The half-life / shelf-life does not depend on the initial concentration of the reactant. This statement is TRUE for

Question

BlackTom AI · Accepted Answer

First-order reactions have a constant half-life, t1/2 = ln(2)/k, which does not depend on the initial concentration.

Relevant concepts:
- Half-life (t1/2) is the time required for a reactant concentration to fall to half its initial value.
- Integrated rate laws differ by reaction order; t1/2 expressions come from substituting [A] = [A]0/2 into those integrated laws.

Step-by-step derivations and reasoning:

1) Zero-order (rate = k, integrated law: [A] = [A]0 - kt)
- Set [A] = [A]0/2 at t = t1/2:
  [A]0/2 = [A]0 - k t1/2
  k t1/2 = [A]0 - [A]0/2 = [A]0/2
  t1/2 = [A]0 / (2k)
- This expression depends linearly on the initial concentration [A]0, so zero-order half-life is NOT independent of [A]0.

2) First-order (rate = k[A], integrated law: ln[A] = ln[A]0 - kt)
- Set [A] = [A]0/2:
  ln([A]0/2) = ln[A]0 - k t1/2
  ln([A]0) - ln(2) = ln[A]0 - k t1/2
  -ln(2) = -k t1/2  =>  t1/2 = ln(2) / k
- No [A]0 appears; t1/2 is constant and independent of the initial concentration. This is the key property that makes first-order processes unique in this respect.

3) Second-order (rate = k[A]^2, integrated law: 1/[A] = 1/[A]0 + k t)
- Set [A] = [A]0/2:
  1/([A]0/2) = 1/[A]0 + k t1/2
  2/[A]0 = 1/[A]0 + k t1/2
  k t1/2 = 1/[A]0  =>  t1/2 = 1 / (k [A]0)
- This depends on [A]0 (inversely), so second-order half-life is also not independent of initial concentration.

Why the correct answer is correct:
Only the first-order derivation eliminates [A]0, yielding t1/2 = ln(2)/k independent of initial concentration. Options a and c are incorrect because their t1/2 expressions include [A]0; option d is incorrect because not all orders share the property.

Conclusion: The statement is TRUE for first-order processes (option b), matching the provided answer.

The half-life / shelf-life does not depend on the initial concentration of the reactant. This statement is TRUE forSingle choice

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