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.