Time Required for Ninety-Nine Percent Completion of First Order Reaction

For a first-order reaction, the relationship between the rate constant $k$ and the half-life $t_{1/2}$ is given by: $k = \frac{\ln 2}{t_{1/2}} = \frac{2.303 \times \log 2}{t_{1/2}}$

The integrated rate equation for a first-order reaction is expressed as: $t = \frac{2.303}{k} \log \left(\frac{[A]_0}{[A]_t}\right)$

where:

• $[A]_0$ is the initial concentration of the reactant.
• $[A]_t$ is the concentration of the reactant remaining at time $t$.

For $99\%$ completion of the reaction:

• $[A]_0 = 100$
• $[A]_t = 100 - 99 = 1$

Substituting these values into the integrated rate equation for $t_{99\%}$: $t_{99\%} = \frac{2.303}{k} \log \left(\frac{100}{1}\right)$ $t_{99\%} = \frac{2.303}{k} \times 2 = \frac{2 \times 2.303}{k}$

Now, substitute the value of $k = \frac{2.303 \times \log 2}{t_{1/2}}$ into the expression for $t_{99\%}$: $t_{99\%} = \frac{2 \times 2.303}{\frac{2.303 \times \log 2}{t_{1/2}}} = \frac{2 \times t_{1/2}}{\log 2}$

Given values:

• $t_{1/2} = 6.93 \text{ minutes}$
• $\log 2 = 0.3010$

Substituting these values into the equation: $t_{99\%} = \frac{2 \times 6.93}{0.3010} = \frac{13.86}{0.3010} \approx 46.05 \text{ minutes}$

Rounding to the nearest integer, the time required for $99\%$ completion of the reaction is $46$ minutes.