Half-life equation for first-order reactions: t 1/2=0.693 k where t 1/2 is the h

Question

Half-life equation for first-order reactions:
t1/2=0.693k  
where t1/2 is the half-life in seconds (s), and k is the rate constant in inverse seconds (s1).

Part A

What is the half-life of a first-order reaction with a rate constant of 5.40×104  s1?

Express your answer with the appropriate units.

Part B

What is the rate constant of a first-order reaction that takes 416 seconds for the reactant concentration to drop to half of its initial value?

Express your answer with the appropriate units.

Part C

A certain first-order reaction has a rate constant of 4.50×103 s1. How long will it take for the reactant concentration to drop to 18 of its initial value?

Express your answer with the appropriate units.

Explanation / Answer

a) t 1/2 = 0.693 / 5.4 x 10-4 = 1283.33 s

b) ln 0.5 = - k x 416
      k = - ln0.5/416   

        =1.6662x10-3

c) ln 18 = - 4.50 x 10-3 t
t = -ln18/4.50 x 10-3

=-2.8903/4.50 x 10-3

=642.2888 s

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

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