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The integrated rate law is a formula that expresses how the concentration of a reactant changes over time. In the case of a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. This implies that the reaction proceeds at a constant rate equal to the rate constant, denoted by \(k\). The integrated rate law for a zero-order reaction is represented as:
\([\mathrm{A}] = -kt + [\mathrm{A}]_0\)
Here, \([\mathrm{A}]\) is the concentration of the reactant A at time \(t\), \([\mathrm{A}]_0\) is the initial concentration of A, and \(k\) is the rate constant. To derive this equation, we apply integration to the rate law formula. This results in a linear equation where the slope of the graph \(–kt\) represents the change in concentration over time.

In simpler terms, for zero-order reactions, the decrease in concentration of the reactant is uniform over time regardless of how much of the reactant remains. This is unlike first or second-order reactions, where the rate depends on the concentration of the reactants.

The half-life of a reaction is the time required for the concentration of a reactant to reduce to half of its initial value. In zero-order reactions, the half-life can be calculated using the following formula derived from the integrated rate law:
\(t_{1/2} = \frac{[\mathrm{A}]_0}{2k}\)
Unlike in first-order reactions, the half-life of a zero-order process is not constant. Instead, it is directly proportional to the initial concentration \([\mathrm{A}]_0\) of the reactant. This implies that if the initial concentration is higher, the half-life increases.

It's important to note:

• The half-life for zero-order reactions increases as the initial amount of reactant increases.
• Calculating half-life helps predict how long it will take for half of the reactant to be consumed.
• For the given example, substituting the given values into the formula results in a half-life of \(1.0 \times 10^{-2} \mathrm{s}\).

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. In a zero-order reaction like the one given:
\(\mathrm{A} \longrightarrow \mathrm{B} + \mathrm{C}\)
One mole of A produces exactly one mole of B and C as they are formed simultaneously. This 1:1 stoichiometric ratio means that changes in their concentrations are mutual.

Understanding reaction stoichiometry helps to predict:

• The quantities of products formed based on the consumed amounts of reactants.
• How the concentration of one product or reactant affects the others due to their direct connection in the reaction equation.
• In the given exercise, the change in the concentration of B is directly calculated by subtracting the remaining concentration of A from its initial value.

By keeping these relationships in mind, stoichiometry becomes a powerful tool for predicting chemical behavior.

The rate constant, \(k\), is a crucial component of zero-order kinetics and helps determine the speed at which a reaction proceeds. For the zero-order reaction presented, the rate constant is given as \(5.0 \times 10^{-2} \mathrm{mol}/\mathrm{L} \cdot \mathrm{s}\).

Key facts about the rate constant in zero-order reactions include:

• It is independent of the concentration of reactants.
• A higher \(k\) value indicates a faster reaction and vice versa.
• In practice, it determines how much concentration decreases per unit time.
• It remains constant as long as the reaction conditions, like temperature, remain unchanged.

The rate constant is vital for calculations involving time, concentration, or determining other important reaction parameters like half-life. Understanding the rate constant's role enables us to predict the course and speed of a reaction accurately.