The reactant concentration in a first-order reaction was 7.10×10−2 M after 30.0 s and 6.00×10−3 M after 60.0 s . ?

Method 1:

Integrated form of first-order reaction:
ln[A] = -kt + ln[A]ₒ

[A]ₒ = 7.10 × 10⁻² M
When t = (60.0 - 30.0)s = 30.0 s, [A] = 6.00 × 10⁻³ M
ln(6.00 × 10⁻³) = -k (30 s) + ln(6.00 × 10⁻³)
k (30.0 s) = ln(7.10 × 10⁻²) - ln(6.00 × 10⁻³)
k = [ln(7.10 × 10⁻²) - ln(6.00 × 10⁻³)] / (30.0 s)
Rate constant, k = 8.24 × 10⁻² s⁻¹

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Method 2:

Let n be the number of half-lives from 30.0 s to 60.0 s.

(6.00 × 10⁻³)/(7.10 × 10⁻²) = (1/2)ⁿ
(1/2)ⁿ = 6/71
log(1/2)ⁿ = log(6/71)
n log(1/2) = log(6/71)
n = log(6/71) / log(1/2) = 3.565

Half-life, t½ = [(60 - 30) s] / 3.565 = 8.415 s
Rate constant, k = ln(2) / t½ = ln(2) / (8.415 s) = 8.24 × 10⁻² s⁻¹

Here is a solution courtesy of the Chemteam

Set the first concentration to be Ao and the second to be A. The time will be 30. seconds.

ln A = -kt + ln Ao
ln 6.00 x 10-3 = - (k) (30. s) + ln 7.10 x 10-2

-5.11599581 = - (k) (30. s) + -2.645075402

2.471 = (k) (30. s) 

k =0.0823 s-1 (to three sig figs)