Q. In a reaction, $A + B \to$ product, rate is doubled when the concentration of $B$ is doubled, and rate increases by a factor of $8$ when the concentration of both the reactants ($A$ and $B$) are doubled, rate law for the reaction can be written as
UP CPMTUP CPMT 2013Aldehydes Ketones and Carboxylic Acids
Solution:
$[A]\quad$
$[B]$
Rate
$x$
$y$
$R$
$\quad \ldots(i)$
$x$
$2y$
$2R$
$\quad \ldots(ii)$
$2x$
$2y$
$8R$
$\quad \ldots(iii)$
Let the rate law; rate $= k[A]^a [B]^b$
From data given, $(x)^a(y)^b = R\quad \ldots (iv)$
$(x)^a(2y)^b = 2R\quad \ldots (v)$
Dividing eqn. $(v)$ by $(iv)$,
$\frac{\left(2y\right)^{b}}{\left(y\right)^{b}}=\frac{2R}{R}$ or $\left(2\right)^{b}=2$
Thus, $b = 1$
From data of $\left(iii\right)$ experiment,
$\left(2x\right)^{a}\left(2y\right)^{b}=8R\quad\ldots\left(vi\right)$
Dividing eqn. $\left(vi\right)$ by $\left(v\right)$,
$\frac{\left(2x\right)^{a}}{\left(x\right)^{a}}=\frac{8R}{2R}$ or $\left(2\right)^{a}=4$
Thus, $a = 2$
By replacing the values of $a$ and $b$ in rate law;
rate $= k[A]^2 \,[B]$
| $[A]\quad$ | $[B]$ | Rate | |
|---|---|---|---|
| $x$ | $y$ | $R$ | $\quad \ldots(i)$ |
| $x$ | $2y$ | $2R$ | $\quad \ldots(ii)$ |
| $2x$ | $2y$ | $8R$ | $\quad \ldots(iii)$ |