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Chapter 14: Problem 118

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}^{-}(\( alc \))\( has an activation energy of \)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 \(0.500 \mathrm{~g} \mathrm{KOH}\) in ethanol to form $500 \mathrm{~mL}\( of solution. Similarly, \)1.500 \mathrm{~g}\( of \)\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\( is dissolved in ethanol to form \)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 \(30^{\circ} \mathrm{C} ?(\mathbf{c})\) 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 $40^{\circ} \mathrm{C} .$

Short Answer

Expert verified
The rate constants for the reaction at 30°C and 40°C are approximately 5.56 x 10^7 M^(-1) s^(-1) and 2.09 x 10^8 M^(-1) s^(-1), respectively. The initial rate of the reaction at 30°C is approximately 1.90 x 10^6 M/s. Neither reactant is clearly limiting in this case, as they both have nearly the same number of moles and a stoichiometric ratio of 1:1.

Step by step solution

01

Part (a): Rate constant at 30°C

Given: Activation energy (Ea) = 86.8 kJ/mol, Frequency factor (A) = 2.1 x 10^11 M^(-1) s^(-1) Firstly, convert the temperature from Celsius to Kelvin: T = 30 + 273.15 = 303.15 K Now, we will make use of Arrhenius equation to find the rate constant, k: \[k = Ae^{\frac{-Ea}{RT}}\] Here, R is the universal gas constant, R = 8.314 J/(mol.K) Converting Ea to J/mol: 86.8 kJ/mol x 1000 = 86800 J/mol \[k = (2.1 \times 10^{11}) e^{\frac{-86800}{(8.314)(303.15)}}\] Calculate the value of k: k ≈ 5.56 x 10^7 M^(-1) s^(-1) The rate constant for the reaction at 30°C is approximately 5.56 x 10^7 M^(-1) s^(-1).
02

Part (b): Initial rate at 30°C

First, we need to calculate the molar concentration of each reactant. For KOH: Mass = 0.500 g, Molar mass = 56.11 g/mol Moles of KOH = \(\frac{0.500}{56.11}\) = 0.00891 mol Volume of solution = 500 mL = 0.500 L Concentration of OH^(-) = \(\frac{0.00891}{0.500}\) = 0.0178 M For C2H5I: Mass = 1.500 g, Molar mass = 155.97 g/mol Moles of C2H5I = \(\frac{1.500}{155.97}\) = 0.00962 mol Volume of solution = 500 mL = 0.500 L Concentration of C2H5I = \(\frac{0.00962}{0.500}\) = 0.0192 M When equal volumes of the two solutions are mixed, the volume doubles but the concentration remains the same. The rate law for the reaction is: Rate = k [C2H5I][OH^-] Using the rate constant at 30°C (k ≈ 5.56 x 10^7 M^(-1) s^(-1), and concentrations as calculated above: Initial rate ≈ (5.56 x 10^7)(0.0192)(0.0178) ≈ 1.90 x 10^6 M/s The initial rate of the reaction at 30°C is approximately 1.90 x 10^6 M/s.
03

Part (c): Limiting reagent

To determine the limiting reagent, compare the mole ratio of reactants: \[\frac{moles\:of\:C_2H_5I}{moles\:of\:OH^-} = \frac{0.00962}{0.00891} = 1.08\] Since the stoichiometric ratio is 1:1, C2H5I and OH^- have nearly the same number of moles, meaning the reaction will consume both reactants almost evenly. Neither reagent is clearly limiting in this case.
04

Part (d): Rate constant at 40°C

Similar to part (a), we will use the Arrhenius equation to find the rate constant at 40°C: T = 40 + 273.15 = 313.15 K \[k = (2.1 \times 10^{11}) e^{\frac{-86800}{(8.314)(313.15)}}\] Calculate the value of k: k ≈ 2.09 x 10^8 M^(-1) s^(-1) The rate constant for the reaction at 40°C is approximately 2.09 x 10^8 M^(-1) s^(-1).

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Most popular questions from this chapter

Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(392.4 \mathrm{pm} .(\mathbf{a})\) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is $(4 / 3) \pi r^{3} .\( Recall that \)1 \mathrm{pm}=1 \times 10^{-12} \mathrm{~m}\( and \)1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m} .(\mathbf{b})$ Estimate how many platinum atoms are on the surface of a 2.0-nm Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of \(280 \mathrm{pm}\) (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a 5.0-nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a \(1 \times 10^{5}\) -fold increase in the reaction rate?

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in $0.1 \mathrm{M} \mathrm{HCl}$ occurs according to the reaction $$ \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) $$ \text { is } 8.56 \times 10^{-5} \mathrm{M} / \mathrm{s} $$ (a) What is the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after $4.00 \times 10^{3} \mathrm{~s}\( if the starting concentration is \)0.500 \mathrm{M}$ ? (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C}\) ?

(a) What are the units usually used to express the rates of reactions occurring in solution? (b) As the temperature increases, does the reaction rate increase or decrease? (c) As a reaction proceeds, does the instantaneous reaction rate increase or decrease?

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: $\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-} .$ This rapid reaction gives the following rate data: $$ \begin{array}{ccc} \hline\left[\mathrm{OCI}^{-}\right](M) & {\left[\mathrm{I}^{-}\right](M)} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1.5 \times 10^{-3} & 1.5 \times 10^{-3} & 1.36 \times 10^{-4} \\ 3.0 \times 10^{-3} & 1.5 \times 10^{-3} & 2.72 \times 10^{-4} \\ 1.5 \times 10^{-3} & 3.0 \times 10^{-3} & 2.72 \times 10^{-4} \\ \hline \end{array} $$ (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when $\left[\mathrm{OCl}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\( and \)\left[\mathrm{I}^{-}\right]=5.0 \times 10^{-4} \mathrm{M}$
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