Question

The rate constant of a first order reaction is \(4.606 \times 10^{-3} \, {s}^{-1}\). The time taken to reduce 20 g of reactant into 2 g is:

• A. 300 s
• B. 500 s
• C. 150 s
• D. 400 s
• E. 250 s

Hint:

For first-order reactions, the time to reach a certain concentration is proportional to the natural logarithm of the ratio of the initial and final concentrations.

Answer 1

The Correct Option is B

For a first-order reaction, the integrated rate law is: \[ \ln \left( \frac{[A]_0}{[A]} \right) = k t \] where \( [A]_0 \) is the initial concentration, \( [A] \) is the concentration at time \( t \), and \( k \) is the rate constant. 
The initial concentration of the reactant is 20 g, and it decreases to 2 g. 
So: \[ \ln \left( \frac{20}{2} \right) = 4.606 \times 10^{-3} \times t \] \[ \ln(10) = 4.606 \times 10^{-3} \times t \implies 2.3026 = 4.606 \times 10^{-3} \times t \] Solving for \( t \): \[ t = \frac{2.3026}{4.606 \times 10^{-3}} = 500 \, {s} \]