Question

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 1

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

Answer 2

The integrated rate law for a second-order reaction is given by:

$\frac{1}{[A]t} = \frac{1}{[A]0} + kt$

where, [A]t= the concentration of A at time t,

[A]0= the concentration of A at time t=0

k = the rate constant for the reaction

Given: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min

Hence, $\frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)$

                                        = 4.858

Therefore, [A]t= 0.2058 M.

Answer: Concentration of A, after 180 min, is 0.2058 M