Question

A graph between log $t_{1/2}$ and log a (abscissa), 'a' is the initial concentration of A in the reaction, is given. For the reaction, A $\to$ Product, the rate law is:

• A. $\frac{-d\left[A\right]}{dt}=K$
• B. $\frac{-d\left[A\right]}{dt}=K\left[A\right]$
• C. $\frac{-d\left[A\right]}{dt}=K[A{]}^2$
• D. $\frac{-d\left[A\right]}{dt}=K[A{]}^3$

Answer 1

Answer: (C)

$\frac{-d\left[A\right]}{dt}=K[A{]}^2$

We know,

$t_{1/2}=K{a}^{1-n},$ n is the order and K is the rate constant.
$\log \left(t_{1/2}\right)=\log K+(1-n)\log a$
When $\log \left(t_{1/2}\right)$ is plotted against log a, the slope is $(1-n)$. But it is equal to -1.
Hence, $1-n=-1$ or $n=2$.
Thus, the reaction is second order and the rate law is $\frac{-d\left[A\right]}{dt}=K[A{]}^2$.