The rate constant for a second-order reaction is 0.54 M-1s-1. What is the half-life of this reaction if the initial concentration is 0.27 M
what is k?
To calculate the half-life of a second-order reaction, we can use the equation:
t1/2 = 1 / (k * [A]0)
Where:
t1/2 = half-life of the reaction
k = rate constant of the reaction
[A]0 = initial concentration of the reactant
Given:
k = 0.54 M-1s-1
[A]0 = 0.27 M
Substituting the given values into the equation, we have:
t1/2 = 1 / (0.54 M-1s-1 * 0.27 M)
t1/2 = 1 / (0.1458 M2s-1)
Simplifying, we get:
t1/2 ≈ 6.87 seconds (rounded to two decimal places)
Therefore, the half-life of the reaction is approximately 6.87 seconds.
To determine the half-life of a second-order reaction, we can use the following equation:
t_1/2 = 1 / (k * [A]_0)
Where:
t_1/2 is the half-life
k is the rate constant
[A]_0 is the initial concentration of the reactant
In this case, the rate constant (k) is given as 0.54 M^(-1)s^(-1) and the initial concentration, [A]_0, is given as 0.27 M.
Plugging these values into the equation, we get:
t_1/2 = 1 / (0.54 M^(-1)s^(-1) * 0.27 M)
Now, let's calculate the half-life:
t_1/2 = 1 / (0.1458 M^(-2)s^(-1))
Using a calculator, we find that:
t_1/2 ≈ 6.87 seconds
Therefore, the half-life of the reaction is approximately 6.87 seconds.
t1/2 = 1/ak[A]o
Substitute 0.27 for [A]o. The reaction is not given; usually in those circumstances a is assumed to be 1.