[La-Lu]

hello there

so here I am trying to solve this silly problem and I'm totally stuck and frustrated

we've got

FREE ENERGY 7.5 kJmol-1
T 37C

so happy I go

K(eq) = DeltaG / RT

and I get 2.9 x 10^-3

which is not among the multiple choice question's answers so I'm guessing I get it wrong.

Every book or website I search gives me the same equation so am I actually using it wrong or what??

please somebody help 

[MrTeo]

K(eq) = DeltaG / RT

Wrong.
http://en.wikipedia.org/wiki/Chemical_equilibrium#Thermodynamics

Every book or website I search gives me the same equation [...]

...

[La-Lu]

K(eq) = DeltaG / RT

Wrong.
http://en.wikipedia.org/wiki/Chemical_equilibrium#Thermodynamics

Every book or website I search gives me the same equation [...]

leading to:

DeltaG = - RT LN Keq

Keq = e^-DG/RT

which gives me another wrong result. this is what I can read out of it. that's why I'm asking what I'm seeing wrong, my knowledge of the topic is obviously very poor since I just started with this so giving me the link to the wikipedia page is not helping. I haven't managed to figure the problem out looking at my books, which is kind of the same but a bit more reliable, if I may.

so thank you but I'm still helpless here!



...

[MrTeo]

Just a wild guess: did you convert ºC to K before applying the formula?

[La-Lu]

Just a wild guess: did you convert ºC to K before applying the formula?

hey there, yes I convert 37C to 310.15K (right?)

[MrTeo]

Just a wild guess: did you convert ºC to K before applying the formula?

hey there, yes I convert 37C to 310.15K (right?)

Did you convert kJ to J? Did you use the right value of the gas constant (R = 8.314 J · mol^-1 · K^-1)?

[Corribus]

Why don't you just show all your work, instead of making us guess what you did wrong.

[danteOne]

The equation that you should use is
∆G=∆G° + RT*ln(Q)
∆G° is the Gibbs free energy at standard conditions
R is the gas constant 0.008314 kJ/(K*mol)
T is the temperature in kelvin
Q is [concentration or reactants]/[concentration of products]

At equilibrium ∆G = 0.
Therefore
0 = ∆G° + RT*ln(Q)
which leads to
∆G° = -RT*ln(Q)
Type out every you took to solve this problem using that final equation or take a picture of your neatly written work so that we can see where you went wrong.