The time required for the completion of 99.9% of a first order reaction is equal to how many times that of its half life?

(A)2
(B)3
(C)5
(D)10

2^n = 0.001

n = ?

it's D(10)

To answer this question, we need to understand the relationship between the time required for the completion of a first-order reaction and its half-life.

In a first-order reaction, the time required for the completion of a reaction can be calculated using the equation:

t = (ln(1 / (1 - x))) / k

Where:
- t is the time required for completion
- x is the fraction reacted (in decimal form)
- k is the rate constant of the reaction

The half-life of a first-order reaction is defined as the time it takes for the initial concentration of the reactant to decrease by half. The equation for the half-life of a first-order reaction is:

t1/2 = (ln(2)) / k

Now, we can compare the time required for the completion of the reaction to its half-life:

We are given that the completion is at 99.9% (x = 0.999). Substituting this value into the equation for the time required for completion:

t = (ln(1 / (1 - 0.999))) / k
t = (ln(1 / 0.001)) / k
t = ln(1000) / k

Similarly, the half-life equation can be rewritten as:

t1/2 = (ln(2)) / k

To find the ratio between the time required for completion (t) and the half-life (t1/2), we can divide t by t1/2:

t / t1/2 = (ln(1000) / k) / ((ln(2)) / k)
t / t1/2 = (ln(1000) / ln(2))

Using logarithmic properties, we can simplify this expression:

t / t1/2 = ln(1000) / ln(2)
t / t1/2 = ln(10^3) / ln(2)
t / t1/2 = 3 / ln(2)

Therefore, the time required for completion of 99.9% of a first-order reaction is equal to 3 times its half-life.

So, the answer to the question is (B) 3.