The t1/2 for a first order rxn is 20 s.the time required for 99.9 percent decomposition is

What's k? k = 0.693/t1/2

Then ln(No/N) = kt
No = 100
N = 0.1
k from above
Solve for t in seconds.

To find the time required for 99.9% decomposition in a first-order reaction, we can use the half-life (t1/2) value.

1. First, let's determine the number of half-life periods required for 99.9% decomposition.
Since each half-life reduces the initial amount by half, the fraction remaining after n half-lives can be calculated using the formula:

Fraction remaining = (1/2)^n

Here, we need to determine the value of n that leaves 0.1% (0.001) of the original substance.

(1/2)^n = 0.001

2. Taking the logarithm of both sides of the equation will enable us to solve for n:

n * log(1/2) = log(0.001)

3. The logarithm of 1/2 can be expressed as a negative value:

n * (-0.301) = -3

4. Now, divide both sides of the equation by -0.301 to solve for n:

n = -3 / -0.301
n ≈ 9.9678

5. The calculated value of n represents the number of half-life periods required for 99.9% decomposition.

6. Finally, multiply the number of half-life periods (n) by the half-life (t1/2) to determine the time required for 99.9% decomposition:

Time required = n * t1/2
Time required ≈ 9.9678 * 20 s
Time required ≈ 199.356 s

Therefore, the time required for 99.9% decomposition in a first-order reaction with a half-life of 20 seconds is approximately 199.356 seconds.