Question

a first order reaction is 50% complete in 25 minutes calculate the time for 80% completion reaction​

Answer 1

Answer : The time for 80% completion reaction will be, 58 minutes.

Solution :

As we know that the radioactive decay follow the first order kinetics.

Formula used :

$\frac{N}N_o}=(\frac{1}{2})^{\frac{t}{t_{1/2}}}$

by taking 'log' on both side, we get

$\log (\frac{N}{N_o})=\frac{t}{t_{1/2}}\times \log (\frac{1}{2})$

where,

N = amount left after decay process = 100 - 80 = 20

$N_o$ = initial amount = 100

t = time for 80% completion reaction​ = ?

$t_{1/2}$ = half life = 25 minutes

Now put all the given values in this formula, we get the time for 80% completion reaction.

$\log (\frac{20}{100})=\frac{t}{25min}\times \log (\frac{1}{2})$

$t=57.97=58min$

Therefore, the time for 80% completion reaction will be, 58 minutes.

Answer 2

Answer : The time for 80% completion reaction will be, 58 minutes.

Solution :

As we know that the radioactive decay follow the first order kinetics

N = amount left after decay process = 100 - 80 = 20

= initial amount = 100

t = time for 80% completion reaction​ = ?

= half life = 25 minutes

Now put all the given values in this formula, we get the time for 80% completion reaction.

Therefore, the time for 80% completion reaction will be, 58 minutes.