Reaction $\ce{A + B ->[$k$] C + D}$ follows the rate law $$r = k [\ce{A}]^{1/2}[\ce{B}]^{1/2}$$ where $k$ is given. Starting with $1 \,\pu{M}$ of $\ce{A}$ and $\ce{B}$ each, what is the time taken for concentration of $\ce A$ to become $0.1 \,\pu{M}$? Do we use the integrated law expression of first order (overall order) or half order (order w.r.t. $\ce{A}$)?

  • $\begingroup$ In this special case, r=k.[A] $\endgroup$
    – Poutnik
    Jul 18, 2023 at 10:28
  • $\begingroup$ Why? What’s special in this particular rxn? $\endgroup$
    – Āñé
    Jul 18, 2023 at 11:14
  • 2
    $\begingroup$ It is not about chemistry, but simple algebra. $x=y \implies x^{1/2}y^{1/2} = x$. // As $[\ce{A}] = [\ce{B}]$ $\endgroup$
    – Poutnik
    Jul 18, 2023 at 12:04

1 Answer 1


*At High School Level
The order of a Reaction is the Sum of powers of concentration terms in The Rate Law Expression. Therefore, acc. to the rate law expression provided in the question $$r = k[A]^{1/2}[B]^{1/2}$$ The order of Reaction is 1.
Let the Concentration of reactants at any point of time during reaction $$\ce{ A +B→C +D}$$ be $xM$
Now the rate law becomes $$ r = k x $$
Now you can integrate the rate law obtained as usual.

  • $\begingroup$ I think the correct transformation follows a similar patter as the second order reaction, i.e. $\frac{\mathrm{d} x}{\mathrm{d} t} = -k\sqrt{(A_0-x)(B_0 -x)}$, where $A_0$ and $B_0$ stands for the initial concentrations. $\endgroup$
    – liuzp
    Jul 18, 2023 at 14:05
  • 1
    $\begingroup$ But it is given that bot reactants start witl $1M$ comcentration. And you would get the same results from you equation any my equation if the limits in integration are put properly. $\endgroup$ Jul 19, 2023 at 2:00

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