Question

The following reaction was monitored as a function of time:
AB→A+B
A plot of 1/[AB] versus time yields a straight line with slope 5.8×10⁻² (M⋅s)−1 .

1)What is the half-life when the initial concentration is 0.59 M ?

2)if the initial concentration of AB is 0.290 M , and the reaction mixture initially contains no products, what are the concentrations of A and Bafter 80  s ?

Answer 1

The following reaction was monitored as a function of time:
AB -> A + B

For zeroth order reaction, [AB] vs t is a straight line
For first order reaction, log[AB] vs t is a straight line

For second order reaction, 1/[AB] vs t is a straight line and slope is k as
the integral expression of second order reaction is 1/[AB] = k*t + 1/[AB]0

A plot of y = 1/[AB] versus x = time yields a straight line with slope 5.8E−2 (M⋅s)−1 .

So reaction is second order with rate constant, k = 5.8E−2 (M⋅s)−1 = 0.058 (M⋅s)−1

1)What is the half-life when the initial concentration is 0.59 M?

[AB]0 = 0.59 M
t1/2 = 1/(k*[AB]0) = 1/(0.058*0.59) = 29.22 s

2)if the initial concentration of AB is 0.290 M , and the reaction mixture
initially contains no products, what are the concentrations of A and B after 80 s?

[AB]0 = 0.290 M
t = 80 s
k = 0.058 (M⋅s)−1
1/[AB] = k*t + 1/[AB]0 = 0.058*80 + 1/0.29 = 8.1 M-1
[AB] = 1/8.1 = 0.124 M
Amount of [AB] reacted, D[AB] = [AB]0 - [AB] = 0.29 - 0.124 = 0.166 M
AB -> A + B
amount of [A] produced = amount of [B] produced = amount of [AB] reacted = 0.166 M