A reaction is first order overall. For a given sample, if its initial rate is 0.0200 mol L-1
s
-1
and 25.0 days later its rate dropped to 6.25 × 10-4 mol L-1
s
-1
, what is its half-life?
To determine the half-life of a reaction, we need to use the first-order rate equation. The equation is as follows:
rate = k * [A]
Where:
- rate is the reaction rate
- k is the rate constant
- [A] is the concentration of reactant A
Given that the reaction is first order overall, we know that the reaction rate is directly proportional to the concentration of the reactant. We can write the equation as:
rate1 = k * [A]1
rate2 = k * [A]2
Where:
- rate1 is the initial rate
- [A]1 is the initial concentration
- rate2 is the rate after a certain time
- [A]2 is the concentration after a certain time
We are given the initial rate (rate1 = 0.0200 mol L-1 s-1) and the rate after 25.0 days (rate2 = 6.25 × 10-4 mol L-1 s-1).
Now we need to convert the time from days to seconds. There are 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute. So, 25.0 days is equal to:
25.0 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute = 2,160,000 seconds
We can substitute the values into the rate equations:
rate1 = k * [A]1
0.0200 mol L-1 s-1 = k * [A]1
rate2 = k * [A]2
6.25 × 10-4 mol L-1 s-1 = k * [A]2
By dividing the second equation by the first equation, we can eliminate the rate constant (k):
(6.25 × 10-4 mol L-1 s-1) / (0.0200 mol L-1 s-1) = [A]2 / [A]1
Simplifying the equation, we get:
0.03125 = [A]2 / [A]1
Since [A]1 is the initial concentration and [A]2 is the concentration after a certain time, the ratio [A]2 / [A]1 represents the fraction of the reactant remaining after that time. In other words, 0.03125 represents the fraction of the reactant remaining after 2,160,000 seconds.
The half-life is defined as the time it takes for half of the reactant to be consumed. Therefore, we need to find the time it takes to reduce the reactant concentration to half.
0.03125 = 1/2 = (1/2)^n
Solving for n (the number of half-lives), we get n ≈ 4.959.
Now, we can determine the half-life of the reaction by multiplying the time it took for the rate to drop (2,160,000 seconds) by the number of half-lives:
Half-life = (2,160,000 seconds) * n
Half-life ≈ 10,699,040 seconds
Therefore, the half-life of the reaction is approximately 10,699,040 seconds, or about 124 days.
To determine the half-life of a first-order reaction, we can use the integrated rate law for a first-order reaction:
ln([A]t/[A]0) = -kt
Where:
[A]t is the concentration of the reactant at time t
[A]0 is the initial concentration of the reactant
k is the rate constant
t is the time
From the given information, we can find the value of k using the initial rate and the rate after 25.0 days.
Given:
Initial rate (r0) = 0.0200 mol L^-1 s^-1
Rate after 25.0 days (r) = 6.25 × 10^-4 mol L^-1 s^-1
Plugging these values into the equation, we have:
ln(r/r0) = -k * t
ln(6.25 × 10^-4 mol L^-1 s^-1 / 0.0200 mol L^-1 s^-1) = -k * 25.0 days
Taking the natural logarithm of both sides:
-4.137 = -k * 25.0
Rearranging the equation to solve for k:
k = -4.137 / 25.0
k = -0.16548 s^-1
The rate constant (k) for the reaction is approximately -0.16548 s^-1.
To find the half-life (t½), we can use the equation:
t½ = ln(2) / k
t½ = ln(2) / -0.16548 s^-1
Using this equation, we find that the half-life of the reaction is approximately 4.20 s.