Question

In a chemical reaction $A + 2B \rightarrow$ 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