For a first order reaction; t_99.9 = x t_50% Find out value of x

The correct answer is 10.

To find the value of x, we can use the integrated rate equation for a first-order chemical reaction:
$k=\frac{2.303}{t}{\log }_{10}\left(\frac{{\left[A\right]}_0}{{\left[A\right]}_t}\right)$
where:
- k is the rate constant of the reaction,
- t is the time taken,
- [A]₀ is the initial concentration of the reactant,
- [A]_t is the concentration of the reactant remaining at time t.

Step 1: Calculate the half-life period (t_50%)
For a 50% completion of the reaction, the remaining concentration of reactant [A]_t is half of the initial concentration, [A]₀:
${\left[A\right]}_t=0.5{\left[A\right]}_0$
Substituting this value into the first-order rate equation:
$t_{50\%}=\frac{2.303}{k}{\log }_{10}\left(\frac{{\left[A\right]}_0}{0.5{\left[A\right]}_0}\right)$
$t_{50\%}=\frac{2.303}{k}{\log }_{10}\left(2\right)$
Since log₁₀(2) = 0.3010:
$t_{50\%}=\frac{2.303}{k}\times 0.3010$ ---- (Equation 1)

Step 2: Calculate the time required for 99.9% completion (t_99.9)
For a 99.9% completion of the reaction, the reactant remaining is 100% - 99.9% = 0.1% of the initial concentration:
${\left[A\right]}_t=0.001{\left[A\right]}_0={10}^{-3}{\left[A\right]}_0$
Substituting this value into the first-order rate equation:
$t_{99.9}=\frac{2.303}{k}{\log }_{10}\left(\frac{{\left[A\right]}_0}{{10}^{-3}{\left[A\right]}_0}\right)$
$t_{99.9}=\frac{2.303}{k}{\log }_{10}\left({10}^3\right)$
Since log₁₀(10³) = 3:
$t_{99.9}=\frac{2.303}{k}\times 3$ ---- (Equation 2)

Step 3: Compare both time values to find x
Divide Equation 2 by Equation 1:
$\frac{t_{99.9}}{t_{50\%}}=\frac{\frac{2.303}{k}\times 3}{\frac{2.303}{k}\times 0.3010}$
$\frac{t_{99.9}}{t_{50\%}}=\frac{3}{0.3010}\approx 9.967\approx 10$

Thus, we get:
$t_{99.9}\approx 10\times t_{50\%}$
Comparing with t_99.9 = x t_50%, the value of x is 10.