For the reaction, $A+B \rightarrow$ products, it is observed that (1) On doubling the initial concentration…
For the reaction, $A+B \rightarrow$ products, it is observed that
(1) On doubling the initial concentration of $A$ only, the rate of reaction is also doubled and
(2) On doubling the initial concentrations of both $A$ and $B$, there is a change by a factor of 8 in the rate of the reaction.
The rate of this reaction is, given by
rate $=k[A]^2[\mathrm{~B}]$
rate $=k[A][B]^2$
rate $=\mathrm{k}[\mathrm{A}]^2[\mathrm{~B}]^2$
rate $=k[\mathrm{~A}][\mathrm{B}]$
Solution
For the reaction, $\mathrm{A}+\mathrm{B} \longrightarrow$ Products
On doubling the initial concentration of A only, the rate of reaction is also doubled, therefore
$\text {Rate } \propto[\mathrm{A}]^1$ ...(i)
Let initially rate law is
$\text {Rate }=k[A][B]^y$ ...(ii)
If concentration of and both are doubled, the rate gets changed by a factor of 8 .
$\begin{array}{r}
8 \times \text { Rate }=\mathrm{k}[2 \mathrm{~A}][2 \mathrm{~B}]^{\mathrm{y}} \quad \ldots(\text { iii) } \\
{\left[\because \text { Rate } \propto[\mathrm{A}]^1\right]}
\end{array}$
Dividing Eq. (iii) by Eq. (ii),
$\begin{aligned}
& 8=2 \times 2^y \\
& 4=2^y \\
& \therefore \quad \mathrm{y}=2 \\
&
\end{aligned}$
Hence, rate law is, rate $=\mathrm{k}[\mathrm{A}][\mathrm{B}]^2$