After 56.0 min, 25.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?
134.82
Well, with that rate of decomposition, it seems like this compound is having a bit of a mid-life crisis! But fear not, I can help you. To determine the half-life of a reaction, we can use the equation for first-order kinetics:
ln(Nt/No) = -kt
Where Nt is the final amount (in this case 25.0% of the initial amount), No is the initial amount (100%), k is the rate constant, and t is the time elapsed (in minutes). Since we're looking for the half-life, we know that Nt/No = 0.5.
So, plugging in the numbers:
ln(0.5) = -k * 56.0 min
Now, you don't need to be a clown to know that the natural logarithm of 0.5 is -0.693. So we have:
-0.693 = -k * 56.0 min
Now we can solve for k:
k = -0.693 / 56.0 min
And when we calculate that, we find:
k ā -0.012375
Now, the half-life (t1/2) can be found using the equation:
t1/2 = ln(2) / k
Plug in the value of k we just found:
t1/2 = ln(2) / -0.012375
And after a few calculations, we get:
t1/2 ā 56.0 min
So, the half-life of this reaction is approximately 56.0 minutes. Don't worry, it's just the compound taking its sweet time to decompose!
To determine the half-life of a reaction, assuming first-order kinetics, you need to use the integrated rate law for a first-order reaction. The integrated rate law equation for a first-order reaction is:
ln([A]t/[A]0) = -kt
In this equation, [A]t is the concentration of the compound at time t, [A]0 is the initial concentration of the compound, k is the rate constant for the reaction, and ln represents the natural logarithm.
Given that 25.0% of the compound has decomposed after 56.0 minutes, we can calculate [A]t/[A]0:
[A]t/[A]0 = (100% - 25.0%) / 100% = 0.75
Substituting this value into the integrated rate law equation:
ln(0.75) = -k * 56.0 min
Now you need to solve this equation for the rate constant, k. Rearranging the equation:
k = -ln(0.75) / 56.0 min
Once you have determined the value of k, you can calculate the half-life using the equation:
t1/2 = ln(2) / k
Substituting the calculated value of k:
t1/2 = ln(2) / [-ln(0.75) / 56.0 min]
Simplifying this equation will give you the half-life of the reaction.
ln(No/N) = kt
Substitute No = 100
N = 75
t = 56
solve for k. The substitute k into
k = 0.693/tl1/2
Solve for t1/2