Question

Which chemical process is associated with the lattice energy for sodium chloride?

NaCl(g) → Na+(g) + Cl-(g)
NaCl(s) + H2O(l) → Na+(aq) + Cl-(aq)
Na(s) + 1/2 Cl2(g) → NaCl(s)
NaCl(s) → Na+(g) + Cl-(g)

Answer 1

@Whitemonsterbunny17 @AccessDenied @sammixboo can someone please help me...I have to finish this today..

Answer 2

Oh, I'm soooooo sorry, but I don't know Chemistry. :/
maybe @elementwielder can help though. *^.^*

Answer 3

ok, thank you anyways :)

Answer 4

Im not strong in this subject but if i were to guess, it would prob would be A or D

Answer 5

I would think it was D, but shouldn't the Na+(g) + Cl-(g) part be in the front? idk if that matters or not.

Answer 6

i have no idea... im a freshmen so im not keen in chemistry yet, but i remember a little from beginners course in 7th grade

Answer 7

ah ok :) thank you though.

Answer 8

yw

Answer 9

if you need help in physics or biology i can probably help a lot more lol so tag me if you need help and ill try to help

Answer 10

k :)

Answer 11

I'm stuck between (D) because it is the reverse process which should release the same energy in the opposite direction (store=release) and (B) because it is also associated with ionization / solubility in this case:
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch7/lattice.html
at the end.

Sorry to throw another possibility here when I am not sure either. I'll do a bit more looking though.

Answer 12

I think I do lean towards (D) though as it is the closest related.

Answer 13

thank you :) It was D.

Answer 14

could you by any chance help me with one other question?

Answer 15

I can try. I haven't done much with Lattice Energy in particular, but I can do my best (or research too).

Answer 16

Ok :)

Answer 17

Calculate the electron affinity for the formation of the hydride ion from the following information:

H(g)+e−→H−(g) ?
1/2H2(g)→H(g) +217.9kJ/mol
Na(s)→Na(g) +107.3kJ/mol
Na(g)→Na+(g)+e− +495.8kJ/mol
Na(s)+1/2H2(g)→NaH(s) −60.0kJ/mol
NaH(s)→Na+(g)+H−(g) +810.9kJ/mol

−816kJ/mol
−70.1kJ/mol
−50.1kJ/mol
−1632kJ/mol

Answer 18

It seems like our goal is Hess' Law, we want to add all these equations in a way to get the equation with:
H(g) + e- yields H- (g) Delta H = sum of Delta H of added equations.

That part sounds familiar?

Answer 19

sort of..., sorry I was away for a few minutes

Answer 20

I tried adding everything together but switching the signs...I don't think that worked

Answer 21

It's a bit tough to arrange them all, but what I would do is start with the last reaction:
NaH(s)→Na+(g)+H−(g) +810.9kJ/mol
And then go upwards, trying to use each equation to eliminate something that might get us closer to the equation we want.
For example:
NaH(s)→Na+(g)+H−(g) +810.9kJ/mol
Na(s)+1/2H2(g)→NaH(s) −60.0kJ/mol
We add these together, and the NaH(s) on each side cancel out.
We would then have:
Na(s) + 1/2H2(g) → Na+(g) + H-(g) 810.9-60.0 kJ/mol
Now the above two equations seem to deal with Na's. We could use the Na(s) →Na(g) equation first to convert the Na(s) in ours. To do so, we reverse the equation and the heat of reaction sign:
Na(s) + 1/2H2(g) → Na+(g) + H-(g) 810.9-60.0 kJ/mol
Na(g)→Na(s) -107.3kJ/mol (from Na(s)→Na(g) +107.3kJ/mol)
Adds to give:
Na(g) + 1/2H2(g) → Na+(g) + H-(g) 810.9-60.0-107.3 kJ/mol.

Can you see how we can eventually get the equation H + e- yields H- in this route? Next we cancel Na+(g) / Na(g), and then 1/2H2(g) for H(g).

Answer 22

That post was a lot longer than I thought. >.> lol

Answer 23

oooh, I think I am getting it a little bit...when you cancel something out and its on the opposite side that you need it, you change the values to the opposite sign as well?

Answer 24

You change signs of the heat where you flip the whole equation... like
Na(s)→Na(g) +107.3kJ/mol
We needed the other way around, so we reverse the equation. Reverse the sign.
Na(g)→Na(s) -107.3kJ/mol
That is clear?

Answer 25

yes :) This might be stupid...but how exactly do you add two of the reactions together. Do you add to both sides or what? Could you just give me an example with that first equation you did to get rid of NaH(s)?

Answer 26

Questions are fine! We add the reactants together and the products together. Then we can get rid of anything that is in the same on both sides... so like this:
NaH(s)→Na+(g)+H−(g)
Na(s)+1/2H2(g)→NaH(s)

An intermediate step would be like this:
NaH(s) + Na(s) + 1/2H2(g)→Na+(g) + H-(g) + NaH(s)
We have NaH(s) on both sides. So it has not been altered in the reaction, we can omit it.
Na(s) + 1/2H2(g) )→Na+(g) + H-(g)

Answer 27

The intermediate step, being when I add them together at first before cancelling. But I generally omit it because I understand what my goal was.

Answer 28

Oh ok :) that makes total sense :)

Answer 29

I will try to figure it out now...maybe you can check when I am done if u dont mind :)

Answer 30

Yep!
I got an answer myself (after I made a mistake with signs, be careful), so I can cross check with that. :D

Answer 31

70.1 :)

Answer 32

with a negative

Answer 33

Looks good! Yeah, negative 70.1 is what I got as well. :D

Answer 34

ok awesome :) Thank you SOOO much! I finally understand this!

Answer 35

Glad to help!