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Reactant Concentrations and Time
Rate equations (laws) relate rate of reaction to rate constant and reactant concentrations, which enable us to calculate the rate of reaction from the rate constant and reactant concentrations. But it would be useful to know how the concentrations of reactants change during the course of a reaction. Let us consider the simplest case- first-order reaction- to illustrate this application.
First-Order Reaction
In the first-order reaction, only one reactant is involved and the rate depends on the 1st power of the reactant concentration.
A Æ products
The rate in terms of rate of disappearance of reactant is
[ A ] rate t
From the rate law, we know that
rate = k [ A ]
Therefore,
rate = k[ A ] =
[ A ]
t
By rearranging this, we get
[ ] [ ]
A
A
− = k ∆t
This is a differential equation and can be integrated between the time limit of t=0 and t=t to produce
[ ] ln [ ]
t o
A
k t A
= − or [ A ]t = [ A ] 0 e -k t^ = [ A ] 0 exp(-k t ) (1)
Where [ A ] 0 is the concentration of A at zero time and [ A ]t is the concentration of A at time t=t. The above equation can be rearranged to give the equation of straight line (y = a
- b x, where a is an intercept, b is a slope, x is independent variable, and y is dependent variable):
ln [A]t = ln A] 0 – k t (y = a +b x)
When we plot ln[A]t against t, we get a straight with slope = k and intercept = ln [A] 0. This is another way of calculating the rate constant k. The Equation (1) is very useful and has some interesting applications, such as,
(a) To calculate the concentration after certain time (b) To determine how long it takes for a concentration to decrease to a certain amount (c) To compute the time required to convert certain percentage of original concentration (d) To calculate the half-life of a reaction
Let us illustrate above four applications with a specific example.
Example
The dissociation of hydrogen peroxide (disinfectant) to water and oxygen
2 H 2 O 2 (l) Æ 2 H 2 O (l) + O 2 (g)
is a first order reaction with rate constant k = 7 x 10 -4^ /s.
(a) If the starting (initial) concentration of H 2 O 2 solution is 0.75 M, what will be the concentration after 15 min?
Solution
Here [A] 0 =0.75 M, k = 7 x 10 -5^ /s, and t= 15 min x 60 s/min = 900 s. Substitute these values into Equation (1) and solve for [A] (^) t
[A]t = 0.75 M exp (- 7 x 10 -4^ /s x 900 s) = 0.75 M exp(-0.63) = 0.39 M
(b) How long will it takes for concentration of H 2 O 2 decreases from 0.75 M to 0.1M?
Solution
Here [A] 0 = 0.75 M, [A]t = 0.1 M, and k = 7 x 10-5^ /s. Substitute these values into
Equation (1) and solve for t.
ln(0.1 M / 0.75 M) = - 7 x 10 -4^ (M/s) x t (s) -2.015 = - 7 x 10-4 x^ t Therefore t= 2878.5 s = 48 min
- Since t (^) 1/2 is inversely proportional to k, it gives some idea about the magnitude of the rate constant – longer the t (^) 1/2 , smaller the k , and vice verse.
Example
The half-life for isomerization of gaseous cyclobutane to butadiene is 35 min. Calculate its rate constant.
Solution
The half-life is given in minutes and hence it needs to converted into seconds
t (^) 1/2 = 35 min = 35 min x 60 s/min = 2100 s
Substitute these values into Equation (3) to calculate k
1/ 2
k t
= = 0.693 / 2100 s = 3.3 x 10 -3 (^) / s
Something You Should Remember
The first-order rate Equation (1) and half-life Equation (3) have important application in nuclear chemistry in determining the age of an ancient object (Carbon dating technique).
