Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
The experimental rate of a reaction
$$\ce{A + 2B -> C + D}$$
$$\frac{\mathrm dx}{\mathrm dt} = K[\ce{A}][\ce{B}]^2$$
Calculate the order of the reaction when (i) $[\ce{A}]\gg[\ce{B}]$ (ii) $[\ce{B}]\gg[\ce{A}]$
My efforts :
Should we ignore the concentration of the reactant that is present negligibly? And thus the orders will be:
(i) Order = 1
(ii) Order =2
Please guide.
Let's compare the rate when $A$ and $B$ are present at their initial amounts, say $A_0$ and $B_0$, to when the reaction is 99% complete.
When $[\ce{A}]\gg[\ce{B}]$, the reaction will be 99% complete when 99% of $B$ is gone, i.e. when $B = 0.01B_0$.
At that point, $$\frac{dx}{dt} = k[A][B]^2 = k(A_0 - \frac{0.99}{2}B_0)(0.01B_0)^2 $$
Now since $A\gg B$ also means $A_0\gg B_0$, then it follows that $(A_0 - \frac{0.99}{2}B_0) \approx A_0$. Thus, when the reaction is 99% complete and $B$ is limiting,
$$\frac{dx}{dt} \approx kA_0(0.01B_0)^2 = kA_0 (0.0001)B_0^2$$
If we lump $k$ and $A_0$ into a single parameter, $k_{app}$, then we can write $\frac{dx}{dt} = k_{app}B^2$, which shows that when $B$ is limiting, the reaction is 2nd order.
The case for $A$ being limiting shows the reverse.
The order overall is 3 , first in A and second in B.
When [A]>>[B], A can be considered as constant, because all of B can be consumed but [A] is hardly changed. In this situation the concentration of A ([A]) can be lumped in with the 3rd order rate constant $k_3$ to give a pseudo-second order reaction, $ k_{psuedo}=k_3[A]$. This has 2nd order units $dm^3mol^{-1}s^{-1}$. A similar argument applies when [B]>>[A] but with a different pseudo order rate constant.
