The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.) C 2 H 5 Br(aq) + OH - (aq) → C 2 H 5 OH(l) + Br - (aq) Temperature (°C) k (L,mol •s) 25 8.81⨉10 -5 35 0.000285 45 0.000854 55 0.00239 65 0.00633 b. Determine the rate constant at 15 °C.

hi everyone for this problem we're told the table below shows the results of a kinetic study for the alkaline hydrogel icis of ethyl acetate. The reaction is second order overall calculate the value of the rate constant at 10 degrees Celsius. So our goal here is to calculate the value of our rate constant at 10 degrees Celsius. To solve this problem, we need to plot the given data and from there we can find the value of our activation energy and frequency factor and then use that to find the value of K. At 10 degrees Celsius. So one thing we're going to need to recall here is the Iranian equate and log form which is Ln of K is equal to negative activation energy over r universal gas constant times one over r temperature in kelvin units plus Ln of a. What this translates to is Y equals M X plus B. So when we plot this data, we're going to get an equation of the line made using this raw data and this is something you can plot either an excel or on your calculator. And when we find out what that equation is, it's going to be why equals negative 6876 X plus 20.997. Okay, so from this this equation that we derived, we have the value of the slope which is M and that is represented by negative 6876. And we have the value for the intercept B, which is 20.997. So we just need to find the value of X. Which is one over T four, So let's go ahead and do that. So we know that our temperature is 10 degrees Celsius but we need this in kelvin, so to go from degrees Celsius to kelvin, we add 273.15 and this is going to give us 283.15 kelvin. So now we can solve for X, which is one over T. So one over T is going to equal one over a temperature that we just converted to Kelvin, which is 283.15. So our one over T is going to equal 0.0035317. So now we can use our equation to solve for K. So why? Which is Ln of K is going to equal M. X plus B. Okay, so we have everything, we have em we have X and we have B. So we're going to solve for K. Which is what we're asked to calculate here the value of the rate constant. So we're going to get L n f k is equal to our slope M which is negative 6876 times the X. We just solved for or the X. Which is our one over T. It is 0. plus B. And we know what our B value was. So it's 20.997. So this simplified gives us L n f k is equal to negative 3.287. So remember we're solving for K here. So to get rid of that, L N. We're going to raise both sides to E. So que is going to equal E raised to negative 3.287. And when we do that we get a final answer of K is equal to 3.737 times to the negative two. And our units for K is leaders times mole inverse, times second inverse. So this is going to be our final answer for the Value of the rate constant at 10°C. That's the end of this problem. I hope this was helpful.

Ch.14 - Chemical Kinetics

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Tro 4th Edition

Ch.14 - Chemical Kinetics

Problem 99b

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Chapter 14, Problem 99b

The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.)
C₂H₅Br(aq) + OH^- (aq) → C₂H₅OH(l) + Br^- (aq)
Temperature (°C) k (L,mol •s)
25 8.81⨉10^-5
35 0.000285
45 0.000854
55 0.00239
65 0.00633
b. Determine the rate constant at 15 °C.

Verified step by step guidance

Identify that the problem involves determining the rate constant at a different temperature using the Arrhenius equation.

The Arrhenius equation is given by: k = A * e^(-Ea/(RT)), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/mol*K), and T is the temperature in Kelvin.

To find the rate constant at 15 °C, first convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature.

Use the given rate constants at different temperatures to plot ln(k) versus 1/T (in Kelvin) and determine the slope, which is equal to -Ea/R.

Once the activation energy (Ea) is determined, use the Arrhenius equation to calculate the rate constant at 15 °C by substituting the values of Ea, R, and the temperature in Kelvin into the equation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Reaction Order

The order of a reaction refers to the power to which the concentration of a reactant is raised in the rate law. In this case, the reaction is first order in each reactant, meaning the rate is directly proportional to the concentration of each reactant. Understanding reaction order is crucial for determining how changes in concentration affect the reaction rate.

Arrhenius Equation

The Arrhenius equation describes how the rate constant (k) of a reaction changes with temperature. It is expressed as k = A * e^(-Ea/RT), where A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. This relationship allows us to estimate the rate constant at different temperatures, which is essential for solving the given problem.

Rate Constant (k)

The rate constant (k) is a proportionality factor in the rate law that relates the rate of a reaction to the concentrations of the reactants. It varies with temperature and is specific to each reaction. In this question, determining the rate constant at 15 °C requires understanding how k changes with temperature, which can be inferred from the provided data.

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Equilibrium Constant K

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Open Question

What fraction of the film is left after 10 s, assuming the same initial coverage as in part a, given that the evaporation of a 120-nm film of n-pentane from a single crystal of aluminum oxide is zero order with a rate constant of 1.92 * 10^13 molecules/cm^2 * s at 120 K?

Textbook Question

The evaporation of a 120-nm film of n-pentane from a single crystal of aluminum oxide is zero order with a rate constant of 1.92⨉1013 molecules/cm²•s at 120 K. a. If the initial surface coverage is 8.9⨉1016 molecules/cm², how long will it take for one-half of the film to evaporate?

The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.)

C₂H₅Br(aq) + OH^- (aq) → C₂H₅OH(l) + Br^- (aq)

Temperature (°C) k (L,mol •s)

25 8.81⨉10^-5

35 0.000285

45 0.000854

55 0.00239

65 0.00633

a. Determine the activation energy and frequency factor for the reaction.

Textbook Question

The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.)

C₂H₅Br(aq) + OH^- (aq) → C₂H₅OH(l) + Br^- (aq)

Temperature (°C) k (L,mol •s)

25 8.81⨉10^-5

35 0.000285

45 0.000854

55 0.00239

65 0.00633

c. If a reaction mixture is 0.155 M in C₂H₅Brand 0.250 M in OH^-, what is the initial rate of the reaction at 75 °C?

Textbook Question

The reaction 2 N₂O₅ → 2 N₂O₄ + O₂ takes place at around room temperature in solvents such as CCl₄. The rate constant at 293 K is found to be 2.35⨉10^-4 s^-1, and at 303 K the rate constant is found to be 9.15⨉10^-4 s^-1.Calculate the frequency factor for the reaction.

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Open Question

a. This reaction has an activation energy of zero in the gas phase: CH3 + CH3 → C2H6. b. Why might the activation energy be zero? c. What other types of reactions would you expect to have little or no activation energy?