Question

Relation between rate constant and half-life for a first-order reaction is:

• A. $ t_{1/2} = \frac{0.693}{K} $
• B. $ K = \frac{t_{1/2}}{0.693} $
• C. $ t_{1/2} = 0.693 + K $
• D. $ t_{1/2} = \frac{K}{0.693} $

Hint:

Key points:

• Unique to first-order reactions
• Half-life constant regardless of concentration
• $K$ and $t_{1/2}$ are inversely proportional
• Used in carbon dating and pharmacokinetics

Answer 1

The Correct Option is A

Step 1: Recall first-order kinetics
For a first-order reaction: \[ \text{Rate} = K[A] \] where $K$ is the rate constant and $[A]$ is the concentration.
Step 2: Derive half-life expression
The integrated rate law gives: \[ \ln\frac{[A]_0}{[A]} = Kt \] At half-life ($t_{1/2}$), $[A] = \frac{[A]_0}{2}$: \[ \ln 2 = K t_{1/2} \] \[ t_{1/2} = \frac{\ln 2}{K} = \frac{0.693}{K} \]
Step 3: Verify options
Only option (a) matches the derived relationship: \[ t_{1/2} = \frac{0.693}{K} \]
Step 4: Characteristics

• Half-life is independent of initial concentration
• $0.693$ is $\ln 2$ (natural log of 2)
• Units: $K$ in $\text{s}^{-1}$, $t_{1/2}$ in seconds