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 concentrations 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)

Let the order of reaction with respect to $A$ and $B$ is x and y respectively.
So, the rate law can be given as
$R=k[A{]}^x[B{]}^y...(i)$
When the concentration of only $B$ is doubled, the rate is doubled, so
$R_1=k[A{]}^x[2B{]}^y=2R...(ii)$
If concentrations of both the reactants $A$ and $B$ are doubled,
the rate increases by a factor of $8$, so
$R"=k[2A{]}^x[2B{]}^y=8R...(iii)$
$\Rightarrow k{2}^x{2}^y[A{]}^x[B{]}^y=8R...(iv)$
From Eqs. (i) and (ii), we get
$\Rightarrow \frac{2R}{R}=\frac{[A{]}^x[2B{]}^y}{[A{]}^x[B{]}^y}$
$2=2^y$
$\therefore y=1$
From Eqs. (i) and (iv), we get
$\Rightarrow \frac{8R}{R}=\frac{2^x{2}^y[A{]}^x[B{]}^y}{[A{]}^x[B{]}^y}$
or $8=2^x{2}^y$
Substitution of the value ofy gives,
$8=2^x{2}^1$
$4=2^x$
$(2{)}^2=(2{)}^x$
$\therefore x=2$
Substitution of the value of $x$ and $y$ in Eq. (i) gives,
$R=k[A{]}^2[B]$