What is the half-life of a first-order reaction if it takes 4.4 $\times$ 10^-2 seconds for the concentration to decrease from 0.50 M to 0.20 M?

Question

Quizplus · Accepted Answer

The half-life of a first-order reaction can be calculated using the following equation:
t1/2 = (0.693/k)
where k is the rate constant of the reaction. 
Given that the concentration decreases from 0.50 M to 0.20 M in 4.4 × 10-2 s, we can calculate the rate constant as follows:
ln (0.20 M/0.50 M) = -k(4.4 × 10-2 s)
k = 0.693/4.4 × 10-2 s
k = 15.75 s-1
Using this value of k, we can calculate the half-life as follows:
t1/2 = (0.693/k)
t1/2 = (0.693/15.75 s-1)
t1/2 = 0.044 s
Therefore, the answer is B, 3.3 × 10-2 s (which is the closest option to 0.044 s).

What Is the Half-Life of a First-Order Reaction If It ×\times×

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