I am having trouble wit the following question on my problem sheet:

"The reaction $\ce{Br- + ClO^- \to BrO^- + Cl^-}$ occurs with a rate constant of $\text{0.35 dm}^3 \text{mol}^{-1} \text{s}^{-1}$

at 298 K. For the following initial conditions calculate the time taken for the Br^- concentration to fall to half its original value:

i) $[\ce{Br-}]=[\ce{ClO-}]=0.0017$

ii) $[\ce{Br-}]=0.0017$ and $[\ce{ClO-}]=0.084$"

I am sure I could do this question if I knew the rate equation but it isn't given. From the units of the rate constant the reaction must be second order but how am I supposed to know whether it's second order with respect to one of the reactants (and thus zeroth order w.r.t the other) or first order with respect to both? Also, what's the significance of the concentrations being equal (there's another question that stresses that the concentrations of both the reactants are equal).

  • 1
    $\begingroup$ I don't see a reasonable mechanism by whom the rate equation would have second order with respect to one reactant. $\endgroup$
    – EJC
    Feb 1, 2015 at 16:37
  • $\begingroup$ Expanding on @Marko 's point, if there were a couple of potential mechanisms, one for first order in both reactants and one for zeroth/second order, then you wouldn't know. There's no particular reason the concentrations should be equal, or reason why it's important (unless perhaps there is a method of initial rates question elsewhere in your problem set.) $\endgroup$ Feb 2, 2015 at 3:10

2 Answers 2


You have to assume it is first order in each reactant.

Zero-order kinetics is always an artifact of the conditions under which the reaction is carried out.

The reaction can't really be zero-order in a reactant, otherwise it would imply the reaction continues after there is no reactant remaining.

  • $\begingroup$ how can any reactant be zeroth order then? $\endgroup$
    – RobChem
    Feb 1, 2015 at 17:19
  • $\begingroup$ it can't, that's what the link is explaining $\endgroup$
    – DavePhD
    Feb 1, 2015 at 17:20
  • $\begingroup$ @RobChem Rewrote my comment to better explain what I meant. Suppose that the kinetics of a reaction are usually decided by a particular slow step yielding rate = k1[A], when [A] and [B] are roughly equal to one another. If we set up the reaction with very, very small amounts of [B] instead, the rate of another reaction in the mechanism that involves B will eventually become slow enough that it becomes the slow/rate determining step. $\endgroup$ Feb 2, 2015 at 3:07

Being the concentrations equal makes it easier to integrate the second order law (first in each reagents). Concentrations are the same at every t and therefore you have only one variable to take into account. The resulting integrated law will be analogous to that for a second order kinetics in one reagent.


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