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:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]5.50\times 10^{-3}s^{-1}[/tex]
t = time passed by the sample = ?
a = let initial amount of the reactant = x
a - x = amount left after decay process = [tex]\frac{x}{8}[/tex]
Now put all the given values in above equation, we get
[tex]t=\frac{2.303}{5.50\times 10^{-3}}\log\frac{x}{(\frac{x}{8})}[/tex]
[tex]t=378.1s[/tex]
Thus, the time taken will be, 378.1 s
