For each of the scenarios, identify the order with respect to the reactant, a. a⟶products the half‑life of a increases as the initial 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:

Answers are in the explanation.

Explanation:

The half‑life of A increases as the initial concentration of A decreases. order: 2. 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: 2. The rate law of second-order is: rate = k[A]²A three‑fold increase in the initial concentration of A leads to a 1.73‑fold increase in the initial rate. order: 1/2. 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: 1. 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: 0. The rate law is: rate = k -where k is a constant-

B? i feel like this is a trick