calculate the half-life of a first- order reaction if the concentration of the reactant is 0.0396 M at 12 seconds after the reaction starts and is 0.00579 M at 47 after the reaction starts. How long does it take for the reactant concentration to decrease to 0.00269 M
ln(No/N) = kt
No = 0.0396
N = 0.00579
k = solve
t = 47-12 = 35 seconds.
Solve for k, then
k = 0.693/t1/2 and solve for t1/2
Then
ln(No/N) = kt and plug in the new numbers to find time for the 0.00269M to be reached.
To determine the half-life of a first-order reaction, you can use the formula:
t1/2 = (0.693 / k)
Where:
t1/2 is the half-life of the reaction,
k is the rate constant of the reaction.
To find the rate constant (k), you need to use the concentration data at two different times and plug them into the following equation:
k = (ln(C1/C2)) / (t2 - t1)
Where:
C1 is the initial concentration,
C2 is the final concentration,
t1 is the time at the initial concentration,
t2 is the time at the final concentration.
Let's calculate the rate constant (k) using the provided data:
C1 = 0.0396 M, t1 = 12 s,
C2 = 0.00579 M, t2 = 47 s.
k = (ln(0.0396/0.00579)) / (47 - 12)
k = (ln(6.833)) / 35
k ≈ 0.0729 s^-1
Now, we can use the rate constant (k) to calculate the half-life (t1/2):
t1/2 = (0.693 / 0.0729)
t1/2 ≈ 9.5 seconds
To determine how long it takes for the reactant concentration to decrease to 0.00269 M, you can rearrange the equation for the first-order reaction:
ln(C1/C2) = -kt
Where:
C1 is the initial concentration,
C2 is the final concentration,
k is the rate constant,
t is the time.
Rearranging the equation:
t = (ln(C1/C2)) / k
Plugging in the values:
C1 = 0.0396 M, C2 = 0.00269 M, k ≈ 0.0729 s^-1
t = (ln(0.0396/0.00269)) / 0.0729
t ≈ 119.2 seconds
Therefore, it takes approximately 119.2 seconds for the reactant concentration to decrease to 0.00269 M.