A researcher is investigating the following overall reaction:

$$\ce{2C + D -> E}$$

The researcher claims the rate law for the reaction is written as follows: $\text{Rate} = k[\ce{C}][\ce{D}]$.

  1. Is the rate law equation possible for the given reaction?
  2. If so, suggest a mechanism that would match the rate law. If not, explain why not.

The answer is yes. I don't know how to arrive at this answer. From what I have learned, the following must be true for a) to be true:

  1. The rate law for the slow step agrees with the experimental rate law
  2. The elementary steps of add up to the overall reaction.

I would assume the following elementary steps from the rate the researcher has confirmed:

$$ \begin{align} \ce{C + D &-> CD}\\ \ce{C + CD &-> E} \end{align} $$

But since $\ce{CD}$ has a larger size than $\ce{D},$ wouldn't it be less likely to have the proper alignment in a collision to enter the active state? Thus, wouldn't it be a slower reaction than $\ce{C + D}?$

Please inform me where my logic is wrong and why the answer is yes.

  • 5
    $\begingroup$ Steric factors are not the only thing that controls the rate of each elementary step. Perhaps most crucially you have to look at the activation energy as well. That second step may have a smaller steric factor than the first step but that doesn't necessarily mean it's slower. $\endgroup$ Commented Oct 25, 2016 at 23:33
  • 1
    $\begingroup$ The rate law does not usually (i.e. almost never) indicate what the actual mechanism is. The only way to determine mechanism is by experiment. The classic example is $\ce{H2 +Br2 = 2HBr}$ which occurs via five recactions involving $\ce{H\cdot , Br\cdot}$ radicals with rate $\ce{d[HBr]/dt= a[H2][Br2]^{1/2}/(1+b[HBr]/c[Br_2])}$ where a, b, c are rate constants (or combinations thereof) of the mechanism. $\endgroup$
    – porphyrin
    Commented Oct 26, 2016 at 20:27
  • $\begingroup$ To comment on your particular reaction (a) if the rate quoted is true then this C+D represents the rate limiting step, i.e. slowest and this means that CD reacts with C more rapidly than D reacts with C, i.e. CD+C has a lower activation barrier. (b) Alignment has a very small effect compared to activation energy $\Delta E$ , look up Arrhenius equation $k=A.\exp(-\Delta E/(RT))$ where $A$ is rate constant at infinite temperature (or zero activation energy) $\endgroup$
    – porphyrin
    Commented Oct 26, 2016 at 20:38
  • $\begingroup$ We know norhing about molecular mechanism we are just provided that c reacting d will result in cd and c reacting with cd will result in e. $\endgroup$ Commented Mar 28, 2019 at 16:51

1 Answer 1


You should use steady state approximation which leads to the mentioned answer:

$$\frac{\mathrm{d}[\ce{CD}]}{\mathrm{d}t} = 0$$

$$ \begin{align} \ce{C + D &->[$k_1$] CD}\\ \ce{C + CD &->[$k_2$] E} \end{align} $$

$$\ce{->[QSSA]}\quad k_1[\ce{C}][\ce{D}] = k_2[\ce{C}][\ce{CD}] \quad\to\quad k_1[\ce{D}] = k_2[\ce{CD}]$$

$$\frac{\mathrm{d}[\ce{E}]}{\mathrm{d}t} = k_2[\ce{C}][\ce{CD}] = k_1[\ce{C}][\ce{D}]$$

  • $\begingroup$ QSSA: quasi-steady-state assumption $\endgroup$
    – Karsten
    Commented Sep 19, 2020 at 1:34

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