Question

A graph between log $t_{1/2}$ and (log a) (abscissa) a being the initial concentration of A in the reaction. For 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. $\frac{-d\left[A\right]}{dt}=k{\left[A\right]}^2$
• D. $\frac{-d\left[A\right]}{dt}=k{\left[A\right]}^3$

Answer 1

The correct option is C $\frac{-d\left[A\right]}{dt}=k{\left[A\right]}^2$

$\log t_{1/2}=-\log a+c$

$\log t_{1/2}+\log a=c$

$\log t_{1/2}\times a=c$

$t_{1/2}\times a=cost$

$t_{1/2}=\frac{\cos t}{a}\Rightarrow t_{1/2}=(csc{)}^{-1}$

So unit of $t_{1/2}=(comcentration{)}^{1-n}$

$1-n=-1$

$n=2$

So, $\left(c\right)$