Question

The rate of the reaction $A\to$ products, at the initial concentration of $3.24\times {10}^{-2}M$ is nine times its rate at another initial concentration of $1.2\times {10}^{-3}M$ . The order of the reaction is

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{3}{2}$
• D. $\frac{2}{3}$
• E. $\frac{1}{3}$

Answer 1

Answer: (D)

For the reaction $A\to$ products; let the order of reaction is n. Then $r=k{[A]}^n$
$=k\times {(1.2\times {10}^{-3})}^n$ ...(i)
At the initial concentration of $3.24\times {10}^{-2}M,$ the rate is $\theta$ times.
Hence, $9r=k\times {(3.24\times {10}^{-2})}^n$ ...(ii)
On dividing, $\frac{9r}{r}=\frac{k\times {(3.24\times {10}^{-2})}^n}{k\times {(1.2\times {10}^{-3})}^n}$
$9={(27)}^n$
$9={(3^3)}^{2/3}$
$\therefore$ $n=\frac{2}{3}$