b) Show that time required for a first order reaction to complete 99.9% reaction is approximately 10 times its half life period.

For a first order reaction, the rate of reaction is directly proportional to the concentration of the reactants. The mathematical expression for a first order reaction is:

k = 1/t * ln([A]0/[A])

where: k = rate constant t = time [A]0 = initial concentration [A] = concentration at time t

The half-life (t1/2) of a first order reaction is given by:

t1/2 = 0.693/k

Now, let's calculate the time required for the reaction to be 99.9% complete. This means that [A] = 0.001[A]0. Substituting these values into the first equation gives:

t = 1/k * ln([A]0/0.001[A]0) t = 1/k * ln(1000) t = 1/k * 6.907

Comparing this with the half-life equation, we can see that:

t = 6.907 * t1/2

So, the time required for a first order reaction to be 99.9% complete is approximately 7 times its half-life. However, this is a rough approximation. The exact value depends on the specific rate constant k for the reaction.

For a first order reaction, the time required for a reaction to complete to a certain extent can be calculated using the formula:

t = (1/k) * ln([A]0/[A])

where:

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

The half-life (t1/2) of a first order reaction is given by the formula:

t1/2 = 0.693/k

Now, let's calculate the time required for the reaction to complete 99.9% (i.e., the concentration of A has decreased to 0.1% of the initial concentration).

t = (1/k) * ln([A]0/0.001[A]0) = (1/k) * ln(1000) = (1/k) * 6.91

Comparing this with the half-life, we get:

t/t1/2 = 6.91/0.693 = 9.97

This is approximately equal to 10.

Therefore, the time required for a first order reaction to complete 99.9% is approximately 10 times its half-life period.