Question

A first-order reaction has a half-life of 29.8 s . How long does it take for the concentration of the reactant in the reaction to fall to one-fourth of its initial value?

Answer 1

Answer : The time taken for the concentration of the reactant in the reaction to fall to one-fourth of its initial value is, 12.4 seconds.

Explanation :

Half-life = 29.8 s

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{29.8s}$

$k=0.0232s^{-1}$

Now we have to calculate the time passed.

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  = $0.0232s^{-1}$

t = time passed by the sample  = ?

a = let initial amount of the reactant  = x M

a - x = amount left after decay process = $x-\frac{1}{4}\times (x)=\frac{3}{4}\times (X)=\frac{3x}{4}M$

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

$t=\frac{2.303}{0.0232s^{-1}}\log\frac{x}{(\frac{3x}{4})}$

$t=12.4s$

Therefore, the time taken for the concentration of the reactant in the reaction to fall to one-fourth of its initial value is, 12.4 seconds.