Question

The \(t_{1 / 2}\) for the hydrolytic degradation of the carbamate insecticide carbaryl is reported to be 31 days at \(6^{\circ} \mathrm{C}\) and 11 days at \(22^{\circ} \mathrm{C}\). Calculate the activation energy for the reaction. Write an equation for the hydrolysis of this pesticide.

Short answer

The activation energy for carbaryl's hydrolysis is calculated using the Arrhenius equation. The reaction equation is: Carbaryl + H₂O → Naphthol + Methylamine.

Step-by-step solution

Understanding the Problem

We are tasked with calculating the activation energy for the hydrolytic degradation of carbaryl, a carbamate insecticide. We are given the half-life \(t_{1/2}\) at two different temperatures: 31 days at \(6^{\circ} \mathrm{C}\) and 11 days at \(22^{\circ} \mathrm{C}\). The problem requires us to find the activation energy (\(E_a\)) and write the hydrolysis equation.

Identifying the Relevant Equation

To calculate the activation energy, we can use the Arrhenius equation in the form: \[\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)\] where \(k\) is the rate constant, \(E_a\) is the activation energy, \(R\) is the gas constant \(8.314 \, \mathrm{J/mol\,K}\), and \(T\) is the absolute temperature in Kelvin. We need to convert the half-lives to rate constants.

Converting Half-Lives to Rate Constants

For a first-order reaction, the rate constant \(k\) is related to the half-life by the equation: \[k = \frac{\ln 2}{t_{1/2}}\] Calculating for both temperatures: - For \(T_1 = 6^{\circ} \mathrm{C} = 279 \mathrm{K}\), \[k_1 = \frac{\ln 2}{31 \text{ days}}\]- For \(T_2 = 22^{\circ} \mathrm{C} = 295 \mathrm{K}\), \[k_2 = \frac{\ln 2}{11 \text{ days}}\] Convert days to seconds for the constants.

Calculating Rate Constants

Convert half-lives from days to seconds: \[ 1 \text{ day} = 86400 \text{ seconds} \] - For \(T_1 = 279 K\), \[k_1 = \frac{\ln 2}{31 \times 86400 \text{ seconds}}\]- For \(T_2 = 295 K\), \[k_2 = \frac{\ln 2}{11 \times 86400 \text{ seconds}}\] Compute \(k_1\) and \(k_2\).

Substituting Values in the Arrhenius Equation

Substitute the calculated rate constants and temperatures into the Arrhenius equation:\[\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{8.314} \left( \frac{1}{279} - \frac{1}{295} \right)\] Solve for \(E_a\).

Solving for Activation Energy

Perform the calculations:1. Calculate \(\ln \left(\frac{k_2}{k_1}\right)\).2. Calculate \( \left( \frac{1}{279} - \frac{1}{295} \right) \).3. Solve for \(E_a\) using these values.

Writing the Hydrolysis Equation

The hydrolysis of carbaryl involves breaking its bond with water to form naphthol and methylamine. The simplified chemical equation is: \[\text{{Carbaryl}} + \text{{H}}_2\text{{O}} \rightarrow \text{{Naphthol}} + \text{{Methylamine}}\] This indicates the reactant and the products formed during the hydrolysis.

Key concepts

Arrhenius Equation

The Arrhenius equation is a formula used to calculate the effect of temperature on the rate of a chemical reaction. It reveals the relationship between the rate constant (\( k \)) and temperature (\( T \)). Here is the equation in its common form:

• \[ k = A \times e^{-\frac{E_a}{RT}} \]

where:

• \( A \) is the pre-exponential factor or frequency factor. It represents the likelihood of molecules colliding with proper orientation.
• \( E_a \) is the activation energy, the energy barrier that must be overcome for a reaction to occur.
• \( R \) is the universal gas constant, approximately 8.314 J/mol·K.
• \( T \) is the temperature in Kelvin.

In this specific problem, we substitute the Arrhenius equation into the logarithmic form, which allows us to find the activation energy by comparing rate constants at two different temperatures.

First-order Reaction

A first-order reaction is a chemical reaction where the reaction rate depends linearly on only one reactant's concentration. This means that if you double the concentration of the reactant, the rate of reaction doubles as well.For a first-order reaction, we use the formula to relate the rate constant to the half-life:

• \[ k = \frac{\ln 2}{t_{1/2}} \]

where:

• \( k \) is the rate constant.
• \( t_{1/2} \) is the half-life, or the time it takes for half of the reactant to be consumed.

This problem specifically involves calculating the rate constants (\( k_1 \) and \( k_2 \)) from given half-lives at two different temperatures, then using these in the Arrhenius equation to determine activation energy.

Carbamate Insecticide

Carbamate insecticides are a class of pesticides derived from carbamic acid. They work by inhibiting the enzyme acetylcholinesterase in pests, leading to the interruption of nerve impulses. Carbaryl is a specific type of carbamate insecticide used widely in agriculture to protect crops. It works effectively against a variety of insects by disrupting their nervous systems, ultimately killing them. When carbaryl undergoes hydrolysis, it breaks down into simpler molecules. These transformations can significantly impact its effectiveness and environmental persistence. This is an important consideration, as understanding how long an insecticide remains active is critical in managing its application.

Hydrolysis Reaction

A hydrolysis reaction involves the breaking of a bond in a molecule using water. This type of reaction is crucial for the breakdown of chemicals like pesticides in the environment.In this exercise, the hydrolysis equation for carbaryl is given by:Carbaryl + \( \text{H}_2\text{O} \rightarrow \text{Naphthol} + \text{Methylamine} \)This equation shows that water (\( \text{H}_2\text{O} \)) assists in breaking carbaryl's molecular bond, forming naphthol and methylamine as products. Understanding this reaction is important for predicting the environmental fate and transformation of pesticides.The study of such reactions helps in determining how these compounds impact ecosystems and aids in engineering more environmentally friendly pesticides.