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 (s−1).

A. What is the half-life of a first-order reaction with a rate constant of 5.60×10⁻⁴

s−1?

B. What is the rate constant of a first-order reaction that takes 150 seconds for the reactant concentration to drop to half of its initial value?

C. A certain first-order reaction has a rate constant of 7.20×10⁻³ s−1. How long will it take for the reactant concentration to drop to 18 of its initial value?

Answer 1

As we know for a first order reaction,

Half life (t_1/2) = 0.693 / k

where,

k = Rate constant for a first order reaction.

A)

k = 5.60 * 10^-4 sec^-1

Half life (t_1/2) = 0.693 / 5.60 * 10^-4 sec = 1237.5 seconds

B)

We are given,

Half life (t_1/2) = 150 seconds

As,

Half life (t_1/2) = 0.693 / k

150 = 0.693 / k

k = 0.693 / 150 sec^-1 = 4.62 * 10^-3 sec^-1

C)

According to the integrated rate law the change in concentration of reactant for a first order reaction is given by equation,

ln[A] = ln[A₀]- kt

Note : Here ln is the natural log

where,

[A] = Final Concentration of reactant (after time t)

[A₀] = Initial Concentration of reactant

k = Rate constant

t = Time (in seconds)

We are given,

[A] = [A₀] / 18

k = 7.20 * 10^-3 sec^-1

On substituting the given values in above equation we get,

ln[A₀ /18] = ln[A₀] - (7.20 * 10^-3) * t

ln[A₀ /18] - ln[A₀] = - (7.20 * 10^-3) * t

ln[1/18] = - (7.20 * 10^-3) * t                            ( ln[A] - ln[B] = ln[A / B] )

-2.890 = - (7.20 * 10^-3) * t

t = 401.388 seconds