Question

For the reaction A + B $\to$ products, it is observed that :-
(a) on doubling the initial concentration of A only, the rate of reaction is also doubled and
(b) 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.

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

Answer 1

The correct option is C rate = k[A]$[B{]}^2$

The given reaction is $A+B⟶$ Product

Let us suppose the rate law of this reaction is:-

$Rate=K\left[A{]}^a\right[B{]}^b$

where $K$ is a rate constant.

$a$ and $b$ are order of the reaction with respect to the reactants $A$ and $B$ respectively.

Given that,

When [A] is doubled, the rate of the reaction is also doubled, so the reaction is first order $w.r.t.A$ and hence $a=1$

When $\left[A\right],\left[B\right]$ is doubled, the rate of reaction becomes $8$ times. Now,

$(Rate{)}_{new}=K[2A{]}^1[2B{]}^b$ $-\left(ii\right)$

$Rate=K\left[A{]}^1\right[B{]}^b$ $-\left(iii\right)$

Now, $\because$ New rate of reaction is $8$ times, so dividing $\left(ii\right)$ by $\left(iii\right)$ :-

$\Rightarrow 8={2.2}^b$

$2^3=2^{1+b}$

Equating the exponents:-

$\Rightarrow 3=1+b\Rightarrow b=2$

So, order of reaction $w.r.t$ to $B$ is $2$

So, $Rate=K\left[A\right][B{]}^2$