Question

The rate constant for the first order reaction is 60 s\(^{-1}\). How much time will it take to reduce the concentration of the reactant to \( \frac{1}{16} \)th value?

• A. 4.6 \( \times 10^2 \) s
• B. 4.6 \( \times 10^4 \) s
• C. 4.6 \( \times 10^3 \) s
• D. 4.6 \( \times 10^2 \) s

Hint:

For first-order reactions, the time to reduce the concentration can be calculated using the formula \( \ln([A_0]/[A]) = kt \).

Answer 1

The Correct Option is A

Step 1: Applying the first-order reaction formula.
For a first-order reaction, the relation between the concentration of the reactant and time is given by: \[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \] where \( [A_0] \) is the initial concentration, \( [A] \) is the concentration at time \( t \), and \( k \) is the rate constant.

Step 2: Substituting values.
The concentration reduces to \( \frac{1}{16} \), so: \[ \ln \left( \frac{1}{16} \right) = 60 \times t \] \[ \ln 16 = 60 \times t \] \[ 2.7726 = 60 \times t \] Solving for \( t \): \[ t = \frac{2.7726}{60} = 4.6 \times 10^2 \, \text{s} \]

Step 3: Conclusion.
The time required is \( 4.6 \times 10^2 \) seconds.