Question

The graph between $\log T_{50}$ and $\log \left(\frac{1}{a}\right)$ (where $T_{50}$ and a are the half - life period and the concentration respectively) for a $n^{th}$ order reaction is of the type:

What is the value of n?

Answer 1

$Rate=k[A{]}^n$
Half-life of $n^{th}$ order by solving the integrated form of the above equation
$\frac{A_0^{1-n}}{2^{1-n}\times (1-n)}-\frac{A_0^{1-n}}{(1-n)}=kt$
$(n-1)log\frac{1}{A_0}+log(\frac{1}{\left(2{)}^{1-n}\right(1-n)}-\frac{1}{1-n})=logk+log{t}_{50}$
slope is $n-1$ which is equal to $tan{45}^0$
$n-1=1$
$n=2$