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

Question

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

Aicharya · Accepted Answer

$\frac{\left(t_{1 / 2}ight)_{1}}{\left(t_{1 / 2}ight)_{2}}=\left(\frac{a_{2}}{a_{1}}ight)^{n-1} ; \frac{120}{240}=\left(\frac{4 	imes 10^{-2}}{8 	imes 10^{-2}}ight)^{n-1} ; n=2$

In a reaction $A ightarrow$ Products, when start is made from $8.0 \times 10^{-2} \mathrm{M}$ of $A$,…

Solution