Chem II
posted by Ken .
The equilibrium constant of a reaction is 12.6. If the rate constant of the reverse reaction is 5.1 x 10 2 the rate constant for the forward reaction is _____
0.32
0.16
0.64
0.08
I don't even know where to start on this question. Any direction will help

Ok, never mind with this one, I think I answered it. I did the 1/12.6 and got 0.070365, which I rounded to 0.08. Is that correct?

See how this sounds.
For the reaction A + B ==> C + D
then rate fwd = kf(A)(B)
and rate reverse = kr(C)(D)
At equilibrium rate fwd = rate reverse and
kf(A)(B)=kr(C)(D)
kf/kr = (C)(D)/(A)(B) = Keq
So 12.6 = kf/kr.
Substitute kr and solve for kf.
Check my thinking. Check my algebra. 
If from what Dr. Bob is saying the answer should be 0.6426
meaning if you are solving for Kf. you will need to
12.6= x/5.1 x 10^2
multiply each side by Kr to get Kf by its self solving for X