# How does the concentration of a reactant change? Is it based on individual order or overall order?

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. $$\ce{A}$$)?

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

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 $$r = k x$$
• 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. Commented Jul 18, 2023 at 14:05
• 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. Commented Jul 19, 2023 at 2:00