Problem 118 - Chemistry The Central Science In Si Units

Question

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(a l c)\) has an activation energy of \(86.8 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(2.1 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Predict the rate constant for the reaction at \(30^{\circ} \mathrm{C}\). (b) A solution of KOH in ethanol is made up by dissolving 0.500 g KOH in ethanol to form \(500 \mathrm{~mL}\) of solution. Similarly, \(1.500 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{l}\) is dissolved in ethanol to form \(500 \mathrm{~mL}\). of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(30^{\circ} \mathrm{C} ?\) (c) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? ((d) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(40^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

(a) Calculate \( k \) at 30°C using the Arrhenius equation: \( k \approx 1.46 \times 10^{-3} \ \text{M}^{-1} \text{s}^{-1} \). (b) Initial rate: \( ext{Rate} \approx 9.5 \times 10^{-4} \ \text{M s}^{-1} \). (c) \( \text{C}_2 \text{H}_5 \text{I} \) is limiting. (d) At 40°C: \( k \approx 2.31 \times 10^{-3} \ \text{M}^{-1} \text{s}^{-1} \). "}

Step 1: Use the Arrhenius Equation

The Arrhenius equation is used to calculate the rate constant \( k \) given the activation energy \( E_a \) and the frequency factor \( A \). The equation is \( k = A e^{-E_a / (RT)} \). We need to convert the given temperature of 30°C to Kelvin, which is 303 K.

Step 2: Convert Activation Energy

The activation energy is given in kJ/mol. We must convert it to J/mol for consistency in the equation. \( E_a = 86.8 \) kJ/mol \( = 86800 \) J/mol.

Step 3: Calculate the Rate Constant at 30°C

Substitute the values into the Arrhenius equation: \[ k = 2.1 \times 10^{11} \exp\left(-\frac{86800}{8.314 \times 303}\right) \]. Calculate \( k \) to get the rate constant at 30°C.

Step 4: Find Initial Concentrations

Determine the initial concentrations of KOH and \( \text{C}_2 \text{H}_5 \text{I} \) in the mixed solution. Each reactant is dissolved in 500 mL ethanol to create the solutions, combined to form 1 L after mixing. Use molar mass to convert grams to moles, then divide by total volume to find concentrations.

Step 5: Calculate the Initial Rate

The reaction is first order in each reactant, so the rate is given by \( ext{Rate} = k [ ext{KOH}][ ext{C}_2 ext{H}_5 ext{I}] \) at 30°C. Substitute the concentrations and the rate constant calculated in Step 3 to find the initial rate.

Step 6: Determine the Limiting Reagent

Compare the moles of KOH and \( \text{C}_2 \text{H}_5 \text{I} \) based on initial concentrations. The reagent with fewer moles will be the limiting reagent.

Step 7: Calculate Rate Constant at 40°C

To find \( k \) at 40°C, repeat Steps 1-3 with the new temperature of 313 K (40°C in Kelvin). Substitute into the Arrhenius equation: \[ k = 2.1 \times 10^{11} \exp\left(-\frac{86800}{8.314 \times 313}\right) \]. Compute \( k \) for the new temperature.

Key Concepts

Understanding Activation Energy

Activation energy is a crucial concept in chemical kinetics. It represents the minimum energy that reactant molecules need to undergo a successful chemical reaction. Imagine trying to roll a ball over a hill; the energy required to push the ball to the top is akin to activation energy. In our specific reaction involving ethyl iodide and hydroxide ions, the activation energy is given as 86.8 kJ/mol.

This energy barrier ensures that only those molecules with sufficient energy can effectively react. Without it, reactions could proceed spontaneously at observable rates, even under mild conditions, leading to potentially disastrous consequences.

• Activation energy is expressed in kJ/mol.
• It influences how often reacting molecules successfully collide to form products.

Converting this energy into other units, like J/mol, is often necessary for calculations involving the Arrhenius equation, which allows us to discover reaction rates at various temperatures.

The Role of the Arrhenius Equation in Reaction Rates

The Arrhenius equation is a fundamental tool in understanding how temperature influences reaction rates. Named after the Swedish scientist Svante Arrhenius, this equation links the rate constant of a chemical reaction to its activation energy and temperature:

• Formula: \[ k = A e^{-E_a / (RT)} \]
• \( k \): Rate constant
• \( A \): Frequency factor, indicating how often molecules collide with the correct orientation
• \( E_a \): Activation energy
• \( R \): Universal gas constant, \( 8.314 \, \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \)
• \( T \): Temperature in Kelvin

For example, if we want to find the rate constant at 30°C for the ethyl iodide and hydroxide ion reaction, we can transform °C into Kelvin, giving 303 K. The Arrhenius equation then helps calculate the rate constant by solving:

• Convert 30°C to 303 K
• Substitute values into the equation

By using the equation, scientists can predict how reaction speed changes with temperature or alterations in the energy landscape.

Identifying the Limiting Reagent in Reactions

The concept of a limiting reagent is pivotal in stoichiometry, as it determines the amount of product formed in a reaction. It is the reactant that will be completely used up first, stopping the reaction from continuing. Think of a simple analogy: if you are making sandwiches, and you have more slices of ham than slices of bread, the number of sandwiches you can make is limited by the bread.

• The limiting reagent is the one with the smallest mole ratio compared to the balanced equation.
• Identifying the limiting reagent helps in calculating maximum product yields.

In our reaction between ethyl iodide and hydroxide ion:

• Calculate initial moles of each reactant by converting grams to moles.
• Compare the mole counts based on the reaction stoichiometry.
• The reactant with fewer moles will be the limiting reagent.

By identifying the limiting reagent, we can predict how long the reaction proceeds and how much of the product can be feasibly achieved. This fundamental chemical concept ensures efficient use of reactants, minimizes waste, and aids in cost-effective practices in both academic and industrial chemical processes.