Question

If half life of a first order reaction is $2.31\times {10}^3$ min, the time it takes for one-fifth of the reaction to be left behind in min. is__________. (give the answer in 1000x form)

Answer 1

I. First method
k = $\frac{0.693}{2.31\times {10}^{3}min}=\frac{2.303}{t}log\frac{a}{a/5}$
Solve for t :
t = 5366 min
II. Second method
Use the relation
$\frac{t_{1/5}\left(left\right)}{t_{1/2}}=\left(\frac{log\frac{a}{a-x}}{0.3}\right)$
$t_{1/5\left(left\right)}=t_{1/2}\times \left(\frac{log\frac{1}{1/5}}{0.3}\right)$
= $\frac{2.31\times {10}^{3}min\times log5}{0.3}$
= $\frac{2.31\times {10}^3\times 0.7}{0.3}$= 5390 min
[Approximate value of log 5 = 0.7 has been taken, so slight variation in the answer by two methods]