[Ann1234]

Hello =)

I am working on this problem, which I have the answers for it, but I don't understand how they came up with those answers:

For the reaction: Na(s) = Na(l) at 25 degree C. b.) Calculate the free energy of the reaction.b.2) How does the enthalpy affect the overall spontaneity of the reaction, Choose 1: Increase, decrease, no change, cannot tell.

b) Answer: +0.5 kj/mol
b2) answer: no change



They gave us a table of delta G, S, and delta H for Na (s) and Na(L). So I used the formula
DELTA G = DELTA H - T DELTA S, as T=298K, and I got delta H and delta S from the values of the table as (products-reactants)

They used delta G= DELTA G NA(L) - DELTA G NA(S) = 0.5 - 0 = 0.5. I used the formula with enthalpy and entropy. shouldn't they give the same answers? I got 0.34 as delta G and I was wrong, delta G was 0.5 and they calculated as products - reactants.

then, they asked how enthalpy affects the spontaneity, the answer is no change. and I can't figure out why =(

can someone help me to figure out how they came up with these answers?

thanks..

[JustinCh3m]

This is the one you want to use:

delta G= DELTA G NA(L) - DELTA G NA(S)

The value for delta G is "so very close to 0", so that's why it's essentially "no change in the spontaneity of the reaction."

--

Not certain about the 0.34 vs. 0.5.  You double-checked your work I suppose. 

[Ann1234]

This is the one you want to use:

delta G= DELTA G NA(L) - DELTA G NA(S)

The value for delta G is "so very close to 0", so that's why it's essentially "no change in the spontaneity of the reaction."

--

Not certain about the 0.34 vs. 0.5.  You double-checked your work I suppose. 

Hello, this is the table they gave me:

substance   delta H    delta G    S

Na(l)            2.4         0.5      5.8

Na(s)            0            0        5.1

The answer is delta G =0.5 (confirmed) so they did that (delta G products - delta G reactants). My question is, why delta G=delta H-Tdelta S gives other value?

I did delta H= delta H Na(l) - delta H Na(s) = 2.4 - 0 = 2.4kj

and I got delta S = S Na (l) - S Na(s) = 5.8 - 5.1 = 0.7 j/k

then I calculated delta G= 2.4 -(298 x 0.007) = 0.314 kj

so with this equation I got 0.314 (* last time I made a typo, it was 0.314 instead of 0.34).

Why the difference between the delta G? with both formulas you get different numbers, but the one they reported as correct was 0.5. I don't know why the other one gives another answer....thanks for your reply.

[cheese (MSW)]

Here's some nice symbols for you: ΔG = ΔH – TΔS     
S°(Na(s)) = 51.3 J K^-1 mol^-1 http://www.webelements.com/sodium/thermochemistry.html
 (confirmed CRC Handbook 5-16) that is, NOT 5.1 as quoted.
I hunted round for S Na(l) take it to be 57.7 J K^-1 mol^-1.
You only have one sig fig for ΔG so consider the agreement in light of this observation.
A little bell should have rung with your answer with 3 sig figs but ΔS with only one.
Conclusion: whoever put this question together was sloppy!
And I also do not understand question 2? 

[Ann1234]

Here's some nice symbols for you: ΔG = ΔH – TΔS     
S°(Na(s)) = 51.3 J K^-1 mol^-1 http://www.webelements.com/sodium/thermochemistry.html
 (confirmed CRC Handbook 5-16) that is, NOT 5.1 as quoted.
I hunted round for S Na(l) take it to be 57.7 J K^-1 mol^-1.
You only have one sig fig for ΔG so consider the agreement in light of this observation.
A little bell should have rung with your answer with 3 sig figs but ΔS with only one.
Conclusion: whoever put this question together was sloppy!
And I also do not understand question 2? 

okay...yeah that makes some sense now...too bad the numbers are wrong  thanks