The reaction 2a → a2 was experimentally determined to be second order with a rate constant, k, equal to 0.0265 m–1min–1. if the initial concentration of a was 4.00 m, what was the concentration of a (in m) after 180.0 min?

Answer: 0.199M

Explanation:

1) An initial clarification: the unit of the concentration is M (molarity) and not m as it was written in the question (m is used for molality, and that is other unit of concentration).

2) Second order reaction means that the rate of reaction is given by:

r = - d[A]/dt = [A]²

3) By integration you get:

1 / [A] - 1[Ao] = kt

=> 1 / [A] = 1 / [Ao] + kt

4) Plug in the data; [Ao] = 4.00M; k = 0.0265 (M⁻¹) (min)⁻¹; t = 180 min

=> 1 / [A] = 1 / 4.00M + 0.0265 (M⁻¹)(min⁻¹) (180min) = 5.02 (M⁻¹)

=> [A] = 1 / (5.02(M⁻¹) = 0.199 M

The reaction 2a → a2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 m–1min–1. if the initial concentration of a was 4.00 m, what was the concentration of a (in m) after 180.0 min?

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The reaction 2a → a2 was experimentally determined to be second order with a rate constant, k, equal to 0.0265 m–1min–1. if the initial concentration of a was 4.00 m, what was the concentration of a (in m) after 180.0 min?