Question

For a \(\text{I}^{\text{st}}\) order reaction, 99% complete in 20 minutes, then the \(\text{t}_{1/2}\) will be : (log2 = 0.3)

• A. 10 min
• B. 5 min
• C. 3 min
• D. 2 min

Hint:

Learn to derive and use the direct ratio: \( \frac{t_{99%}}{t_{50%}} = \frac{\log(100/1)}{\log(100/50)} = \frac{\log 100}{\log 2} = \frac{2}{0.3} = \frac{20}{3} \).
So, \( t_{99%} = \frac{20}{3} \times t_{1/2} \implies 20 = \frac{20}{3} \times t_{1/2} \implies t_{1/2} = 3 \text{ min} \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a first-order reaction, the integrated rate law connects the time elapsed to the initial and current concentrations of the reactant. A reaction that is 99% complete means that 99% of the initial reactant has been consumed, and only 1% remains.
Step 2: Key Formula or Approach:
We use the integrated rate equation for a first-order reaction:
\[ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right) \]
And the relation between the rate constant (\(k\)) and half-life (\(t_{1/2}\)):
\[ k = \frac{2.303 \log 2}{t_{1/2}} \]
Step 3: Detailed Explanation:
Let the initial concentration \([A]_0 = 100\).
Since the reaction is 99% complete, the amount reacted is 99.
The remaining concentration at time \(t=20 \text{ min}\) is \([A]_t = 100 - 99 = 1\).
Substitute these values into the integrated rate equation:
\[ t_{99%} = \frac{2.303}{k} \log \left( \frac{100}{1} \right) \]
\[ 20 = \frac{2.303}{k} \log (100) \]
Since \(\log(100) = 2\):
\[ 20 = \frac{2.303 \times 2}{k} \]
From this, express the rate constant \(k\):
\[ k = \frac{4.606}{20} \text{ min}^{-1} \]
Now, we find the half-life \(t_{1/2}\) using the formula:
\[ t_{1/2} = \frac{2.303 \log 2}{k} \]
Substitute the given value \(\log 2 = 0.3\):
\[ t_{1/2} = \frac{2.303 \times 0.3}{k} \]
Now substitute the expression for \(k\) we derived earlier:
\[ t_{1/2} = \frac{2.303 \times 0.3}{\left(\frac{4.606}{20}\right)} \]
\[ t_{1/2} = \frac{2.303 \times 0.3 \times 20}{4.606} \]
Recognize that \(4.606 = 2 \times 2.303\):
\[ t_{1/2} = \frac{2.303 \times 6}{2 \times 2.303} \]
Cancel out \(2.303\):
\[ t_{1/2} = \frac{6}{2} = 3 \text{ minutes} \]
Step 4: Final Answer:
The half-life of the reaction is 3 minutes.