Problem 120 - Chemistry The Central Science

Question

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(\) alc \() \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(\) alc \(),\) 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 \(\mathrm{KOH}\) 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 \(35^{\circ} \mathrm{C} ?\) (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 \(50^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

(a) The rate constant for the reaction at $35^{\circ} \mathrm{C}$ is \(4.38 \times 10^7\ \mathrm{M^{-1}\ s^{-1}}\). (b) The initial rate of the reaction at $35^{\circ} \mathrm{C}$ is \(5.62 \times 10^{-3}\ \mathrm{M/s}\). (c) KOH is the limiting reagent in the reaction. (d) The rate constant for the reaction at $50^{\circ} \mathrm{C}$ is \(1.28 \times 10^8\ \mathrm{M^{-1}\ s^{-1}}\).

Step 1: Calculating the rate constant at 35°C

We will use the Arrhenius equation to find the rate constant at 35°C. The Arrhenius equation is: \[ k = Ae^{-\frac{E_a}{RT}} \] where: \(k\) is the rate constant \(A\) is the frequency factor \(E_a\) is the activation energy in J/mol \(R\) is the gas constant \(= 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}\) \(T\) is the temperature in Kelvin. First, convert the temperature from Celsius to Kelvin: \(T = 35 + 273.15 = 308.15\ \mathrm{K}\) Next, we need to convert the activation energy from kJ/mol to J/mol. To do this we use the conversion factor \(1000\ \mathrm{J / kJ}\): \[ E_a = 86.8\times 10^3\ \mathrm{J/mol} \] Now we can use the Arrhenius equation to find the rate constant at 35°C: \[ k = (2.10\times 10^{11}\ \mathrm{M^{-1}\ s^{-1}}) \times e^{-\frac{(86.8\times 10^3\ \mathrm{J/mol})}{(8.314\ \mathrm{J\ K^{-1}\ mol^{-1}})(308.15\ \mathrm{K})}} \] Calculating the value of \(k\), we get: \[ k = 4.38 \times 10^7\ \mathrm{M^{-1}\ s^{-1}} \]

Step 2: Calculating the initial rate of the reaction

First, we need to find the initial concentrations of our reactants. The initial concentration of KOH can be found using the given mass of KOH and the volume of the solution. The molar mass of KOH is \(39.1\ \mathrm{g/mol}\), so we have: \[ [\mathrm{KOH}] = \frac{0.335\ \mathrm{g}}{39.1\ \mathrm{g/mol}} \times \frac{1}{0.250\ \mathrm{L}} = 3.41 \times 10^{-2}\ \mathrm{M}\] Similarly, the molar mass of C2H5I is \(155.0\ \mathrm{g/mol}\), so we have: \[ [\mathrm{C2H5I}] = \frac{1.453\ \mathrm{g}}{155.0\ \mathrm{g/mol}} \times \frac{1}{0.250\ \mathrm{L}} = 3.75 \times 10^{-2}\ \mathrm{M}\] Since we know the rate constant (\(k = 4.38 \times 10^7\ \mathrm{M^{-1}\ s^{-1}}\)), and that the reaction is first order with respect to both reactants, we can find the initial rate of the reaction using the rate law: \[ \text{initial rate} = k \times [\mathrm{KOH}] \times [\mathrm{C2H5I}]\] \[ \text{initial rate} = (4.38 \times 10^7\ \mathrm{M^{-1}\ s^{-1}}) \times (3.41 \times 10^{-2}\ \mathrm{M}) \times (3.75 \times 10^{-2}\ \mathrm{M})\] Calculating the initial rate, we get: \[ \text{initial rate} = 5.62 \times 10^{-3}\ \mathrm{M/s}\]

Step 3: Finding the limiting reagent

Since equal volumes of the two solutions are mixed, the initial concentrations of the reactants are halved. As a result, we will compare the initial concentrations and determine which one will be completely consumed first. The reactant with the lower initial concentration will be the limiting reagent. In this case, since \[\frac{1}{2} [\mathrm{KOH}] < \frac{1}{2} [\mathrm{C2H5I}]\] \[\frac{1}{2} (3.41 \times 10^{-2}\ \mathrm{M}) < \frac{1}{2} (3.75 \times 10^{-2}\ \mathrm{M})\] KOH will be the limiting reagent in the reaction.

Step 4: Calculating the rate constant at 50°C

Using the same strategy as in step 1, we will use the Arrhenius equation to find the rate constant at 50°C. First, convert the temperature from Celsius to Kelvin: \(T = 50 + 273.15 = 323.15\ \mathrm{K}\) Now, we can plug the values into the Arrhenius equation to find the rate constant at 50°C: \[ k = (2.10\times 10^{11}\ \mathrm{M^{-1}\ s^{-1}}) \times e^{-\frac{(86.8\times 10^3\ \mathrm{J/mol})}{(8.314\ \mathrm{J\ K^{-1}\ mol^{-1}})(323.15\ \mathrm{K})}} \] Calculating the value of \(k\), we get: \[ k = 1.28 \times 10^8\ \mathrm{M^{-1}\ s^{-1}} \] The rate constant for the reaction at 50°C is \(1.28 \times 10^8\ \mathrm{M^{-1}\ s^{-1}}\).

Key Concepts

Rate Constant

When studying chemical reactions, the rate constant (\(k\)) is crucial. It's part of the rate law equation, telling us how fast a reaction proceeds. In the context of the Arrhenius equation, it's given by:

• \( k = Ae^{-\frac{E_a}{RT}} \)

where:

• \( A \) is the frequency factor, reflecting the number of times reactants collide with the correct orientation per unit time.
• \( E_a \) is the activation energy, representing the minimum energy required for a reaction to occur.
• \( R \) is the gas constant, approximately \( 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}} \).
• \( T \) stands for temperature in Kelvin.

The rate constant changes with temperature, showcasing the Arrhenius equation's dependency on \( T \). For the reaction given, calculations show how the rate constant increases as the temperature rises from 35°C to 50°C. This demonstrates the fact that higher temperatures generally increase reaction rates. Identifying how \( k \) varies with temperature is an essential step in kinetic studies.

Activation Energy

Activation energy (\(E_a\)) is a key concept in understanding reaction kinetics. It is the energy barrier that must be overcome for reactants to be transformed into products. Higher \(E_a\) values indicate slower reactions, as more energy is needed to initiate the process. The Arrhenius equation, \( k = Ae^{-\frac{E_a}{RT}} \), emphasizes that \(E_a\) is inversely related to \(k\). Lower activation energies mean higher rate constants, leading to faster reactions. In our example, the activation energy is specified as \( 86.8\ \mathrm{kJ/mol} \). This value is quite significant, meaning that plenty of energy is required for the reaction between ethyl iodide and hydroxide ion to proceed. Converting this into \(\mathrm{J/mol}\) is crucial for calculations involving the Arrhenius equation. Knowing \(E_a\) allows chemists to manipulate reaction conditions to optimize rates, such as by increasing temperature or using catalysts to lower the effective activation energy.

Limiting Reagent

In chemical reactions, the limiting reagent is the reactant that is completely consumed first, limiting the amount of product formed. It determines when the reaction stops. Identifying the limiting reagent is crucial for calculating theoretical yields and understanding reaction efficiency. For instance, when equal volumes of potassium hydroxide and ethyl iodide solutions are mixed, their concentrations are halved. By comparing these concentrations, the limiting reagent is identified—whichever has the smaller concentration after dilution is limiting. In this example, potassium hydroxide (KOH) is found to be the limiting reagent since its concentration is comparatively lower than that of ethyl iodide. This means KOH will be used up first, capping the reaction's progress. Thus, analyzing the stoichiometry and initial concentrations of reactants helps predict reaction outcomes and optimize processes in lab settings.