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%

Explanation:

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

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

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

$t=\frac{2.303}{k}\log\frac{100}{50}$

$t=\frac{0.693}{0.1}$

$t=6.93sec$

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

$t=\frac{2.303}{k}\log\frac{100}{100-75}$

$t=\frac{2.303}{0.1}\times \log\frac{100}{25}$

$t=13.9sec$

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