

Hope that helps.

[Jabus]

Hmm but then I'd end up with

C2H4 + 3O2 + C2H4 + H2 + H2 +1/2O2 --> 2CO2 + 2H2O + C2H6 + H2O

The only problem with this is that none of the elements cancel out and then I would just add all the Hf together, but somehow that seems too easy or wrong...
Maybe I just need to sleep on it I'm probably missing something, I'll give it another go tomorrow.

[enahs]

Look at this past recent thread for Hess' law, see if it helps then ask more questions:
http://www.chemicalforums.com/index.php?topic=11096

[Jabus]

Ok, hmm, so using the example in the link above

CO2 (g) + 2 H2O (g)   ----->     CH4 (g)    +   2 O2 (g)              + 802.7 kJ
_______________________________________________________________
3 H2 (g) + CO (g)       ----->      CH4 (g)    +   H2O  (g)

That is already given so would it be fair to say that the rest of it should look somewhat like this:

2CO (g)                     ----->     2C (s) + O2 (g)                       +221.0 kJ
C (s) + O2 (g)            ----->     CO2 (g)                                  -393.5 kJ
2H2 (g) + O2 (g)        ----->     2 H2O (g)                                -483.6 kJ
CO2 (g) + 2 H2O (g)   ----->     CH4 (g)    +   2 O2 (g)              + 802.7 kJ
_______________________________________________________________
3 H2 (g) + CO (g)       ----->      CH4 (g)    +   H2O  (g)

So then to cancel elements out...

2CO (g)     ----->     C (s) + O2 (g)           +221.0 kJ
               ----->                                    -393.5 kJ
2H2 (g) +  ----->                                   -483.6 kJ
                ----->     CH4 (g)                   + 802.7 kJ
_______________________________________________________________
3 H2 (g) + CO (g)       ----->      CH4 (g)    +   H2O  (g)

So I would end up with 2CO + 2H2 -----> C + O2 + CH4  at 146.6kJ ?

How did I do on that one? 

[jennielynn_1980]

Your goal by adding the equations together and cancelling out common ions/molecules is to come up with your combustion equation.  Remember that you can manipulate your equations in such a way as to come up with what you need.

So if you have

2A + B --> C
A + 3B + D --> E
B + E -->  A  + 205 kJ

And you need to have 2E and 5 B you can multiply equation 3 by 2 to get what you need.

So 2x B + E --> A   +205kJ  would become:

2B + 2E --> 2A  + 410kJ

Also remember you can reverse the equations so equation 2 could become E --> A + 3B + D

[Jabus]

Your goal by adding the equations together and cancelling out common ions/molecules is to come up with your combustion equation.  Remember that you can manipulate your equations in such a way as to come up with what you need.

So if you have

2A + B --> C
A + 3B + D --> E
B + E -->  A  + 205 kJ

And you need to have 2E and 5 B you can multiply equation 3 by 2 to get what you need.

So 2x B + E --> A   +205kJ  would become:

2B + 2E --> 2A  + 410kJ

Also remember you can reverse the equations so equation 2 could become E --> A + 3B + D

But wouldn't I get 6B because of the B in the first equation, the 3B in the second equation, and the B*2 (2B) in the third equation? Am I ignoring equation 1 and if so why?
Also if I switched the second equation to E --> A + 3B + D would I still be able to count the 3B from the product side to the reactant side? I must be missing something...because something isn't clicking.

[jennielynn_1980]

Sorry, it is probably bad math on my part.  I was just trying to show you that you can multiply your 3 equations to get what you need in your final combustion equation that you came up with.  Check in your textbook on pg. 331 and look at step 3.  That was all I was trying to show you.

[Jabus]

C2H4  + 3O2 --> 2CO2 + 2H2O                    -1409.5 kJ
C2H4 + H2 --> C2H6                                   -136.7 kJ
H2 + 1/2O2 --> H2O                                   -285.5 kJ
_________________________________
C2H6 + 7/2O2 -> 2CO2 +3H2O

So

C2H4  + 3O2 --> 2CO2 + 2H2O                    -1409.5 kJ
C2H6  --> C2H4 + H2                                  +136.7 kJ
H2 + 1/2O2 --> H2O                                   -285.5 kJ
_________________________________
C2H6 + 7/2O2 -> 2CO2 +3H2O

So

I'd only multiply by 1 in order to get that combustion equation...

So

C2H4 + 3O2 + C2H6 + H2 + 1/2O2 --> 2CO2 + 3H2O + C2H4 + H2     -1260.7kJ
Simplified to
3O2 + C2H6 + 1/2O2 --> 2CO2 + 3H2O

That is what I get if I follow the book though I did that earlier and somehow came up with a 1600 number but this looks like i've followed it proper...
Is this better or worse than my previous attempts lol

[jennielynn_1980]

Now you have the idea!  I got a different number when I added the heats of reaction though.

[DevaDevil]

C2H4  + 3O2 --> 2CO2 + 2H2O                    -1409.5 kJ
C2H6  --> C2H4 + H2                                  +136.7 kJ
H2 + 1/2O2 --> H2O                                   -285.5 kJ
_________________________________
C2H6 + 7/2O2 -> 2CO2 +3H2O

C2H4 + 3O2 + C2H6 + H2 + 1/2O2 --> 2CO2 + 3H2O + C2H4 + H2     -1260.7kJ

look at the - sign before 1409.5 and then recalculate, you should get the right answer then:
-1409.5 + 136.7 - 285.5 = ...

[Jabus]

Ahh yes, me and my silly calculator fingers. -1558.3 i think i added the 285 rather tahn subtracted. hmm. Well then, thank you all so very much for the *delete me*

Just a quick check then in regards to that one other question. Just as some extra practice:

As an alternative to combustion, coal gas can undergo a process called methanation.

3H2 (g) + CO (g)       -->       CH4 (g) + H2O (g)

Determine the standard enthalpy change for this methanation reaction using the following chemical equations and the value for the standard enthalpy changes.

2H2 (g) + O2 (g)      -->      2 H2O (g)                                deltaH = -483.6 kJ
2C (s) + O2 (g)        -->      2CO (g)                                   deltaH = -221.0 kJ
CH4 (g) + 2O2 (g)   -->       CO2 (g) + 2 H2O (g)                 deltaH = -802.7 kJ
C (s) + O2 (g)         -->       CO2 (g)                                   deltaH = -393.5 kJ

2H2 (g) + O2 (g)         -->       2 H2O (g)               -483.6 kJ
CO2 (g) + 2 H2O (g)    -->       CH4 (g) + 2O2 (g)   +802.7 kJ
________________________________________
3H2 (g) + CO (g)       -->       CH4 (g) + H2O (g)


2H2 + CO2 --> CH4  @ /\Hf 319.1kJ

How does that one look? I wasn't sure if I could use the first equation as a substitution for both CH4 AND CO because it had both chemicals one on reactant and one on product side. Am I allowed to 'kill 2 birds with one stone' or would I have to add in the third equation for the CO?

Also the book says no diatomic gases (or elements) count towards heat formation and are considered zero. Is there any reason for this? and does this mean anything that is 'double' is negated down to 0?