Question

For a reaction, \( A \to B \), rate equation is \( r = k[A]^0 \). If initial concentration of reactant is \( a \) mol dm\(^{-3}\), find the half-life time of the reaction.

• A. \( a/k \)
• B. \( k/a \)
• C. \( \frac{a}{k} \)
• D. \( \frac{k}{a} \)

Hint:

For zero-order reactions, the half-life is directly proportional to the initial concentration and inversely proportional to the rate constant \( k \).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The rate law \( \text{r} = \text{k}[\text{A}]^0 \) signifies a zero-order reaction. We need to find the formula for its half-life (\( \text{t}_{1/2} \)).
Step 2: Key Formula or Approach:
For a zero-order reaction, the integrated rate equation is:
\[ [\text{A}]_t = -kt + [\text{A}]_0 \] At half-life \( \text{t} = \text{t}_{1/2} \), the concentration is half the initial concentration: \( [\text{A}]_t = \frac{[\text{A}]_0}{2} \).
Step 3: Detailed Explanation:
Substitute \( \text{t} = \text{t}_{1/2} \) and \( [\text{A}]_t = \frac{\text{a}}{2} \) (where \( [\text{A}]_0 = \text{a} \)) into the equation:
\[ \frac{\text{a}}{2} = -k\text{t}_{1/2} + \text{a} \] \[ k\text{t}_{1/2} = \text{a} - \frac{\text{a}}{2} \] \[ k\text{t}_{1/2} = \frac{\text{a}}{2} \] \[ \text{t}_{1/2} = \frac{\text{a}}{2\text{k}} \] Step 4: Final Answer:
The half-life time for a zero-order reaction is \( \frac{\text{a}}{2\text{k}} \).