Step 1: Understanding the Concept:
The half-life (\(t_{1/2}\)) of a chemical reaction is the time required for the concentration of a reactant to decrease to exactly half of its initial value.
The relationship between half-life and the initial concentration depends on the order of the reaction.
Key Formula or Approach:
For a first-order reaction, the integrated rate equation is:
\[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \]
At \(t = t_{1/2}\), \([A] = \frac{[A]_0}{2}\).
Substituting this:
\[ k = \frac{2.303}{t_{1/2}} \log(2) = \frac{0.693}{t_{1/2}} \]
Rearranging for half-life:
\[ t_{1/2} = \frac{0.693}{k} \]
Step 2: Detailed Explanation:
Looking at the final expression \(t_{1/2} = \frac{0.693}{k}\):
1. The expression contains only the constant 0.693 and the rate constant \(k\).
2. There is no term for initial concentration (\([A]_0\)). This is a unique and defining property of first-order reactions: the time taken to consume half the material is the same regardless of how much you start with. This eliminates options (A) and (C).
3. Since \(k\) is in the denominator, half-life is inversely proportional to \(k\). If the rate constant is large (fast reaction), the half-life is short. This confirms option (B).
4. The rate constant \(k\) is a constant for a given reaction at a specific temperature. Therefore, the half-life does not change as the reaction proceeds. This eliminates option (D).
The mathematical relationship clearly shows \(t_{1/2} \propto \frac{1}{k}\).
Step 3: Final Answer:
The half-life of a first-order reaction is inversely proportional to the rate constant \(k\).