Question

In a graph between $\log t_{1/2}$ and $\log a$ (abscissa), ' $a$ ' being the initial conc. of $A$ in the reaction, $A\to$ product, the rate law is

• A. $-\frac{d[A]}{dt}=k[A{]}^{\circ }$
• B. $-\frac{d[A]}{dt}=k[A]$
• C. $-\frac{d[A]}{dt}=k[A{]}^2$
• D. None of these

Answer 1

Answer: (C)

$t_{1/2}=k_a^{(1-n)}$

$\log t_{1/2}=\log k+(1-n)\log a$

$\therefore 1-n=1,$

$\Rightarrow n=2$ (Second-order reaction)

$\therefore -\frac{d[A]}{dt}=k[A{]}^2$