Question

In a reaction $A\to$ Products, when start is made from $8.0\times {10}^{-2}M$ of $A$, half-life is found to be $120$ minute. For the initial concentration $4.0\times {10}^{-2}M$, the half-life of the reaction becomes $240$ minute. The order of the reaction is:

• A. $0$
• B. $1$
• C. $2$
• D. $0.5$

Answer 1

The correct option is C $2$

From the half life reaction,

$\frac{{\left(t_{1/2}\right)}_1}{{\left(t_{1/2}\right)}_2}={\left(\frac{a_2}{a_1}\right)}^{n-1}$

$\frac{120}{140}={\left(\frac{4\times {10}^{-2}}{8\times {10}^{-2}}\right)}^{n-1}$

$n=2$

Hence, option C is correct.