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Question
Consider a certain reaction \[\ce{A -> Products}\] with k = 2.0 × 10−2 s−1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L−1.
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Solution 1
Given: k = 2.0 × 10−2 s−1
T1 = 0
[A]1 = 1.0 M
T2 = 100 s
[A]2 = ?
k = `1/(T_2 - T_1) ln ([A]_1)/([A]_2)`
2.0 × 10−2 = `1/(100 - 0) ln 1.0/([A]_2)`
(2.0 × 10−2) × 100 = `ln 1.0/([A]_2)`
2 = `ln 1/([A]_2)`
7.389 = `1/([A]_2)`
∴ [A]2 = `1/7.389`
= 0.135 mol L−1
Solution 2
The units of k (s−1) indicate that the reaction is of first order. For a first-order reaction,
k = `2.303/t log_10 [A]_0/([A])`
∴ 2.0 × 10−2 = `2.303/100 log_10 1.0/([A])`
2.0 × 10−2 = `-2.303/100 log_10 [A]`
⇒ log10 [A] = `-(2.0 xx 10^-2 xx 100)/2.303`
= `-2/2.303`
⇒ log10 [A] = −0.8684
⇒ [A] = antilog10 (−0.8684)
= 0.1354 mol L−1
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