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 18 of its initial value?

Answer 1

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:

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

where,

k = rate constant  = $5.50\times 10^{-3}s^{-1}$

t = time passed by the sample  = ?

a = let initial amount of the reactant = x

a - x = amount left after decay process = $\frac{x}{8}$

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

$t=\frac{2.303}{5.50\times 10^{-3}}\log\frac{x}{(\frac{x}{8})}$

$t=378.1s$

Thus, the time taken will be, 378.1 s