Half-life (kinetics) for First Order Reactions The integrated rate law allows ch

Question

Half-life (kinetics) for First Order Reactions The integrated rate law allows chemists to predict the reactant concentration after a certain amount of time, or the time it would take for a certain concentration to be reached. The integrated rate law for a first-order reaction is: [A] = [A]0e^-kt Now say we are particularly interested in the time it would take for the concentration to become one-half of its initial value. Then we could substitute [A]0/2 for [A] and rearrange the equation to: t1/2 = 0.693/k This equation calculates the time required for the reactant concentration to drop to half its initial value. In other words, it calculates the half-life. Part A What is the half-life of a first-order reaction with a rate constant of 8.70x10^-4 s^-1? Express your answer with the appropriate units. Part B What is the rate constant of a first-order reaction that takes 502seconds 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 6.80x 10^-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

For a First order reaction

t1/2 = 0.693/ k                 {where t1/2 is the half-life of the reaction and k is the rate constant}

PART A

t1/2 = 0.693 /8.70 x 10-4 s-1 = 796.55 s

PART B

t1/2 = 0.693/ k

therefore

k = 0.693/ t1/2 = 0.693/ 502 s = 1.38 x 10-3 s-1

PART C

For a first order reaction

[A] = [A0]e-kt

where k is the rate constant, [A0] is the initial concentration and [A] is the concentration after time t

As per the problem [A] = [A0]/8. Substituting [A] = [A0]/8 in [A] = [A0]e-kt we get

[A0]/8 = [A0]e-kt

or 1/8 = e-kt

or ln(1/8) = -kt

-2.08 = - 6.80 x 10-3 x t

therefore

t = 2.08 / 6.80 x 10-3 = 305.88 s

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