How do I Calculate the Equilibrium Constant for the Formation of AlF6^3-?

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Equilibrium Frustration...

Ive been going over a chem paper and cannot seem to get any of the multiple choice given answers,

the question is as follows,

The free energy change for the formation of the complex ion $$AlF_6^3-$$ is -140. kJ at 25ºC. What is the equilibrium constant, K, for the reaction?

Ive got:

deltaG = -RT InK

hence,

-140 = -(8.3145)(298.15) InK

hence,

InK = -140 + (8.3145)(298.15)

InK = 2338.0

but this gives a value of K=0, which can't be correct, or if it is, the answer does not appear in the choices in my paper.

Any ideas,

Can anyone see where I am going wrong?

Thanks for your time.

 

[siddharth]

Match the units you use.
The value of the gas constant here is in Joules/(K*mol)
In the left hand side of the equation you used the value of deltaG in Kilo Joules.

 

[gravenewworld]

Like Sid said, ALWAYS USE UNITS! UNITS! UNITS! The problem you are having is easy to spot when you simply use units.

 

[Gokul43201]

And in addition to the problem with the units, how do you get K=0 from ln(K) = 2338 ?

 

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I got k=0 because,

$$In(K) = 2338$$
and,
e^In(K) = K
hence;
e^2330 = 0

Anyways,

I relooked over the problem, and it seems that i was simply not rearranging the values correctly. (Boy, did I feel super dopey!)
I ended up with the right answer in the end.
K = 3.5*10^25
I think it was from memory.

Thanks for the help everyone.