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antoniya [11.8K]
3 years ago
12

For each of the scenarios, identify the order with respect to the reactant, A. A⟶products The half‑life of A increases as the in

itial concentration of A decreases. order: A three‑fold increase in the initial concentration of A leads to a nine‑fold increase in the initial rate. order: A three‑fold increase in the initial concentration of A leads to a 1.73‑fold increase in the initial rate. order: The time required for [A] to decrease from [A]0 to [A]0/2 is equal to the time required for [A] to decrease from [A]0/2 to [A]0/4. order: The rate of decrease of [A] is a constant. order:
Chemistry
1 answer:
nirvana33 [79]3 years ago
4 0

Answer:

Answers are in the explanation.

Explanation:

  • The half‑life of A increases as the initial concentration of A decreases. order: <em>2. </em>In the half-life of second-order reactions, the half-life is inversely proportional to initial concentration.
  • A three‑fold increase in the initial concentration of A leads to a nine‑fold increase in the initial rate. order: <em>2. </em>The rate law of second-order is: rate = k[A]²
  • A three‑fold increase in the initial concentration of A leads to a 1.73‑fold increase in the initial rate. order: <em>1/2. </em>The rate law for this reaction is: rate = k √[A]
  • The time required for [A] to decrease from [A]₀ to [A]₀/2 is equal to the time required for [A] to decrease from [A]₀/2 to [A]₀/4. order: <em>1. </em>The concentration-time equation for first-order reaction is: ln[A] = ln[A]₀ - kt. That means the [A] decreasing logarithmically.
  • The rate of decrease of [A] is a constant. order: <em>0. </em>The rate law is: rate = k -<em>where k is a constant-</em>
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