Question

The following graph shows how $t_{1/2}$ (half-life) of a reactant $R$ changes with the initial reactant concentration $a_0$. The order of the reaction will be

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

Half-life period $t_{1/2}\propto \frac{1}{a_0^{n-1}}\dots$(i)
where, $a_0=$ initial concentration and $n=$ order of reaction since graph betweent ${}_{1/2}$ and $\frac{1}{a_0}$ is a straight line,
$t_{1/2}\propto \frac{1}{a_0}\dots$(ii)
From Eq (i) and (ii),
$\Rightarrow \frac{1}{a_0^{n-1}}=\frac{1}{a_0}$
or $n-1=1$
$\therefore n=2$
Hence, the reaction is of second order.