Question

The half-life equation for first order kinetics is

• A. \( \frac{a}{2k} \)
• B. \( \frac{0.693}{k} \)
• C. \( \frac{1}{ak} \)
• D. \( 0.5k \)

Hint:

For first-order reactions, the half-life is constant and independent of the initial concentration. Use \( t_{1/2} = \frac{0.693}{k} \) for accurate calculations.

Answer 1

The Correct Option is B

Step 1: Understanding the half-life equation for first order kinetics.
The half-life (\( t_{1/2} \)) for first-order reactions is independent of the initial concentration and can be calculated using the equation: \[ t_{1/2} = \frac{0.693}{k} \] where \( k \) is the rate constant of the reaction.
Step 2: Analyzing the options.
(1) \( \frac{a}{2k} \): Incorrect, this is not the correct formula for half-life in first-order reactions.
(2) \( \frac{0.693}{k} \): Correct — This is the correct formula for the half-life of a first-order reaction.
(3) \( \frac{1}{ak} \): Incorrect, this is not related to the first-order reaction half-life.
(4) \( 0.5k \): Incorrect, this is not the formula for half-life.
Step 3: Conclusion.
The correct half-life equation for first-order kinetics is (2) \( \frac{0.693}{k} \).

Answer 2

Rather than quoting the half-life formula directly, let's derive it from the first-order integrated rate law and then check which option matches.

1. \( \frac{a}{2k} \): This mixes an initial concentration term \( a \) into the half-life expression. But the derivation of first-order half-life shows the concentration terms cancel out, so a formula that keeps \( a \) in it cannot be correct.
2. \( \frac{0.693}{k} \): Starting from the integrated first-order rate law \( \ln \frac{[A]_0}{[A]} = kt \), substituting \( [A] = \frac{[A]_0}{2} \) at the half-life gives \( \ln 2 = kt_{1/2} \), so \( t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k} \). This matches the derived result exactly, with the initial concentration cancelling out as expected for a first-order process.
3. \( \frac{1}{ak} \): This form still carries a concentration-dependent term \( a \) in the denominator, which is inconsistent with the fact that first-order half-life is concentration-independent.
4. \( 0.5k \): This has the rate constant multiplying rather than dividing, and does not follow from the integrated rate law derivation at all.

The derivation from the first-order integrated rate law confirms that concentration cancels out, leaving only the rate constant in the half-life expression.

Therefore, the correct answer is \( \frac{0.693}{k} \).