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A = CH2(CN)2
B = CH(CN)2-
C = BrCH(CN)2
The rate law above is derived from applying the steady state approximation.

I am trying to work out how to modify the rate law to accomodate the following conditions:

1) [Br2] is low relative to [H+]

2) [Br2] is high relative to [H+]

And in the 2nd case how this affects the RDS and order of reaction.

It seems to me like these conditions would just affect the rate coefficient, not actually change the rate law.


$$ \frac{d [CH(CN)_{2}^{-}]}{dt} = k_{1}[CH_{2}(CN)_{2}] - k2[CH(CN)_{2}^{-}][Br_{2}] - K_{-1}[CH(CN)_{2}^{-}][H^{+}] $$

$$ Applying\ Steady\ State\ Approximation: $$ $$ k_{1}[CH_{2}(CN)_{2}] - k2[CH(CN)_{2}^{-}][Br_{2}] - K_{-1}[CH(CN)_{2}^{-}][H^{+}] = 0 $$

$$ k2[CH(CN)_{2}^{-}][Br_{2}] + K_{-1}[CH(CN)_{2}^{-}][H^{+}] = k_{1}[CH_{2}(CN)_{2}] $$ $$ Dividing\ by\ [CH(CN)_{2}^{-}]: $$

$$ k2[Br_{2}] + K_{-1}[H^{+}] = \frac{k_{1}[CH_{2}(CN)_{2}]}{[CH(CN)_{2}^{-}]} $$

$$ [CH(CN)_{2}^{-}](k2[Br_{2}] + K_{-1}[H^{+}]) = k_{1}[CH_{2}(CN)_{2}] $$

$$ [CH(CN)_{2}^{-}] = \frac{k_{1}[CH_{2}(CN)_{2}]}{(k2[Br_{2}] + K_{-1}[H^{+}])} $$

$$ \frac{d [BrCH(CN)_{2}]}{dt} = {k_{2}[Br_{2}][CH(CN)_{2}^{-}]}{} $$

$$Rate\ Equation:$$

$$ \frac{d [BrCH(CN)_{2}]}{dt} = \frac{k_{2}[Br_{2}]k_{1}[CH_{2}(CN)_{2}]}{{k_{2}[Br_{2}]}+k_{-1}[H^{+}]} $$

So if $ [Br_{2}]$ is low compared to $ [H^{+}]$ then I would remove $ [Br_{2}]$ from the rate equation giving:

$$ \frac{d [BrCH(CN)_{2}]}{dt} = \frac{k_{1}[CH_{2}(CN)_{2}]}{k_{-1}[H^{+}]} $$

If $ [H^{+}]$ is low compared to $ [Br_{2}]$ then I would remove $ [H^{+}]$ from the rate equation giving:

$$ \frac{d [BrCH(CN)_{2}]}{dt} = k_{1}[CH_{2}(CN)_{2}] $$

So in this case the reaction is 1st order and the RDS no longer proceeds backwards with rate coefficient $k_{-1}$.

Is this right or have I made a mistake somewhere?

  • $\begingroup$ How do Br an H interact? Are C, B or A = Br? What does Br2 have to to with this. This question is pretty mangled right now. $\endgroup$
    – Lighthart
    Apr 26, 2016 at 14:54
  • $\begingroup$ My apologies I've fixed the error and added identities of A, B and C. $\endgroup$
    – draksi
    Apr 26, 2016 at 15:30
  • 2
    $\begingroup$ Rate law above is not correct. Try to derive it yourself via steady state approximation and you will quickly figure out some answers. $\endgroup$
    – Rok Narobe
    Apr 26, 2016 at 20:12

1 Answer 1


The steady state approximation allows us to assume that the concentration of $\ce{B}$ does not change.


Thus the rates of production and consumption of $\ce{CH(CN)2-}$ must be equal.

$\ce{CH(CN)2-}$ can be produced by one pathway:

$$\ce{CH2(CN)2 ->[k_1] CH(CN)2- + H+} \\ \mathrm{rate}=k_1[\ce{A}]$$

$\ce{CH(CN)2-}$ can be consumed by two pathways:

$$\ce{CH(CN)2- + Br2 ->[k_2] BrCH(CN)2 + Br-} \\ \ce{CH(CN)2- + H+ ->[k_{-1}] CH2(CN)2} \\ \mathrm{rate}=k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}]$$


$$\dfrac{d[\ce{CH(CN)2-}]}{dt}=k_1[\ce{CH2(CN)2}] -\left( k_2 [\ce{CH(CN)2-}][\ce{Br2}]+K_{-1}[\ce{CH(CN)2-}][\ce{H+}] \right)= 0$$

Now you can can solve for $[\ce{CH(CN)2-}]$ and substitute into the rate law equation for the second step:


You should get a final rate law that contains $[\ce{CH2(CN)2}],\ [\ce{Br2}],\ \&\ [\ce{H+}]$, and then you can answer your questions.

Update based on revised question.

Your new rate late is correct:


Your interpretation of the simplification you can do is correct for one case and almost correct for the other.

$[\ce{H+}] \ll [\ce{Br}]$

In this case, we assume that $k_{-1}[\ce{H+}]\ll k_2[\ce{Br}]$ so

$$k_2[\ce{Br2}]+k_{-1}[\ce{H+}] \approx k_2[\ce{Br2}] \\ \therefore \ \dfrac{d[\ce{BrCH(CN)2}]}{dt}\approx \dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_2[\ce{Br2}]}\approx k_1[\ce{CH2(CN)2}$$

which is as you figured.

$[\ce{H+}] \gg [\ce{Br}]$

In this case, we assume that $k_{-1}[\ce{H+}]\gg k_2[\ce{Br}]$ so

$$k_2[\ce{Br2}]+k_{-1}[\ce{H+}] \approx k_{-1}[\ce{H+}] \\ \therefore \ \dfrac{d[\ce{BrCH(CN)2}]}{dt}\approx \dfrac{k_2[\ce{Br2}]k_1[\ce{CH2(CN)2}]}{k_{-1}[\ce{H+}]}$$

Note that this approximation does not remove the $k_2[\ce{Br2}]$ term in the numerator.

  • $\begingroup$ I think I've got the correct rate law now and I've added it to the original post with my answers to the questions. Have I done it right? $\endgroup$
    – draksi
    Apr 27, 2016 at 11:49

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