Chemical Kinetics Result Question 15
15 For the reaction, $2 A+B \longrightarrow$ products
When concentration of both $(A$ and $B)$ becomes double, then rate of reaction increases from $0.3 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ to $2.4 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$
When concentration of only $A$ is doubled, the rate of reaction increases from $0.3 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ to $0.6 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$.
(2019 Main, 9 Jan II)
Which of the following is true?
(a) The whole reaction is of 4th order
(b) The order of reaction w.r.t. $B$ is one
(c) The order of reaction w.r.t. $B$ is 2
(d) The order of reaction w.r.t. $A$ is 2
Show Answer
Answer:
Correct Answer: 15. ( c )
Solution:
For the reaction, $2 A+B \longrightarrow$ products.
Let, the rate expression is
$ \begin{aligned} & r \propto[A]^a[B]^b \\ & Expt \hspace {1mm} 1 \quad \dfrac{r_2}{r_1}=\left(\dfrac{2 A}{A}\right)^a\left(\dfrac{2 B}{B}\right)^b \\ & \Rightarrow \quad \dfrac{2.4}{0.3}=2^a \times 2^b \Rightarrow 2^3=2^{a+b} \\ & \Rightarrow \quad 3=a+b ……(i) \end{aligned} $
$\begin{aligned} & \text { Expt } 2 \quad \dfrac{r_2}{r_1}=\left(\dfrac{2 A}{A}\right)^a\left(\dfrac{B}{B}\right)^b \\ & \Rightarrow \dfrac{0.6}{0.3}=2^a \times 1 \Rightarrow 2^1=2^a \Rightarrow a=1 ……(ii) \end{aligned}$
$\begin{array}{lc}\therefore & \text { From Eq. (i), } 1+b=3 \Rightarrow b=2 \\ \Rightarrow & \text { Order of the reaction }(n)=a+b=1+2=3 \\ \Rightarrow & \text { Order of the reaction wrt. } A=1 \\ \Rightarrow & \text { Order of the reaction wrt. } B=2\end{array}$
BETA
