t99.9% with respect to t90% for a first order reaction is:

The correct option is Three.

To understand why this is correct, let us derive the relationship step-by-step using the integrated rate equation for a first-order chemical reaction.

For a first-order reaction, the rate constant $k$ is given by the integrated rate law:

$k=\frac{2.303}{t}{\log }_{10}\left(\frac{{\left[A\right]}_0}{{\left[A\right]}_t}\right)$

where:
- ${\left[A\right]}_0$ is the initial concentration of the reactant.
- ${\left[A\right]}_t$ is the concentration remaining at time $t$.

Rearranging the equation to solve for time $t$, we get:

$t=\frac{2.303}{k}{\log }_{10}\left(\frac{{\left[A\right]}_0}{{\left[A\right]}_t}\right)$

Step 1: Calculate the time required for 99.9% completion ($t_{\text{99.9\%}}$)
When the reaction is 99.9% complete, the amount of reactant consumed is 99.9% of ${\left[A\right]}_0$. The remaining concentration of the reactant, ${\left[A\right]}_t$, is:

${\left[A\right]}_t={\left[A\right]}_0-0.999{\left[A\right]}_0=0.001{\left[A\right]}_0={10}^{-3}{\left[A\right]}_0$

Substituting this value into the expression for time:

$t_{\text{99.9\%}}=\frac{2.303}{k}{\log }_{10}\left(\frac{{\left[A\right]}_0}{{10}^{-3}{\left[A\right]}_0}\right)$

$t_{\text{99.9\%}}=\frac{2.303}{k}{\log }_{10}\left({10}^3\right)$

Since ${\log }_{10}\left({10}^3\right)=3$, we have:

$t_{\text{99.9\%}}=\frac{2.303}{k}\times 3$ ---- (Equation 1)

Step 2: Calculate the time required for 90% completion ($t_{\text{90\%}}$)
When the reaction is 90% complete, the amount of reactant consumed is 90% of ${\left[A\right]}_0$. The remaining concentration of the reactant, ${\left[A\right]}_t$, is:

${\left[A\right]}_t={\left[A\right]}_0-0.90{\left[A\right]}_0=0.10{\left[A\right]}_0={10}^{-1}{\left[A\right]}_0$

Substituting this value into the expression for time:

$t_{\text{90\%}}=\frac{2.303}{k}{\log }_{10}\left(\frac{{\left[A\right]}_0}{{10}^{-1}{\left[A\right]}_0}\right)$

$t_{\text{90\%}}=\frac{2.303}{k}{\log }_{10}\left(10\right)$

Since ${\log }_{10}\left(10\right)=1$, we have:

$t_{\text{90\%}}=\frac{2.303}{k}\times 1$ ---- (Equation 2)

Step 3: Find the ratio of $t_{\text{99.9\%}}$ to $t_{\text{90\%}}$
Dividing Equation 1 by Equation 2:

$\frac{t_{\text{99.9\%}}}{t_{\text{90\%}}}=\frac{\frac{2.303}{k}\times 3}{\frac{2.303}{k}\times 1}$

$\frac{t_{\text{99.9\%}}}{t_{\text{90\%}}}=3$

Therefore, the time required for 99.9% completion is three times the time required for 90% completion ($t_{\text{99.9\%}}=3\times t_{\text{90\%}}$).