Consider the following hypothetical aqueous reaction: A(aq)→B(aq). A flask is charged with 0.065 mol (not molarity) of A in a total volume of 100.0 mL. The following data are collected: Time (min) 0 10 20 30 40
Moles of A 0.065 0.051 0.042 0.036 0.031

This is a second order reaction.
What is the value of the rate constant for the reaction?

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mauricioalessandrell mauricioalessandrell · Answer 1 · 2019-09-11T02:44:46+02:00

Answer:

The value of the constant is k = 0.042 M⁻¹ min⁻¹

Explanation:

The equation of velocity for this reaction can be written as follows:

v = k[A]²

where:

v = velocity

[A] = concentration of A ([A] = 0.065 mol / 0.1 l = 0.65 M)

The velocity of this reaction can also be written as follows:

v = -d[A]/dt

where d[A] and dt are infinitesimal variation of [A] and t respectively.

Then:

-d[A]/dt = k[A]² (separating variables)

-d[A]/[A]² = k dt

( integrating both sides from [A] (concentration fo A at time t) to [A]₀ (initial concentration) and from t = t to t = 0):

1/[A] - 1/[A]₀ = kt

Solving for k:

(1/[A] - 1/[A]₀) / t = k

Now we can replace with the data provided, for example:

at time t = 10 min, [A] = 0.051 mol / 0.1 l = 0.51 M

Then:

k = (1/0.51 M - 1/0.65 M) / 10 min = 0.042 M⁻¹ min⁻¹

The same should be done with every pair of data and the average of k can be calculated:

k = (1/0.42 M - 1/0.65) / 20 min = 0.042 M⁻¹ min⁻¹

k = (1/0.36 M - 1/0.65) / 30 min = 0.041 M⁻¹ min⁻¹

k = (1/0.31 M - 1/0.65) / 40 min = 0.042 M⁻¹ min⁻¹

The average value of k is:

k = 0.042 M⁻¹ min⁻¹

The same could be done plotting 1/[A] vs t and the slope of the straight line will be k (see attachment):

1/[A] = kt + 1/[A]₀

y = mx + b