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) A 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 reac-solution of \(\mathrm{KOH}\) in ethanol is made up by dissolving 0.335 g 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 \(35^{\circ} \mathrm{C} ?(\mathbf{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(50^{\circ} \mathrm{C}\) .

Short answer

(a) The rate constant at 35°C is calculated using the Arrhenius equation: \(k = (2.10 \times 10^{11} \, M^{-1} s^{-1})e^{-\frac{86.8 \times 10^3 \, J/mol}{(8.314 \, J/mol \cdot K)(308.15 \, K)}} \approx 5.47 \times 10^7 \, M^{-1} s^{-1}\). (b) The initial rate of the reaction at 35°C is calculated using the rate law: \(Rate = k[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}][\mathrm{OH}^-] \approx 5.47 \times 10^7 M^{-1} s^{-1} \cdot 0.01864 \, M \cdot 0.01194 \, M \approx 1.21 \times 10^{-2} \, M s^{-1}\). (c) Since the moles of OH- are less than the moles of C2H5I, OH- is the limiting reagent. (d) The rate constant at 50°C is calculated using the Arrhenius equation: \(k = (2.10 \times 10^{11} \, M^{-1} s^{-1})e^{-\frac{86.8 \times 10^3 \, J/mol}{(8.314 \, J/mol \cdot K)(323.15 \, K)}} \approx 1.72 \times 10^8 \, M^{-1} s^{-1}\).

Step-by-step solution

(a) Calculate the rate constant using the Arrhenius equation

For this exercise, we will use the Arrhenius equation, which connects the rate constant (k) with the activation energy (Ea), the frequency factor (A), the gas constant (R), and the absolute temperature (T): \[k = Ae^{-\frac{Ea}{RT}}\] First, convert the temperature from Celsius to Kelvin: \[T = 35 + 273.15 = 308.15 \, K\] Now, substitute the values given in the problem into the Arrhenius equation: \[k = (2.10 \times 10^{11} \, M^{-1} s^{-1})e^{-\frac{86.8 \times 10^3 \, J/mol}{(8.314 \, J/mol \cdot K)(308.15 \, K)}}\] Calculate the value of k at 35°C.

(b) Calculate the initial rate of reaction

The rate law for the reaction can be expressed as: \[Rate = k[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}][\mathrm{OH}^-]\] Calculate the concentrations of C2H5I and OH- in the solutions: - For KOH: moles = mass / molar mass = 0.335 g / 56.11 g/mol = 0.00597 mol Thus, concentration of OH-: [OH-] = 0.00597 mol / 0.25 L = 0.02388 M - For C2H5I: moles = mass / molar mass = 1.453 g / 155.97 g/mol = 0.00932 mol Thus, concentration of C2H5I: [C2H5I] = 0.00932 mol / 0.25 L = 0.03728 M Since equal volumes are mixed, the final concentrations are halved: - [OH-]final = 0.02388 M / 2 = 0.01194 M - [C2H5I]final = 0.03728 M / 2 = 0.01864 M Now, we can calculate the initial rate using the rate law and the value of k found in part (a): \[Rate = k[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}][\mathrm{OH}^-]\]

(c) Determine the limiting reagent

To determine which reactant is limiting, compare the initial mole ratio of reactants to the stoichiometric ratio of 1:1: - Moles of C2H5I = 0.00932 mol - Moles of OH- = 0.00597 mol Since the moles of OH- are less than the moles of C2H5I, OH- is the limiting reagent. It will be used up first as the reaction proceeds.

(d) Calculate the rate constant at 50°C using the Arrhenius equation

To calculate the rate constant at the new temperature, use the Arrhenius equation again: \[k = Ae^{-\frac{Ea}{RT}}\] First, convert the temperature from Celsius to Kelvin: \[T = 50 + 273.15 = 323.15 \, K\] Now, substitute the values given in the problem into the Arrhenius equation: \[k = (2.10 \times 10^{11} \, M^{-1} s^{-1})e^{-\frac{86.8 \times 10^3 \, J/mol}{(8.314 \, J/mol \cdot K)(323.15 \, K)}}\] Calculate the value of k at 50°C.

Key concepts

Activation Energy

Activation energy, denoted as Ea, is the minimum amount of energy required for a chemical reaction to occur. It acts as an energy barrier that reactants must overcome for a transformation to products. In the context of the Arrhenius equation, activation energy is crucial as it influences the rate constant, k.

For a reaction between ethyl iodide and hydroxide ion, the activation energy is given as 86.8 kJ/mol. This relatively high energy barrier explains why this reaction, like many others, doesn't occur spontaneously at room temperature and requires additional energy, like heat, to proceed. In practice, a higher activation energy usually means a slower reaction rate because fewer molecules have the requisite energy to surpass the barrier at any given time.

Rate Constant

The rate constant, k, is a proportionality factor that connects the reaction rate to the reactant concentrations in the rate law equation. It is dependent on the activation energy and the temperature of the system, as modelled by the Arrhenius equation. Mathematically, k is determined by combining the frequency factor, A, which represents the number of times reactants successfully collide per unit time, with the exponential factor that includes the activation energy and temperature.

The rate constant changes with temperature; as the temperature increases, k generally increases, implying a faster reaction rate. This is because higher temperatures allow more reactant molecules to have equal to or greater energy than the activation energy.

Chemical Kinetics

Chemical kinetics is the study of reaction rates and the factors that affect them, such as temperature, concentration, and the presence of catalysts. It gives us insight into the speed at which a chemical reaction proceeds and how the rate can be manipulated. By applying the Arrhenius equation to a chemical reaction, kineticists can predict how changes in temperature will affect the reaction rate.

Understanding kinetics is vital for controlling reactions in industrial processes, ensuring product quality, and even in predicting the stability of medications. In our example with ethyl iodide and hydroxide ion, kinetic principles help us determine the initial rate and how it will change with temperature.

Limiting Reagent

The limiting reagent in a chemical reaction is the reactant that is completely consumed first, stopping the reaction and limiting the amount of product that can be formed. Identifying the limiting reagent requires comparing the moles of the reactants used with their stoichiometric coefficients in the balanced equation.

In the reaction involving ethyl iodide and hydroxide ions, the limiting reagent is determined by comparing the initial moles of each reactant. As the stoichiometry of the reaction is 1:1, the reactant with fewer moles is the limiting reagent. In this case, KOH supplies the OH- ions, and its initial moles are fewer compared to C2H5I, making OH- the limiting reagent.