Question

The first-order reaction A→products has t1/2= 170 s .

a)What percent of a sample of A remains unreacted 850 s after a reaction has been started? Express your answer using two significant figures.

b)What is the rate of reaction when [A]= 0.50 M ? Express your answer using two significant figures

Answer 1

a) After t½ concentration has dropped to one half of it initial value.
For first order reactions, half life is independent from concentration level. Therefore: after an additional period of t½ concentration has dropped to one half of one half, i.e. one quarter of the initial value.

850s represents quintuple half life. So the concentration has dropped to:
[A]/[A]₀ = (1/2)^5 = 1/32 = 3.215%

More formally you can deduce it from integrated rate law
ln[A] = ln[A]₀ - k∙t
<=>
ln( [A]/[A]₀ ) = - k∙t
<=>
[A]/[A]₀ = e^(- k∙t)
First order rate constant is related to half life as:
k = ln(2)/t½
Hence:
[A]/[A]₀ = e^(- ln(2)∙t/t½) = 2^(-t/t½) = (1/2)^(t/t½)

b)

k = ln(2)/t½
= ln(2)/170s
= 4.07×10⁻³s⁻¹

rate = k∙[A]
= 4.07×10⁻³s⁻¹ x 0.50 M
= 2.035×10⁻³Ms⁻¹