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).

What is the half-life of a first-order reaction with a rate constant of 8.30×104  s1?

What is the rate constant of a first-order reaction that takes 340 seconds for the reactant concentration to drop to half of its initial value?

A certain first-order reaction has a rate constant of 3.60×103 s1. How long will it take for the reactant concentration to drop to 18 of its initial value?

Explanation / Answer

t1/2 = 0.693/k

1). k = 8.30X10-4 s-1

t1/2 = 0.693/8.30X10-4

= 834.93 sec

= 13.91 min

2). t1/2 = 340 sec

k = 0.693/340

k = 2.03X10-3 s-1

3). k = 3.60X10-3 s-1

[A] = [A]o/18

We know that,

ln([A]/[A]o) = - k*t

ln1/18 = - 3.60X10-3 *t

- 2.890 = - 3.60X10-3 *t

t = 2.890/3.60X10-3

t = 802.77 sec

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