# Chem II

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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 -

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 -

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 -

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

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