Question

What is the value of rate constant of first-order reaction, if it takes 15 minutes for consumption of 20% of reactants?

• A. \( 1.84 \times 10^{-2} \, \text{min}^{-1} \)
• B. \( 1.38 \times 10^{-2} \, \text{min}^{-1} \)
• C. \( 1.07 \times 10^{-2} \, \text{min}^{-1} \)
• D. \( 1.48 \times 10^{-2} \, \text{min}^{-1} \)

Hint:

For first-order reactions, the rate constant can be calculated using \( k = \frac{2.303}{t} \log \left( \frac{1}{1 - \text{Fraction of reactants consumed}} \right) \).

Answer 1

The Correct Option is D

Step 1: Formula for rate constant of first-order reaction.
For a first-order reaction, the equation is given by: \[ k = \frac{2.303}{t} \log \left( \frac{1}{1 - \text{Fraction of reactants consumed}} \right) \] Substitute the given values into the formula: \[ k = \frac{2.303}{15} \log \left( \frac{1}{1 - 0.2} \right) \approx 1.48 \times 10^{-2} \, \text{min}^{-1} \]
Step 2: Conclusion.
The correct answer is (D) \( 1.48 \times 10^{-2} \, \text{min}^{-1} \).