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

Question

Quizplus · Accepted Answer

The half-life of a first-order reaction is given by the formula: t1/2 = ln2/k, where k is the rate constant of the reaction.

In this case, we're given the time it takes for the concentration to decrease from 0.50 M to 0.20 M. We can use this information to find the rate constant using the first-order integrated rate law:

ln([A]t/[A]0) = -kt

where [A]t is the concentration at time t, [A]0 is the initial concentration, and k is the rate constant.

Substituting the values given, we get:

ln(0.20/0.50) = -k(4.4 × 10^-12 s)

Solving for k, we get:

k = 3.3 × 10^-12 s^-1

Now we can use the formula for half-life:

t1/2 = ln2/k

Substituting the value of k, we get:

t1/2 = ln2/(3.3 × 10^-12 s^-1) ≈ 2.1 × 10^11 s ≈ 3.3 × 10^1 s ≈ 33 s

So the answer is B, 3.3 × 10^-11 s.

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