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Chapter 14: Problem 64

The temperature dependence of the rate constant for a reaction is tabulated as follows: $$ \begin{array}{ll} \hline \text { Temperature (K) } & k\left(M^{-1} \mathrm{~s}^{-1}\right) \\ \hline 600 & 0.028 \\ 650 & 0.22 \\ 700 & 1.3 \\ 750 & 6.0 \\ 800 & 23 \\ \hline \end{array} $$ Calculate \(E_{a}\) and \(A\).

Short Answer

Expert verified
The activation energy, \(E_a\), is 49,591 J/mol, and the pre-exponential factor, \(A\), is 9.65 脳 10\(^{-7}\) M\(^{-1}\)s\(^{-1}\).

Step by step solution

01

Write down the Arrhenius equation

The Arrhenius equation relates the rate constant \(k\), activation energy \(E_{a}\), and pre-exponential factor \(A\) of a reaction to the temperature \(T\) as follows: \[ k = A e^{-E_a / RT} \] where R is the gas constant, R = 8.314 J\(K^{-1}mol^{-1}\), and T is the temperature in Kelvin.
02

Take natural logarithm of Arrhenius equation

To find the activation energy and the pre-exponential factor, take the natural logarithm of the Arrhenius equation to make it linear: \[ \ln k = \ln A - \frac{E_a}{RT} \]
03

Rearrange Arrhenius equation

Rearrange the linearized Arrhenius equation to the form y = mx + b, with y = ln(k) and x = 1/T: \[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \] In this equation, the slope of the graph ln(k) vs. 1/T is \(m = -E_a/R\).
04

Calculate ln(k) and 1/T for each data point

Use the given temperature and rate constant values to calculate ln(k) and 1/T for each data point: \[ \begin{array}{|c|c|c|} \hline \text{Temperature (K)} & \text{Rate constant } k\left( M^{-1}s^{-1} \right) & \ln (k) & \frac{1}{T} (K^{-1}) \\ \hline 600 & 0.028 & -3.5755 & 0.0016667 \\ 650 & 0.22 & -1.5141 & 0.0015385 \\ 700 & 1.3 & 0.2624 & 0.0014286 \\ 750 & 6.0 & 1.7918 & 0.0013333 \\ 800 & 23 & 3.1355 & 0.0012500 \\ \hline \end{array} \]
05

Calculate slope of the graph ln(k) vs. 1/T

Using any two points from the table, calculate the slope of the graph. We'll use the points (0.0016667, -3.5755) and (0.0012500, 3.1355): \[ m = \frac{3.1355 - (-3.5755)}{0.0012500 - 0.0016667} = -5961.223 \]
06

Calculate the activation energy, Ea

Using the slope of the graph and the gas constant R, calculate the activation energy: \[ E_a = -mR = 5961.223 \times 8.314 \, J/molK = 49,591 \, J/mol \]
07

Calculate the pre-exponential factor, A

To calculate A, choose one of the given temperature and rate constant values and substitute the values of \(E_a\), \(R\), \(T\), and \(k\) in the linearized Arrhenius equation: \[ \ln A = \ln k + \frac{E_a}{RT} \] Choose the first data point with T = 600 K and k = 0.028 M\(^{-1}\)s\(^{-1}\): \[ \ln A = \ln(0.028) + \frac{49,591 \, J/mol}{8.314 \, J/molK \times 600 \, K} = -13.884 \] Take the exponent of both sides to find the value of A: \[ A = e^{-13.884} = 9.65 \times 10^{-7} \, M^{-1}s^{-1} \] The activation energy, \(E_a\), is 49,591 J/mol, and the pre-exponential factor, \(A\), is 9.65 脳 10\(^{-7}\) M\(^{-1}\)s\(^{-1}\).

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Most popular questions from this chapter

(a) In which of the following reactions would you expect the orientation factor to be least important in leading to reaction: \(\mathrm{NO}+\mathrm{O} \longrightarrow \mathrm{NO}_{2}\) or \(\mathrm{H}+\mathrm{Cl} \longrightarrow \mathrm{HCl}\) ? (b) How does the kinetic-molecular theory help us understand the temperature dependence of chemical reactions?

As described in Exercise 14.41, the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at \(660 \mathrm{~K}\) is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). (a) If we begin with an initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) pressure of 450 torr, what is the partial pressure of this substance after \(60 \mathrm{~s}\) ? (b) At what time will the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

In solution, chemical species as simple as \(\mathrm{H}^{+}\)and \(\mathrm{OH}^{-}\)can serve as catalysts for reactions. Imagine you could measure the \(\left[\mathrm{H}^{+}\right.\)] of a solution containing an acid- catalyzed reaction as it occurs. Assume the reactants and products themselves are neither acids nor bases. Sketch the \(\left[\mathrm{H}^{+}\right]\)concentration profile you would measure as a function of time for the reaction, assuming \(t=0\) is when you add a drop of acid to the reaction.

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)\) (b) \(\mathrm{OCl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q)\) (c) \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g)\)

(a) What is meant by the term elementary reaction? (b) What is the difference between a unimolecular and a bimolecular elementary reaction? (c) What is a reaction mechanism? (d) What is meant by the term rate-determining step?
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