Problem 41 - General Chemistry: Principles and Modern Applications

Question

In three different experiments, the following results were obtained for the reaction \(A \longrightarrow\) products: \([\mathrm{A}]_{0}=1.00 \mathrm{M}, t_{1 / 2}=50 \mathrm{min} ;[\mathrm{A}]_{0}=200 \mathrm{M}, t_{1 / 2}=\) \(25 \min ;[\mathrm{A}]_{0}=0.50 \mathrm{M}, t_{1 / 2}=100 \mathrm{min} .\) Write the rate equation for this reaction, and indicate the value of \(k.\)

Step-by-Step Solution

Answer

The rate equation for this reaction is \(rate = k [A]\), and the rate constant (\(k\)) is \(0.0139 min^{-1}\).

Step 1: Determine the Order of Reaction

The half-life of a first-order reaction is constant and it's independent of the initial concentration of the reactant. In our case, when the concentration of reactant \(A\) is doubled, the half-life halves, indicating a first-order reaction.

Step 2: Write the Rate Equation

With the first-order reaction identified, the rate equation can be written as: \(rate = k [A]^{1}\). The exponent '1' indicates that the reaction is indeed first-order.

Step 3: Calculate rate constant (\(k\))

The half-life of a first-order reaction is related to the rate constant through the formula: \(t_{1/2} = 0.693/k\). Rearranging for \(k\), gives: \(k = 0.693 / t_{1/2}\). Plugging in the value from one of the experiments (for example: \(t_{1 / 2}=50 \mathrm{min}\)): \(k = 0.693 / 50 min = 0.0139 min^{-1}\).

Key Concepts

Reaction Order

When talking about reaction kinetics, understanding the **reaction order** is essential. It tells us how the concentration of a reactant influences the rate of a reaction. In simple terms, the reaction order shows whether the concentration of the reactants has a direct effect, or not, on how fast the reaction proceeds. In the case of a **first-order reaction**, the rate of reaction is directly proportional to the concentration of one reactant. This means that if you double the concentration of that reactant, the rate of reaction doubles as well. The experiments provided in the exercise point to a first-order reaction, as changes in concentration resulted in changes in the half-life that reflect such proportionality. Envision a scenario: If the concentration of reactant \( A \) is changed, and you observe that the half-life halves or stays consistent, it often signifies a first-order dependency. This characteristic remains constant, irrespective of the initial concentration, which simplifies analysis and helps predict reaction outcomes efficiently.

Rate Equation

Once the reaction order is determined, formulating the **rate equation** becomes straightforward. A rate equation expresses the reaction rate in terms of the concentration of reactants and the rate constant \( k \). For a first-order reaction, such as the one you've identified in this exercise, the rate equation is written as:

• \( ext{Rate} = k [A]^1 \)

Here, \( [A]^1 \) represents the concentration of reactant \( A \) raised to the power of 1, indicating a first-order relationship.This equation is concise yet powerful, allowing predictions on how the reaction progresses over time. By using the rate equation, you can understand how at any given time, changes in reactant concentration impact the reaction speed. This is crucial for controlling reaction conditions in laboratory and industrial settings, ensuring that reactions proceed at desired rates.

Rate Constant

The **rate constant** \( k \) is a crucial parameter in any rate equation, acting as a proportionality factor that links the reaction rate with the concentration of reactants. Its value reflects how fast the reaction can proceed under certain conditions.To calculate \( k \) for a first-order reaction, you use the half-life formula:

• \( t_{1/2} = \frac{0.693}{k} \)

By rearranging this formula to solve for \( k \), you find:

• \( k = \frac{0.693}{t_{1/2}} \)

Inserting the given half-life from our exercise — for example, \( 50 \) minutes — results in:

• \( k = \frac{0.693}{50 \, \text{min}} = 0.0139 \, \text{min}^{-1} \)

The rate constant is unique to each reaction and depends on factors like temperature and presence of catalysis. Understanding \( k \) allows chemists to make accurate predictions about how a reaction behaves under various conditions, enabling precise control over chemical processes.