Answer :
Answer:
Approximately [tex]1.7 \times 10^3\; \rm s[/tex].
Explanation:
In a first-order reaction, the reaction rate is proportional to the concentration of a single reactant (call it [tex]A[/tex].) The half-life [tex]t_{1/2}[/tex] of that first-order reaction would be the time it takes for [tex][A][/tex] (concentration of [tex]A[/tex]) to drop to [tex]1/2[/tex] its initial value.
The equation for the half-life of a first-order reaction is
[tex]\displaystyle t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}[/tex],
where [tex]k[/tex] is the rate constant of this reaction.
In this case,
[tex]\begin{aligned} t_{1/2} &= \frac{\ln 2}{k} \\ &\approx \frac{0.693}{k} = \frac{0.693}{4.2 \times 10^{-4}\; \rm s^{-1}} \\ &\approx 1.7\times 10^3\; \rm s \end{aligned}[/tex].