[Boo]

Hi there,

Question:
For the hypothetical equilibrium A   B, you obtain the following data.

T(K)       Keq
260          6.2*10^-2
280          1.6*10^-2
300          5.2*10^-3
320          1.9*10^-3

Determine the values of  :delta: G^o,  :delta: H^o, and  :delta: S^o.

My attempt:
First I plotted lnKeq vs. 1/T.

Based on the equation:
lnKeq = -( :delta: H^o/RT) + ( :delta: S^o/R)

So the slope of the line is equal to -( :delta: H^o/R)
and the y-interecept is ( :delta: S^o/R). I used this to figure out :delta: H^o, and  :delta: S^o.

 :delta: H^o = 40 kJ/mol
 :delta: S^o = -177 J/mol

I am confused about how to find :delta: G^o.
At first I thought of using the formula
 :delta: G^o = :delta: H^o - T :delta: S^o,
but the temperature is not constant, so this didn't work.

I then tried using :delta: G^o = -RTlnKeq, and substituting in a point from the data table for T and Keq.  I assumed each point would give the same value for  :delta: G (ie. the data at 260K would be the same as at 280K since the Keq varies according to the temperature.).  However, I got a different value at each point (and none of them were the right answer!).

The right answer for  :delta: G^o is 13kJ/mol.

Am I supposed to be finding the rate of change of G? Do I get this from another graph (if so, I am not sure what I am supposed to be graphing...  :delta: G= :delta: H-T :delta: S?) Or am I just not using the right equation to find  :delta: G?

Thanks for any direction you are able to provide!

[Borek]

What does the index 0 mean?

Are you sure about signs?

[Boo]

It's the standard free energy change (that's how it was written in the question) - sorry if that is incorrect and should have been written without the "O's".

[Borek]

No, you wrote it correctly. But what does "standard" mean?

[Boo]

So we are looking at the change in Gibbs free energy at standard state conditions.  So gases have to be at 100 kPa and aqueous solution at 1M, and it is implied it is being calculated at 298K. 

And I had the sign wrong on my calculation for  :delta: H^o, it equals -40kJ, not +40kJ.
(thanks for noticing that!)

So I calculate:
  :delta: G^o = -40kJ/mol - 298K(-0.177kJ/mol*K)
  :delta: G^o = 12.746 kJ/mol
  :delta: G^o ~ 13 kJ/mol

Thank you very much for your help, I really appreciate it!