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

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 6.60×104 s1?

Part B

What is the rate constant of a first-order reaction that takes 7.70 minutes for the reactant concentration to drop to half of its initial value?

Part C

A certain first-order reaction has a rate constant of 6.70×103 s1. How long will it take for the reactant concentration to drop to 18 of its initial value?

Explanation / Answer

A)

We know that half life (t 1/2) = ln 2 / K

where

K = rate constant

Given,

K = 6.6 x 10^-4 s-1

=> Half life = ln 2 / (6.6 x 10^-4) = 1050.2 s

B)

Half life is defined as the time taken for the concentration to drop to half of its initial value

Given,

t 1/2 = 7.7 min = 462 s

K = ln 2 / (t 1/2)

=> K = ln 2 / 7.7 = 0.09 min-1

or K = ln 2 / 462 = 1.5 x 10^-3 s-1

C)

For a first order reaction

C(t) = Co exp (-Kt)

where,

C(t) = concentration at time t

Co = Initial concentration

Given,

C(t) = 18 % of Co

=> C(t) = 0.18 Co

=> 0.18 Co = Co exp (- 6.7 x 10^-3 t)

=> ln (0.18) = (- 6.7 x 10^-3 t)

=> t = 255.94 s

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