Step-by-Step Solution

Given: The half-life of compound A is inversely proportional to its initial concentration. This implies that the reaction order with respect to A is 2, because for a second-order reaction, the half-life is given by $t_{1/2}=\frac{1}{k\text{'}{[}_A{]}_0}$, which is inversely proportional to [A]₀. Therefore, b = 2.

Step 1: Determine the order with respect to H⁺ (a)

The half-life changes with pH. Since pH = -log[H⁺], a change in pH corresponds to a change in [H⁺].

Given:

• At pH = 3, [H⁺] = 10^-3 M, t_1/2 = 100 min
• At pH = 4, [H⁺] = 10^-4 M, t_1/2 = 1 min

The half-life is inversely proportional to the initial concentration of A, but since [A]₀ is the same in both experiments (only pH is changed), the change in half-life is due to the change in [H⁺].

Step 2: Relate half-life to the rate constant

For a reaction with order b=2 with respect to A, the half-life is t_1/2 = 1 / (k' [A]₀), where k' is the observed rate constant.

The observed rate constant k' depends on [H⁺] as per the rate law: R = k [H⁺]^a [A]^b = k [H⁺]^a [A]².

Therefore, k' = k [H⁺]^a, and so t_1/2 = 1 / (k [H⁺]^a [A]₀).

Step 3: Set up the ratio of half-lives

Since [A]₀ is the same in both cases, the ratio of half-lives is:

Step 4: Substitute the values

t_1/2,1 = 100 min, [H⁺]₁ = 10^-3 M

t_1/2,2 = 1 min, [H⁺]₂ = 10^-4 M

So, 100 = 10^-a

Since 100 = 10², we have:

Therefore, 2 = -a, which gives a = -2.

Final Answer: a = -2, b = +2

Related Topics

• Chemical Kinetics: The study of reaction rates and the factors affecting them.
• Order of Reaction: The power to which the concentration of a reactant is raised in the rate law.
• Acid Catalysis: A process where the reaction rate is accelerated by an acid, which provides protons (H⁺ ions).
• Half-life (t_1/2): The time required for the concentration of a reactant to reduce to half its initial value.

Important Formulae and Theory

1. General Rate Law: For a reaction, Rate = k [A]^m [B]ⁿ, where m and n are the orders with respect to A and B, and k is the rate constant.

2. Half-life for a Second-Order Reaction: For a reaction with rate law Rate = k [A]², the half-life is given by $t_{1/2}=\frac{1}{k{[}_A{]}_0}$. This shows that t_1/2 is inversely proportional to the initial concentration.

3. pH and [H⁺] Relationship: pH = -log₁₀[H⁺], so [H⁺] = 10^-pH.

4. Analyzing Orders from Half-life Data: If half-life is inversely proportional to the initial concentration of a reactant, the order with respect to that reactant is 2. The dependence on other reactants can be found by varying their concentrations while keeping others constant.