Question

Assume that you have developed a tracer with first-order kinetics and k = 0.1 s − 1 . For any given mass, how long will it take the tracer concentration to drop by (a) 50% and (b) by 75%?

Answer 1

Answer: It will take 6.93 sec for tracer concentration to drop by 50% and 13.9 sec for tracer concentration to drop by 75%

Step-by-step explanation:

Expression for rate law for first order kinetics is given by:

`t=(2.303)/(k)\log(a)/(a-x)`

where,

k = rate constant =
`0.1s^(-1)`

t = age of sample

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process

a) for tracer concentration to drop by 50%

a - x = amount left after decay process = 50

`t=(2.303)/(k)\log(100)/(50)`

`t=(0.693)/(0.1)`

`t=6.93sec`

It will take 6.93 sec for tracer concentration to drop by 50%

b) for tracer concentration to drop by 75 %

a - x = amount left after decay process = 25

`t=(2.303)/(k)\log(100)/(100-75)`

`t=(2.303)/(0.1)* \log(100)/(25)`

`t=13.9sec`

It will take 13.9 sec for tracer concentration to drop by 75%.