Question

A compound decomposes by a first-order process. If 37 % of the compound decomposes in 60 minutes, the half-life of the compound is ________minutes.

Answer 1

Answer:

90 minutes

Explanation:

The half life of a first order reaction is;

t1/2 = ln(2) / k

To solve for t;

The integral rate expression for a first order reaction is;

ln[A] = ln[Ao] - kt

t = 60 minutes

Final Concentration = [A] = 100 - 37 = 63

Initial Concentration = [Ao] = 100

ln(63) = ln(100) - k (60)

k (60) = ln (100) - ln(63)

k = 0.4620 / 60 = 0.0077

Inserting the value of k in the half life equatin;

t1/2 = ln(2) / 0.0077 = 90 minutes