Question

A graph between $t_{1/2}$ and concentration for $n^{th}$ order reaction is a straight line. Reaction of this nature is completed 50% in 10 minutes when concentration is 2 mol $L^{-1}.$ This is decomposed 50% in $t$ minutes at 4 mol $L^{-1}$. Then $n$ and $t$ are respectively:

• A. 0, 20 min
• B. 1, 10 min
• C. 1, 20 min
• D. 0, 5 min

Answer 1

The correct option is B 1, 10 min

It can be observed from the given graph that $t_{1/2}$ is constant and is independent of the concentration. We know that,

$t_{1/2}\propto (1/a_o{)}^{n-1}$

where, $a_o$ is the initial concentration of the reactant

$n$ is the order of the reaction

Now for first order reaction $n=1$ and $t_{1/2}$ will be independent of $a_o$ (i.e. concentration) as depicted in the graph.

Hence, the reaction is of first order.

As,$t_{1/2}$ is independent of initial concentration , it will be same whether the concentration is $2$ $Mol$ $/L$ or $4$ $Mol$ $/L$

So, $n=1$ and $t=10$ minutes (as already given in the question)