A second order reaction is 50 % complete in 16 minutes. How long will it take to be 82 % complete
(1/A) - (1/Ao) = kt
Determine k. A = 50. Ao = 100. t = 16 min.
Then use the same equation but A = 18(that's 100-82); Ao = 100; k from above. Solve for t in minutes.
To determine the time it takes for a second-order reaction to be 82% complete, we can use the concept of half-life. The half-life of a second-order reaction can be found using the formula:
t1/2 = 1 / (k * [A]0)
Where:
t1/2 is the half-life of the reaction,
k is the rate constant of the reaction, and
[A]0 is the initial concentration of the reactant.
Since the reaction is 50% complete in 16 minutes, we can use this information to calculate the rate constant (k). Since the reaction is second order, the concentration at any given time can be calculated using the equation:
1 / [A]t = kt + 1 / [A]0
Where:
[A]t is the concentration of the reactant at time t.
[A]0 is the initial concentration of the reactant.
Since the reaction is 50% complete, [A]t / [A]0 = 0.5. Substituting this value into the equation:
1 / 0.5 = k * 16 + 1 / [A]0
Simplifying the equation:
2 = 16k + 1 / [A]0
Now, we need to determine the initial concentration of the reactant [A]0. Since the reaction is 50% complete, the initial concentration is twice the concentration at the 16-minute mark. Therefore, [A]0 = 2 * [A]16.
Substituting this back into the equation:
2 = 16k + 1 / (2 * [A]16)
Now, we can solve for the rate constant (k). Once we have the value of k, we can use it to determine the half-life (t1/2) using the initial concentration [A]0.
Then, we can use the half-life to calculate the time required for the reaction to be 82% complete. Assuming the reaction follows second-order kinetics, we calculate the time using the following formula:
t = (ln([A]0 / [A]t)) / (k)
Substituting [A]t / [A]0 = 0.82, we can solve for t.