Question

For a first-order reaction, the time required to reduce the concentration of the reactant to half its initial value is:

• A. \(0.3010/k \)
• B. \(1/k \)
• C. \(0.693/k \)
• D. \(2.303/k \)

Hint:

For a first-order reaction, the half-life is independent of the initial concentration and is given by: \[ t_{1/2} = \frac{0.693}{k} \] This is a key result frequently used in chemical kinetics problems.

Answer 1

The Correct Option is C

Concept: For a first-order reaction, the rate of reaction depends linearly on the concentration of the reactant. The integrated rate law for a first-order reaction is: \[ \ln \left(\frac{[A]_0}{[A]}\right) = kt \] where
• \( [A]_0 \) = initial concentration
• \( [A] \) = concentration at time \(t\)
• \( k \) = rate constant
• \( t \) = time The half-life \(t_{1/2}\) is the time required for the concentration of the reactant to become half of its initial value.

Step 1: Definition of half-life. For half-life, \[ [A] = \frac{[A]_0}{2} \] Substitute this in the first-order rate equation: \[ \ln \left(\frac{[A]_0}{[A]_0/2}\right) = kt_{1/2} \] \[ \ln (2) = kt_{1/2} \]

Step 2: Evaluating the logarithmic value. \[ \ln 2 = 0.693 \] Therefore, \[ t_{1/2} = \frac{0.693}{k} \]