Question

The following graph shows how $(t_{1/2})$ a (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

The Correct Option is C

For second order reaction,

It can be represented as : $-\frac{dx}{dt}=Kx^{2}$
$\frac{1}{\left[a\right]}=\frac{1}{\left[a_{0}'\right]}+Kt; Kt_{1/2}=\frac{1}{\left[a_{0}'\right]}$
half-life $\left(t_{1/2}\right)=\frac{1}{K\left[a_{0}'\right]}$
So, graph of $t_{1/2}$ and $\frac{1}{a_{0}}$, is linear.