Q25 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Chemistry
Rate constants in the following reaction are
Reaction 1: $\mathrm{A} \stackrel{\text { catalyst } 1}{\longrightarrow} \mathrm{P}_1, \mathrm{k}_1=1 \mathrm{~s}^{-1}$
Reaction 2: A $\stackrel{\text { catalyst } 2}{\longrightarrow} \mathrm{P}_2, \mathrm{k}_2=0.1 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$
Reaction 3: A $\stackrel{\text { catalyst } 3}{\longrightarrow} \mathrm{P}_3, \mathrm{k}_3=0.01 \mathrm{~L}^2 \mathrm{~mol}^{-2} \mathrm{~s}^{-1}$
The correct relations between the rate of the reactions at $1 \mathrm{M}$ of $\mathrm{A}$ are
- A$r_1=\frac{r_3}{100}, r_2=\frac{r_3}{10}$
- B$\mathrm{r}_1=\frac{\mathrm{r}_2}{10}, \mathrm{r}_2=\frac{\mathrm{r}_3}{10}$
- C$r_1=100 r_3, r_2=\frac{r_3}{10}$
- Correct answerD$\mathrm{r}_1=10 \mathrm{r}_2, \mathrm{r}_3=\frac{\mathrm{r}_2}{10}$
Solution
The unit of rate constant indicates the order of reaction
$$
\begin{aligned}
& \mathrm{k}=-\left(\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\right) /[\mathrm{A}]^{\mathrm{n}} \\
\Rightarrow \quad & \mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1} / \mathrm{mol}^{\mathrm{n}} \mathrm{L}^{-\mathrm{n}}=\mathrm{mol}^{1-\mathrm{n}} \mathrm{L}^{\mathrm{n}-1} \mathrm{~s}^{-1} \\
& \mathrm{n}=1, \mathrm{k}=\mathrm{s}^{-1} ; \mathrm{n}=2, \mathrm{k}=\mathrm{mol}^{-1} \mathrm{Ls}^{-1} ; \mathrm{n}=3, \mathrm{k}=\mathrm{mol}^{-2} \mathrm{~L}^2 \mathrm{~s}^{-1} .
\end{aligned}
$$
$\therefore$ Reaction 1:1st order, Reaction 2:2nd order, Reaction $3:$ 3rd order
$$
\begin{aligned}
\therefore \quad \mathrm{r}_1 & =\mathrm{k}_1[\mathrm{~A}] & & \mathrm{r}_2=\mathrm{k}_2[\mathrm{~A}]^2 \mathrm{r}_3=\mathrm{k}_3[\mathrm{~A}]^3 \\
& =1 & & =0.1 \times 1^2=0.01 \times 1^3 \\
& =10 \mathrm{r}_2 & & =\frac{1}{10}=\frac{\mathrm{r}_1}{10}=\frac{1}{100}=\frac{\mathrm{r}_1}{100} \\
& =100 \mathrm{r}_3 & & \therefore \mathrm{r}_2=10 \mathrm{r}_3 \therefore \mathrm{r}_1=100 \mathrm{r}_3
\end{aligned}
$$
Hence, the correct answer is option: 'D'.