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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