Chemical Kinetics Result Question 16
16 The following results were obtained during kinetic studies of the reaction;
(2019 Main, 9 Jan I)
$ 2 A+B \longrightarrow \text { Products } $
$ \begin{array}{|c|c|c|c|} \hline \text { Experiment } & \begin{array}{l} \begin{array}{c} {[A]} \\ \text { (in } \mathrm{mol} \mathrm{L}^{-1} \text { ) } \end{array} \end{array} & \begin{array}{l} \begin{array}{c} {[B]} \\ \text { (in } \mathrm{mol} \mathrm{L}^{-1} \text { ) } \end{array} \end{array} & \begin{array}{l} \text { Initial rate of } \\ \text { reaction } \\ \text { (in } \mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1} \text { ) } \end{array} \\ \hline \text { I. } & 0.10 & 0.20 & 6.93 \times 10^{-3} \\ \hline \text { II. } & 0.10 & 0.25 & 6.93 \times 10^{-3} \\ \hline \text { III. } & 0.20 & 0.30 & 1.386 \times 10^{-2} \\ \hline \end{array} $
The time (in minutes) required to consume half of $A$ is
(a) 5
(b) 10
(c) 100
(d) 1
Show Answer
Answer:
Correct Answer: 16. ( b )
Solution:
Let, the rate expression is $r \propto[A]^a[B]^b$.
From experiment I, $ \begin{array}{cc} \dfrac{r_2}{r_1}=\left(\dfrac{0.1}{0.1}\right)^a \times\left(\dfrac{0.25}{0.20}\right)^b \\ \Rightarrow \quad & \dfrac{6.93 \times 10^{-3}}{6.93 \times 10^{-3}}=1 \times\left(\dfrac{5}{4}\right)^b \\ \Rightarrow \quad 1=\left(\dfrac{5}{4}\right)^b \Rightarrow\left(\dfrac{5}{4}\right)^0=\left(\dfrac{5}{4}\right)^b \Rightarrow b=0 \end{array} $
From experiment II, $\dfrac{r_3}{r_1}=\left(\dfrac{0.2}{0.1}\right)^a \times\left(\dfrac{0.30}{0.20}\right)^b$
$ \begin{aligned} & \Rightarrow \quad \dfrac{1.386 \times 10^{-2}}{0.693 \times 10^{-2}}=(2)^a \times(1.5)^0 \\ & \Rightarrow \quad 2=2^a \times 1 \Rightarrow 2^1=2^a \Rightarrow a=1 \\ & \Rightarrow \text { So, } \quad r \propto[A]^1[B]^0 \Rightarrow r \propto[A] \\ & \end{aligned} $
Order of the reaction $(n)=1$
$\Rightarrow$ Now, let for the 1st experiment,
$ \begin{gathered} r_1=k \cdot[\mathrm{A}] \\ \Rightarrow \quad k=\dfrac{r_1}{[A]}=\dfrac{6.93 \times 10^{-3}}{0.1}=6.93 \times 10^{-2} \mathrm{~s}^{-1} \\ \Rightarrow \quad t_{50}=\dfrac{0.693}{k}=\dfrac{0.693}{6.93 \times 10^{-2}}=10 \mathrm{~s} \end{gathered} $
BETA
