Question

The reaction $$ \mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C} $$ is known to be zero order in A and to have a rate constant of \(5.0 \times 10^{-2} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\) . An experiment was run at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.0 \times 10^{-3} \mathrm{M} .\) a. Write the integrated rate law for this reaction. b. Calculate the half-life for the reaction. c. Calculate the concentration of \(\mathrm{B}\) after \(5.0 \times 10^{-3} \mathrm{s}\) has elapsed assuming \([\mathrm{B}]_{0}=0\)

Short answer

a. The integrated rate law for this zero-order reaction is \([\mathrm{A}] = -kt + [\mathrm{A}]_0\). b. The half-life for the reaction is \(t_{1/2} = 1.0 \times 10^{-2} \mathrm{s}\). c. The concentration of \(\mathrm{B}\) after \(5.0 \times 10^{-3} \mathrm{s}\) has elapsed is \([\mathrm{B}] = 2.5 \times 10^{-4} \mathrm{M}\).

Step-by-step solution

Writing the integrated rate law for a zero-order reaction

For a zero-order reaction, the rate depends only on the rate constant (k) and is independent of the concentration of reactants. The rate can be written as: \(rate = -\frac{d[\mathrm {A}]}{dt} = k\) To find the integrated rate law, we need to integrate this expression with respect to time: \(\int_{[\mathrm{A}]_0}^{[\mathrm{A}]} -\frac{d[\mathrm{A}]}{k} = \int_0^t dt\) Now, let's integrate both sides: \(-\frac{[\mathrm{A}] - [\mathrm{A}]_0}{k} = t\) This is the integrated rate law for a zero-order reaction. We can rewrite it to isolate the final concentration of A: \([\mathrm{A}] = -kt + [\mathrm{A}]_0\)

Calculating the half-life for the reaction

The half-life (t_1/2) is the time it takes for the concentration of a reactant to decrease to half its initial value. We can find the half-life by setting \([\mathrm{A}] = \frac{1}{2}[\mathrm{A}]_0\) in the integrated rate law: \(\frac{1}{2}[\mathrm{A}]_0 = -kt_{1/2} + [\mathrm{A}]_0\) Now, solve for \(t_{1/2}\): \(t_{1/2} = \frac{([\mathrm{A}]_0 - \frac{1}{2}[\mathrm{A}]_0)}{k}\) We are given the rate constant, k, and the initial concentration of A, \([\mathrm{A}]_0\), so we can plug in these values to find the half-life: \(t_{1/2} = \frac{(1.0 \times 10^{-3} \mathrm{M} - \frac{1}{2}(1.0 \times 10^{-3} \mathrm{M}))}{(5.0 \times 10^{-2} \mathrm{mol/L \cdot s})}\) \(t_{1/2} = 1.0 \times 10^{-2} \mathrm{s}\)

Calculate the concentration of B after a given time, assuming [B]₀=0

The stoichiometry of the reaction is such that for every mole of A consumed, one mole of B is produced: \(\mathrm{A} \longrightarrow \mathrm{B} + \mathrm{C}\) This means that the change in the concentration of B is equal to the change in the concentration of A. To find the concentration of B after 5.0 x 10^{-3} s, we first find the remaining concentration of A at this time using the integrated rate law: \([\mathrm{A}] = -kt + [\mathrm{A}]_0 = - (5.0 \times 10^{-2} \mathrm{mol/L \cdot s})(5.0 \times 10^{-3} \mathrm{s}) + (1.0 \times 10^{-3} \mathrm{M})\) \([\mathrm{A}] = 7.5 \times 10^{-4} \mathrm{M}\) Since the change in the concentration of B is equal to the change in the concentration of A, we can find the concentration of B as: \([\mathrm{B}] = [\mathrm{A}]_0 - [\mathrm{A}] = (1.0 \times 10^{-3} \mathrm{M}) - (7.5 \times 10^{-4} \mathrm{M})\) \([\mathrm{B}] = 2.5 \times 10^{-4} \mathrm{M}\)

Key concepts

Integrated Rate Law

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.

Half-Life Calculation

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

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.

Rate Constant

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.