99% of a 1st order reaction completed in 2.303 minutes. What is the rate constant and half-life of the reaction

Solution:

Using formula for first order reactions:

t = (2.303 / k) log([A]₀ / [A])

Here, 99% complete means 1% remaining, so:

log([A]₀ / [A]) = log(100) = 2

Given t = 2.303 minutes, substituting:

2.303 = (2.303 / k) × 2 k = 2 min^-1

Half-life formula for first order reaction:

t_1/2 = 0.693 / k = 0.693 / 2 = 0.3465 minutes

Answer:

Rate constant, k = 2 min^-1
Half-life, t_1/2 = 0.3465 minutes