After 62.0 min, 11.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?
ln(No/N) = kt
No = 100
N = 89
t = 62 min
solve for k.
Then k = 0693/t1/2
To determine the half-life of a reaction assuming first-order kinetics, we need to use the following formula:
t1/2 = (0.693 / k)
where:
t1/2 = half-life of the reaction
k = rate constant of the reaction
In this case, we have been given the percentage of the compound decomposed after a certain time, but not the rate constant. The rate constant can be calculated using the following equation:
k = (0.693 / t)
where:
k = rate constant of the reaction
t = time taken for a certain percentage of the compound to decompose
Given that after 62.0 minutes, 11.0% of the compound has decomposed, we can substitute the values into the equation to find the rate constant:
k = (0.693 / 62.0)
Now, we can substitute the value of k into the first equation to calculate the half-life:
t1/2 = (0.693 / k)
t1/2 = (0.693 / (0.693 / 62.0))
Simplifying the equation gives:
t1/2 = (0.693 * 62.0)
t1/2 = 43.066 minutes
Therefore, the half-life of this reaction, assuming first-order kinetics, is approximately 43.066 minutes.