Question

For a first-order reaction, the time required to reduce the concentration of the reactant to half its initial value is:

• A. \(0.3010/k \)
• B. \(1/k \)
• C. \(0.693/k \)
• D. \(2.303/k \)

Hint:

For a first-order reaction, the half-life is independent of the initial concentration and is given by: \[ t_{1/2} = \frac{0.693}{k} \] This is a key result frequently used in chemical kinetics problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to identify the expression for the half-life (\(t_{1/2}\)) of a first-order reaction, which is the time taken for the reactant concentration to reach 50% of its original value.
Step 2: Key Formula or Approach:
For a first-order reaction, the rate constant \(k\) is related to the initial concentration \([A]_0\) and concentration at time \(t\), \([A]\), by the formula:
\[ k = \frac{2.303}{t} \log_{10} \left( \frac{[A]_0}{[A]} \right) \]
Step 3: Detailed Explanation:
To find the half-life, we set \(t = t_{1/2}\) and \([A] = \frac{[A]_0}{2}\).
Substituting these into the integrated rate law:
\[ k = \frac{2.303}{t_{1/2}} \log_{10} \left( \frac{[A]_0}{[A]_0/2} \right) \]
\[ k = \frac{2.303}{t_{1/2}} \log_{10} (2) \]
Using the value \(\log_{10} 2 \approx 0.3010\):
\[ k = \frac{2.303 \times 0.3010}{t_{1/2}} \]
\[ k = \frac{0.693}{t_{1/2}} \]
Rearranging for \(t_{1/2}\):
\[ t_{1/2} = \frac{0.693}{k} \]
Step 4: Final Answer:
The half-life for a first-order reaction is represented by the expression \(0.693/k\).