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 \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C}\). (b) \(\mathrm{A}\) solution of KOH in ethanol is made up by dissolving \(0.335 \mathrm{~g} \mathrm{KOH}\) in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Similarly, \(1.453 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form \(250.0 \mathrm{~mL}\) 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 \(35^{\circ} \mathrm{C} ?(\mathrm{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion?

Short answer

(a) The rate constant for the reaction at \(35^{\circ}C\) is approximately \(3.53 \times 10^8 M^{-1}s^{-1}\). (b) The initial rate of the reaction at \(35^{\circ}C\) is approximately \(2.1 \times 10^{-3} M/min\). (c) The limiting reagent in the reaction is ethyl iodide (C₂H₅I).

Step-by-step solution

(a) Calculate the rate constant at \(35^{\circ}C\) using Arrhenius equation

The Arrhenius equation is given by: \[k = Ae^{-\frac{E_a}{RT}}\] where: - \(k\) is the rate constant - \(A\) is the frequency factor (\(2.10 \times 10^{11}\)) - \(E_a\) is the activation energy (\(86.8 kJ/mol\)) - \(R\) is the gas constant (8.314 J/mol\cdot K) - \(T\) is the temperature (in Kelvin) First, we need to convert the temperature from Celsius to Kelvin: \( T = 35 + 273.15 = 308.15 K \) Now, let's plug the values into the Arrhenius equation: \[ k = 2.10 \times 10^{11} \times e^{-\frac{86,800}{8.314 \times 308.15}} \] Calculating the equation gives: \[ k \approx 3.53 \times 10^8 \, M^{-1}s^{-1} \]

(b) Determine the initial rate of the reaction

In this step, we need to calculate the initial concentrations of OH⁻ and C₂H₅I: For KOH: - 0.335 g of KOH is dissolved in 250 mL of solution - KOH's molar mass = 39.10 g/mol - Convert volume to liters: 0.250 L \[ \frac{0.335 \, g}{39.10 \, g/mol} \times \frac{1}{0.250 \, L} \approx 0.0341 \, M \] For C₂H₅I: - 1.453 g of C₂H₅I is dissolved in 250 mL of solution - C₂H₅I's molar mass = 155.97 g/mol - Convert volume to liters: 0.250 L \[ \frac{1.453 \, g}{155.97 \, g/mol} \times \frac{1}{0.250 \, L} \approx 0.0187 \, M \] Considering the reaction is first order in each reactant, the initial rate can be calculated as follows: \[ \text{initial rate} = k \times \left[\text{OH}^{-}\right] \times \left[\text{C}_{2}\text{H}_{5} \text{I}\right] \] \[ \text{initial rate} = 3.53 \times 10^8 \, M^{-1}s^{-1} \times 0.0341 \, M \times 0.0187 \, M \] \[ \text{initial rate} \approx 2.1 \times 10^{-3} \, M/min \]

(c) Identify the limiting reagent

We can determine the limiting reagent by comparing the moles of each reactant present when equal volumes of the two solutions are mixed. For OH⁻ ions, 0.335 grams of KOH are dissolved in 250 mL of solution: \[ \frac{0.335 \, g}{39.10 \, g/mol} \approx 0.00857 \, mol \] For C₂H₅I, 1.453 grams of C₂H₅I are dissolved in 250 mL of solution: \[ \frac{1.453 \, g}{155.97 \, g/mol} \approx 0.00931 \, mol \] Since equal volumes of the two solutions are mixed, the volume of the reaction mixture doubles, and the concentration of each reactant halves. Therefore, the final concentrations are 0.0172 M for the OH⁻ ions and 0.00936 M for the C₂H₅I. To determine the limiting reagent, we need to compare the number of moles of each reactant: - Moles of OH⁻ ions: 0.5 × 0.0341 M × 0.5 L = 0.00857 mol - Moles of C₂H₅I: 0.5 × 0.0187 M × 0.5 L = 0.00468 mol Since there are fewer moles of C₂H₅I (0.00468 mol) than OH⁻ ions (0.00857 mol), the limiting reagent is C₂H₅I.

Key concepts

Arrhenius equation

The Arrhenius equation is a crucial part of chemical kinetics, providing insight into how reaction rates vary with temperature. This equation is expressed as:
\[k = Ae^{-\frac{E_a}{RT}}\]
where:

• \(k\) is the rate constant, representing the reaction rate at a particular temperature.
• \(A\) is the frequency factor, which indicates how often molecules collide in the proper orientation.
• \(E_a\) is the activation energy, or the minimum energy required for a reaction to occur.
• \(R\) is the ideal gas constant (8.314 J/mol·K), a universal constant in these calculations.
• \(T\) is the temperature in Kelvin, reflecting the conditions under which the reaction occurs.

Temperature plays a vital role in this equation. As temperature increases, the exponential term \(e^{-\frac{E_a}{RT}}\) decreases, making \(k\) larger. A higher rate constant suggests a faster reaction because more particles have sufficient energy to surpass the activation barrier. So, using the Arrhenius equation allows us to predict how quickly a reaction takes place under different temperature conditions.

Rate constant

The rate constant \(k\) is an essential factor in understanding the speed of chemical reactions. It varies with both temperature and the presence of a catalyst. Engineers and chemists often use this constant to model and design chemical processes.

The magnitude of \(k\) directly relates to how fast a reaction proceeds. In the equation for a simple first-order reaction:
\[ ext{Rate} = k[ ext{Reactant}] \]
we see that the rate constant is the proportionality factor that links the reaction rate to the concentration of the reactant.

Factors influencing \(k\):

• **Temperature**: Increasing temperature generally leads to an increase in \(k\) due to the greater kinetic energy of the molecules.
• **Activation Energy**: A lower activation energy results in a higher \(k\), as more molecules have the necessary energy to react.
• **Catalysts**: The presence of a catalyst lowers \(E_a\), thereby increasing \(k\) without being consumed in the reaction.

Understanding \(k\) allows scientists to predict and control reaction rates, crucial for industries like pharmaceuticals and energy production.

Limiting reagent

In any chemical reaction, the limiting reagent determines the extent of the reaction. It is the reactant that is entirely consumed first, stopping the reaction from proceeding further.

To identify the limiting reagent, compare the mole ratios of the reactants based on the balanced chemical equation. For the reaction:
\[ ext{C}_2 ext{H}_5 ext{I} + ext{OH}^- \rightarrow ext{C}_2 ext{H}_5 ext{OH} + ext{I}^- \]
we mixed equal volumes of two solutions, one containing KOH and the other C₂H₅I. By calculating the moles of each:

• OH⁻: 0.335 g KOH in 250 mL solution; molar mass = 39.10 g/mol; resulting in 0.00857 mol.
• C₂H₅I: 1.453 g in 250 mL solution; molar mass = 155.97 g/mol; resulting in 0.00931 mol.

After mixing,

• OH⁻ concentration becomes 0.0172 M, while
• C₂H₅I becomes 0.00936 M.

The clear limiting reagent here is C₂H₅I, as it has fewer moles compared to OH⁻. As a result, C₂H₅I will be depleted first, limiting the formation of products. Understanding which reactant is the limiting reagent is crucial because it allows chemists to predict the maximum amount of product that can be formed from a given set of reactants.