Question

Half-life (\(t_{1/2}\)) of a first order reaction is 1386 s. The value of rate constant is :

• A. \(0.5 \times 10^{4}\;s^{-1}\)
• B. \(5.0 \times 10^{-4}\;s^{-1}\)
• C. \(0.5 \times 10^{-5}\;s^{-1}\)
• D. \(0.5 \times 10^{-2}\;s^{-1}\)

Hint:

For every first order reaction, always remember: \[ t_{1/2}=\frac{0.693}{k} \] The half-life of a first order reaction does not depend upon the initial concentration of the reactant.

Answer 1

The Correct Option is B

Concept: A first order reaction is a reaction whose rate depends upon the concentration of only one reactant raised to the first power. One of the most important characteristics of a first order reaction is that its half-life is independent of the initial concentration of the reactant. For a first order reaction, the relationship between half-life and rate constant is given by \[ t_{1/2}=\frac{0.693}{k} \] where \[ t_{1/2}=\text{half-life} \] and \[ k=\text{rate constant} \] This formula allows us to determine the rate constant directly if the half-life is known.

Step 1: Writing the formula for the half-life of a first order reaction. For a first order reaction, \[ t_{1/2}=\frac{0.693}{k} \] The given value is \[ t_{1/2}=1386\;s \] Substituting into the formula, \[ 1386=\frac{0.693}{k} \]

Step 2: Rearranging the equation to calculate the rate constant. Solving for \(k\), \[ k=\frac{0.693}{1386} \] \[ k=0.0005 \] \[ k=5.0\times10^{-4}\;s^{-1} \]

Step 3: Matching the obtained value with the given options. The calculated value is \[ k=5.0\times10^{-4}\;s^{-1} \] which corresponds to Option (B). \[ \boxed{k=5.0\times10^{-4}\;s^{-1}} \] Therefore, \[ \boxed{\text{Option (B)}} \]