A certain first-order reaction has a rate constant of 5.50×10−3 s−1. How long will it take for the reactant concentration to drop to 18 of its initial value?

This is an incomplete question, here is a complete question.

A certain first-order reaction has a rate constant of 5.50 × 10⁻³ s⁻¹. How long will it take for the reactant concentration to drop to 1/8 of its initial value?

Answer : The time taken will be, 378.1 s

Explanation :

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

where,

k = rate constant  =

t = time passed by the sample  = ?

a = let initial amount of the reactant = x

a - x = amount left after decay process =

Now put all the given values in above equation, we get

Thus, the time taken will be, 378.1 s