Q47 / 200 · NEET 2024 (Re-NEET) · Chemistry
Following data is for a reaction between reactants A and B :
The order of the reaction with respect to $A$ and $B$, respectively, are
The order of the reaction with respect to $A$ and $B$, respectively, are
- A1,0
- B0,1
- Correct answerC1,2
- D2,1
Solution
Let the rate equation is
Rate $=\mathrm{k}[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}$
Therefore, we can write
$2 \times 10^{-3}=k[0.1] \times\left[0.1 \rrbracket^y\right.$...(i)
$4 \times 10^{-3}=k[0.2]^x[0.1]^y$...(ii)
$1.6 \times 10^{-2}=\mathrm{k}[0.2]^{\times}[0.2]^{\mathrm{y}}$...(iii)
(ii) $\div$ (i);
$\frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\frac{k[0.2]^x[0.1]^y}{k[0.1]^x[0.1]^y}$
$\Rightarrow \quad \frac{2}{1}=\frac{(0.2)^x}{(0.1)^x}=\left(\frac{2}{1}\right)^x$
$\therefore \quad x=1$
(ii) $\div$ (iii);
$\frac{4 \times 10^{-3}}{1.6 \times 10^{-2}}=\frac{k[0.2]^{\mathrm{x}}[0.1]^y}{\mathrm{k}[0.2]^{\mathrm{x}}[0.2]^{\mathrm{y}}}$
$\Rightarrow \quad \frac{1}{4}=\frac{(0.1)^y}{(0.2)^y}=\left(\frac{1}{2}\right)^y$
$\therefore \quad y=2$
$\therefore \quad$ Rate $=k[A]^1[B]^2$
First order with respect to A while second order with respect to B .
Hence, the correct answer is option: 'C'.