Question

The correct relation to find out the half-life of a first order reaction is
(k = rate constant of reaction, [A]\(_0\) = initial concentration of reactant)

• A. \( \frac{\ln 2}{k} \)
• B. \( \frac{[A]_0}{2k} \)
• C. \( \frac{1}{k[A]_0} \)
• D. \( \frac{1}{k} \)

Hint:

For first-order reactions, half-life is constant and independent of initial concentration.

Answer 1

The Correct Option is A

Step 1: Understand first order reaction kinetics.
For a first order reaction, the rate of reaction depends only on the first power of reactant concentration. The integrated rate law is: \[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \] This relation is used to derive half-life expression.

Step 2: Define half-life condition.
Half-life (\(t_{1/2}\)) is the time required for the concentration of reactant to become half of its initial value: \[ [A] = \frac{[A]_0}{2} \]

Step 3: Substitute into integrated rate equation.
\[ k = \frac{2.303}{t_{1/2}} \log \frac{[A]_0}{[A]_0/2} \] \[ k = \frac{2.303}{t_{1/2}} \log 2 \]

Step 4: Simplify logarithmic term.
Since: \[ \log 2 = 0.3010 \] \[ 2.303 \times 0.3010 = 0.693 = \ln 2 \]

Step 5: Rearrange for half-life.
\[ t_{1/2} = \frac{\ln 2}{k} \]

Step 6: Final interpretation.
Half-life of a first order reaction is independent of initial concentration and depends only on rate constant \(k\).
Final Answer: \[ \boxed{\frac{\ln 2}{k}} \]