Question

\(t_{99.9%}\) with respect to \(t_{90%}\) for a first order reaction is:

• A. Two
• B. One
• C. Three
• D. Four

Hint:

For first-order reactions: \[ t_{90%} : t_{99%} : t_{99.9%} = 1 : 2 : 3 \] This shortcut is extremely useful in MCQs.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a first-order reaction, the time required for a specific percentage of completion can be derived from the integrated rate law.
The formula is: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$
where $[A]_0$ is the initial concentration and $[A]_t$ is the concentration at time $t$.
Step 2: Key Formula or Approach:
For 90% completion, the remaining concentration $[A]_t = 100% - 90% = 10%$ of $[A]_0$.
For 99.9% completion, the remaining concentration $[A]_t = 100% - 99.9% = 0.1%$ of $[A]_0$.
Step 3: Detailed Explanation:
Calculating $t_{90%$:}
\[ t_{90%} = \frac{2.303}{k} \log \frac{100}{10} = \frac{2.303}{k} \log 10 \]
Since $\log 10 = 1$:
\[ t_{90%} = \frac{2.303}{k} \] --- (Equation 1)

Calculating $t_{99.9%$:}
\[ t_{99.9%} = \frac{2.303}{k} \log \frac{100}{0.1} = \frac{2.303}{k} \log 1000 \]
Since $\log 1000 = \log 10^3 = 3$:
\[ t_{99.9%} = \frac{2.303}{k} \times 3 \] --- (Equation 2)

Comparison:
Divide Equation 2 by Equation 1:
\[ \frac{t_{99.9%}}{t_{90%}} = \frac{(2.303/k) \times 3}{2.303/k} = 3 \]
Therefore, $t_{99.9%} = 3 \times t_{90%}$.
This result shows that reaching 99.9% completion takes exactly three times as long as reaching 90% completion for any first-order reaction, regardless of the rate constant $k$.
Step 4: Final Answer:
The time $t_{99.9%}$ is three times $t_{90%}$.