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.

Answer 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

Answer 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])`

= `-2.303/100 log_10 [A]`

or, log₁₀ [A] = `-(2.0 xx 10^-2 xx 100)/2.303`

or, log₁₀ [A] = `-2/2.303`

or, log₁₀ [A] = −0.8684

or, [A] = antilog₁₀ (−0.8684)

[A] = 0.1354 mol L⁻¹