Question

In the reaction: $A\to P$, the rate is doubled which the concentration of 'A' is quadrupled. If 50% of the reaction occurs in $8\sqrt{2}hr$, how long (in hours) would it take for completion of next 50% reaction.

Answer 1

Let rate =$k[A{]}^x$

$2\times rate=k[4A{]}^x$

$4^x=2=>x=\frac{1}{2}$

$-\frac{d\left[A\right]}{dt}=k[A{]}^{1/2}$

$-{\int }_a^{at}\left[A{]}^{-1/2}d\right[A]=k{\int }_0^{t}dt$

$-\frac{1}{2}(\sqrt{at}-\sqrt{a})=kt$

$2kt=\sqrt{a}-\sqrt{at}$

For 50% completion, $at=\frac{a}{2}$

$2\times k\times 8\sqrt{2}=\sqrt{a}-\frac{\sqrt{a}}{\sqrt{2}}$

$k=\frac{(\sqrt{2}-1)}{32}\sqrt{a}$

For completion of 100% reaction,

$a_t=0$

$2\times k\times t=\sqrt{a}$

$t=\frac{\sqrt{a}}{2\times (\sqrt{2}-1)}\times 32=\frac{16}{\sqrt{2}-1}=16\sqrt{2}+16$ hrs

time for completion of next 50% reaction

$=(16\sqrt{2}+16)-8\sqrt{2}=8\sqrt{2}+16=27.31$ hrs