Question

In a chemical reaction $A+2B\to$ products, when concentration of $A$ is doubled, rate of the reaction increases $4$ times and when concentration of $B$ alone is doubled rate continues to be the same. The order of the reaction is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Let the order of reaction $w.r.t.A$ is $x$ and w.r.t. $B$ is $y$.
$r_1-k{\left[A\right]}^x{\left[B\right]}^y$ ...(i)
$r_2-k{\left[2A\right]}^x{\left[B\right]}^y$ ...(ii)
$r_3-k{\left[A\right]}^x{\left[2B\right]}^y$ ...(iii)
$\frac{r_1}{r_2}=\frac{k{\left[A\right]}^x{\left[B\right]}^y}{k{\left[2A\right]}^x{\left[B\right]}^y}$
$\Rightarrow \frac{1}{4}={\left(\frac{1}{2}\right)}^x\Rightarrow {\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^x$
$\Rightarrow x=2$
Similarly
$\frac{r_1}{r_3}=\frac{k{\left[A\right]}^x{\left[B\right]}^y}{k{\left[A\right]}^x{\left[2B\right]}^y}$
$\Rightarrow 1={\left(\frac{1}{2}\right)}^y\Rightarrow {\left(\frac{1}{2}\right)}^0={\left(\frac{1}{2}\right)}^y$
$\Rightarrow y=0$
Hence the rate law equation is
Rate = $k[A{]}^2[B{]}^0$
$\Rightarrow$ Order of reaction = 2