Show that the time required for 99.9% completion in a first-order reaction is 10 times of half-lif…

Let’s summarize the given information and derivation:

For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. The integrated rate law for a first-order reaction is:

ln⁡([�][�]0)=−��ln([A]0​[A]​)=−kt

Where:

• $[A]$ is the concentration of the reactant at time $t$,

• [�]0[A]0​ is the initial concentration of the reactant,

• $k$ is the rate constant of the reaction,

• $t$ is the time.

The half-life (�1/2t1/2​) of a first-order reaction is given by:

�1/2=ln⁡2�t1/2​=kln2​

Given that $\log 2=0.3010$, we can rewrite the equation for the half-life as:

�1/2=0.3010�t1/2​=k0.3010​

Now, if we want to find the time required for 99.9% completion of the reaction, which means that only 0.1% of the reactant remains, we can express this as:

ln⁡([�][�]0)=−ln⁡(0.001)ln([A]0​[A]​)=−ln(0.001)

Using the integrated rate law for a first-order reaction, we can set ln⁡([�][�]0)=−��ln([A]0​[A]​)=−kt and solve for $t$:

$-kt=-\ln (0.001)$ $kt=\ln (1000)$ �=ln⁡(1000)�t=kln(1000)​

We know that $\ln (1000)=3\ln (10)=3$, since $\ln (10)=1$ (given that $\log 10=1$).

Therefore, we have: �=3�t=k3​

We also know that �1/2=0.3010�t1/2​=k0.3010​.

So, the time required for 99.9% completion is: �=10×�1/2t=10×t1/2​

Thus, we have demonstrated that the time required for 99.9% completion in a first-order reaction is 10 times the half-life (�1/2t1/2​) of the reaction.