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: Start from the first-order integrated rate law: \( \ln \frac{[A]_0}{[A]} = kt \).
Step 2: At the half-life, \( [A] = \frac{[A]_0}{2} \), so \( \ln \frac{[A]_0}{[A]_0/2} = \ln 2 = k t_{1/2} \).
Step 3: Solving for \( t_{1/2} \) gives \( t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k} \), independent of the initial concentration.
\[ \boxed{t_{1/2} = \frac{0.693}{k}} \]

Answer 2

Another way to check this is by looking at what each option implies about the units and behavior of half-life for a first-order process.

1. \( \frac{a}{2k} \): Since first-order half-life is known experimentally to stay constant regardless of starting concentration (for example, a fixed number of minutes for each halving), any expression containing the initial amount \( a \) fails this test.
2. \( \frac{0.693}{k} \): This expression depends only on the rate constant \( k \), which has units of inverse time for a first-order reaction, so \( \frac{0.693}{k} \) correctly comes out in units of time and stays the same at every halving of concentration, matching the known behavior of first-order half-life.
3. \( \frac{1}{ak} \): This again ties the result to the initial amount \( a \), which contradicts the observed constancy of first-order half-life.
4. \( 0.5k \): Since \( k \) has units of inverse time, multiplying by 0.5 would give units of inverse time rather than time, which is not dimensionally valid for a half-life.

Only the option with \( k \) in the denominator gives a constant, correctly-dimensioned half-life.

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