Half-life equation for first-order reactions: t1/2 = 0.693/k where t1/2 is the h

Question

Half-life equation for first-order reactions: t1/2 = 0.693/k where t1/2 is the half-life in seconds (s), and k is the rate constant in inverse seconds (s-1). What is the half-life of a first-order reaction with a rate constant of 1.00times10-4 s-1? Express your answer with the appropriate units. What is the rate constant of a first-order reaction that takes 538seconds for the reactant concentration to drop to half of its initial value? Express your answer with the appropriate units. A certain first-order reaction has a rate constant of 1.00times10-3 s-1. How long will it take for the reactant concentration to drop to 1/8 of its initial value? Express your answer with the appropriate units.

Explanation / Answer

t1/2 = 0.693/k a) t1/2 = 0.693/0.0001 = 6930 sec b) k = 0.693/t1/2 = 0.693/538 = 1.288*10 ^ -3 c)t1/2 = 0.693/k = 0.693/0.001 = 693 sec to drop to 1/8 , it would take, 3 * 693 = 2079 sec 100% sure of the answers Do rate the asnwers bro !Cheers :)

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.