Question

For a first-order reaction, if the time taken for $90\%$ completion is $t$, what will be the approximate time taken for $99.9\%$ completion of the same reaction?

• A. \(2t \)
• B. \(3t \)
• C. \(4t \)
• D. \(1.5t \)

Hint:

For first-order kinetics, logarithmic ratios simplify to highly structured scaling constants: \(t_{99.9\%} = 3 \times t_{90\%}\) and \(t_{99\%} = 2 \times t_{90\%}\).

Answer 1

The Correct Option is B

Concept: For a first-order kinetic process, the integrated rate equation is defined as: \[ k = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]_t}\right) \] Where \([A]_0\) is the initial concentration and \([A]_t\) is the remaining reactant concentration at time \(t\).

Step 1: Expressing time for $90\%$ completion (\(t_{90\%}\)).
When the reaction is \(90\%\) complete, the remaining concentration is \([A]_t = [A]_0 - 0.90[A]_0 = 0.10[A]_0\). \[ t = \frac{2.303}{k} \log\left(\frac{[A]_0}{0.10[A]_0}\right) = \frac{2.303}{k} \log(10) = \frac{2.303}{k} \quad \cdots (1) \]

Step 2: Expressing time for $99.9\%$ completion (\(t_{99.9\%}\)).
When the reaction is \(99.9\%\) complete, the remaining concentration is \([A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0 = 10^{-3}[A]_0\). \[ t_{99.9\%} = \frac{2.303}{k} \log\left(\frac{[A]_0}{10^{-3}[A]_0}\right) = \frac{2.303}{k} \log(10^3) = 3 \times \frac{2.303}{k} \quad \cdots (2) \] Substituting equation (1) into equation (2) directly yields \(t_{99.9\%} = 3t\).