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: Identify the rate equation for zero order reaction.
For a zero-order reaction, the rate law is:
\[ r = k[A]^0 = k \]
This means the rate of the reaction is constant, independent of the concentration of \( A \). The integrated rate law for a zero-order reaction is:
\[ [A] = [A_0] - kt \]
where \( [A_0] \) is the initial concentration of \( A \), \( [A] \) is the concentration of \( A \) at time \( t \), and \( k \) is the rate constant.

Step 2: Deriving the half-life expression.
The half-life time \( t_{1/2} \) is the time taken for the concentration of the reactant to reduce to half of its initial concentration. At \( t = t_{1/2} \), \( [A] = \frac{[A_0]}{2} \). Substituting into the rate law:
\[ \frac{[A_0]}{2} = [A_0] - kt_{1/2} \]
Simplifying:
\[ \frac{[A_0]}{2} = [A_0] - k t_{1/2} \] \[ k t_{1/2} = \frac{[A_0]}{2} \] \[ t_{1/2} = \frac{[A_0]}{2k} \]

Step 3: Final conclusion.
Thus, the half-life time of the reaction is:
\[ \boxed{\frac{a}{k}} \]