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 1.70×104 s1?

Express your answer with the appropriate units.

Part B

What is the rate constant of a first-order reaction that takes 430 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 7.30×103 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

Part(A)

t1/2 = 0.693/k

given that, k = 1.70 x 10^-4 s-1

=> t1/2 = 0.693/1.70 x 10^-4

=> t1/2 = 4076.5 s

Part(B)

t1/2 = 0.693/k

given that, t1/2 = 430 s

=> k = 0.693/t1/2 = 0.693/430

=> k = 1.61 x 10^-3 s-1

Part(C)

t1/2 = 0.693/k

given that, k = 7.30 x 10^-3 s-1

=> t1/2 = 0.693/7.30 x 10^-3

=> t1/2 = 94.93 s

time take for the reactant concentration to drop to 1/8 of its initial value = 3 x t1/2
= 3 x 94.93
= 284.8 s

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