Question

Half life of first order reaction is 20 minutes. What is the time taken to reduce the initial concentration of the reactant to $\dfrac{1{10}$th?}

• A. $6.6\ \text{min}$
• B. $66.56\ \text{min}$
• C. $150\ \text{min}$
• D. $79.68\ \text{min}$

Hint:

For first order reactions, time required depends on fractional change, not on initial concentration.

Answer 1

The Correct Option is B

Step 1: Relation between half-life and rate constant.
For a first order reaction, the half-life is given by:
\[ t_{1/2} = \frac{0.693}{k} \] Given $t_{1/2} = 20\ \text{min}$,
\[ k = \frac{0.693}{20} = 0.03465\ \text{min}^{-1} \]
Step 2: First order rate equation.
For reduction to $\dfrac{1}{10}$th of initial concentration:
\[ t = \frac{2.303}{k} \log \frac{a}{a/10} \]
Step 3: Substitution of values.
\[ t = \frac{2.303}{0.03465} \log 10 \] \[ t = \frac{2.303}{0.03465} \times 1 = 66.56\ \text{min} \]
Step 4: Conclusion.
Thus, the time required is $66.56\ \text{minutes}$.