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

01

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: \[ k = 4.38 \times 10^7\ \mathrm{M^{-1}\ s^{-1}} \]

02

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: \[ \text{initial rate} = 5.62 \times 10^{-3}\ \mathrm{M/s}\]

03

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.

04

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 \(1.28 \times 10^8\ \mathrm{M^{-1}\ s^{-1}}\).

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