Question

For a single step reaction $2A+B\mapsto$ products, the rate of reaction is $6\times {10}^{-3}mol{e}^{-1}{l}^{-1}{S}^{-1}.$ The rate of reaction changes to $1.2\times {10}^{-2}$ mole $l^{-1}{S}^{-1}$ when :

• A. concentration of B is doubled and A is reduced to half
• B. concentration of A is doubled and B is reduced to half
• C. concentration of B is doubled and A is constant.
• D. both (B) and (C)

Answer 1

Answer: (D)

both (B) and (C)

Since it is a single step reaction, it is a $3^{rd}$ order reaction.
$r=k\left[A{]}^2\right[B]$
The rate of reaction is doubled as:
$\frac{r_2}{r_1}=\frac{1.2\times {10}^{-2}}{6\times {10}^{-3}}=2.$
The reaction is 2^{nd} order wrt A and $1^{st}$ order wrt B.
So if $\left[A\right]$ is doubled and B is reduced to half; rate of reaction will double
or
when $\left[B\right]$ is doubled and [A] it constant, the rate of reaction it doubled.
Hence B & C are correct options.