Question:
For the reaction, $2 \mathrm{~A}+\mathrm{B} \rightarrow$ products, when the concentrations of $A$ and $B$ both were doubled, the rate of the reaction increased from $0.3 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$ to $2.4 \mathrm{~mol}$ $\mathrm{L}^{-1} \mathrm{~s}^{-1}$. When the concentration of A alone is doubled, the rate increased from $0.3$ and $\mathrm{L}^{-1} \mathrm{~s}^{-1}$ to $0.6 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. Which one of the following statements is correct?
Correct Option: , 2
Solution:
$r=K[\mathrm{~A}]^{x}[\mathrm{~B}]^{y}$
$\frac{r_{2}}{r_{1}}=2^{x} \cdot 2^{y}=8 \Rightarrow x+y=3$
$\frac{r_{3}}{r_{1}}=2 x=2 \Rightarrow x=1$
$\therefore \quad y=2$