Question

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

• A. $-\frac{d[ A ]}{d t}=k[ A ]^{\circ}$
• B. $-\frac{d[ A ]}{d t}=k[ A ]$
• C. $-\frac{d[ A ]}{d t}=k[ A ]^{2}$
• 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)

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