Question

Which from following is correct regarding \(t_{1/2}\) of reaction if we double the initial concentration of a reactant in first order reaction?

• A. \(t_{1/2}\) will increase by two times
• B. \(t_{1/2}\) will decrease by four times
• C. \(t_{1/2}\) remains the same
• D. \(t_{1/2}\) will decrease by two times

Hint:

Only first-order reactions have constant half-life.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The half-life (\(t_{1/2}\)) of a reaction is the time required for the concentration of a reactant to decrease to half of its initial value. The dependence of half-life on initial concentration varies depending on the order of the reaction.
Step 2: Key Formula or Approach:
For a first-order reaction, the relationship between half-life and the rate constant (\(k\)) is given by: \[ t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k} \] Step 3: Detailed Explanation:
Looking at the formula for the half-life of a first-order reaction, we see that \(t_{1/2}\) depends only on the rate constant \(k\). It does not contain the initial concentration term (\([A]_0\)).
This means that for a first-order reaction, the time it takes for half of the reactant to be consumed is always the same, regardless of how much reactant you start with.
Therefore, if you double the initial concentration of the reactant, the half-life will remain completely unaffected.
Step 4: Final Answer:
The half-life remains the same, which corresponds to option (C).