Posted by Ken on Monday, June 16, 2008 at 9:57pm.
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

Chem II  Ken, Monday, June 16, 2008 at 11:46pm
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?

Chem II  DrBob222, Tuesday, June 17, 2008 at 12:08am
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. 
Chem II  Dr. Jones, Saturday, November 7, 2009 at 7:33pm
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