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 :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{29.8s}[/tex]
[tex]k=0.0232s^{-1}[/tex]
Now we have to calculate the time passed.
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]0.0232s^{-1}[/tex]
t = time passed by the sample = ?
a = let initial amount of the reactant = x M
a - x = amount left after decay process = [tex]x-\frac{1}{4}\times (x)=\frac{3}{4}\times (X)=\frac{3x}{4}M[/tex]
Now put all the given values in above equation, we get
[tex]t=\frac{2.303}{0.0232s^{-1}}\log\frac{x}{(\frac{3x}{4})}[/tex]
[tex]t=12.4s[/tex]
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.
