Question

Half-life of a first order reaction is 20 seconds and initial concentration of reactant is 0.2M. The concentration of reactant left after 80 seconds is

• A. 0.5 M
• B. 0.0125 M
• C. 0.2 M
• D. 0.1 M

Hint:

For first-order reactions, use the formula \( \ln \left( \frac{[A]_0}{[A]_t} \right) = kt \) to determine the concentration at any given time.

Answer 1

The Correct Option is B

For a first-order reaction, the relationship between the concentration at time \( t \) and the initial concentration is given by: \[ \ln \left( \frac{[A]_0}{[A]_t} \right) = kt \] Where: - \( [A]_0 \) is the initial concentration, - \( [A]_t \) is the concentration at time \( t \), - \( k \) is the rate constant, - \( t \) is the time. The half-life \( t_{1/2} \) for a first-order reaction is given by: \[ t_{1/2} = \frac{0.693}{k} \] Given \( t_{1/2} = 20 \text{ sec} \), we can calculate \( k \): \[ k = \frac{0.693}{20} = 0.03465 \text{ sec}^{-1} \] Now, to find the concentration at 80 seconds: \[ \ln \left( \frac{0.2}{[A]_t} \right) = 0.03465 \times 80 \] \[ \ln \left( \frac{0.2}{[A]_t} \right) = 2.772 \] \[ \frac{0.2}{[A]_t} = e^{2.772} = 16 \] \[ [A]_t = \frac{0.2}{16} = 0.0125 \text{ M} \]