Question

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

• A. rate $=k[A][B{]}^2$
• B. rate $=k[A{]}^2[B{]}^2$
• C. rate $=k[A][B]$
• D. rate $=k[A{]}^2[B]$

Answer 1

Answer: (D)

$[A]$	$[B]$	Rate
$x$	$y$	$R$	$\dots (i)$
$x$	$2y$	$2R$	$\dots (ii)$
$2x$	$2y$	$8R$	$\dots (iii)$

Let the rate law; rate $=k[A{]}^a[B{]}^b$
From data given, $(x{)}^a(y{)}^b=R\dots (iv)$
$(x{)}^a(2y{)}^b=2R\dots (v)$
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,
${\left(2x\right)}^a{\left(2y\right)}^b=8R\dots \left(vi\right)$
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 $=k[A{]}^2[B]$