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The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: $$ \begin{array}{ll} \hline \text { Temperature (K) } & \text { Rate Constant }\left(\mathbf{s}^{-1}\right) \\ \hline 300 & 3.2 \times 10^{-11} \\ 320 & 1.0 \times 10^{-9} \\ 340 & 3.0 \times 10^{-8} \\ 355 & 2.4 \times 10^{-7} \\ \hline \end{array} $$ From these data, calculate the activation energy in units of \(\mathrm{kJ} / \mathrm{mol}\)

Short Answer

Expert verified
The activation energy for the given reaction is approximately 96.8 kJ/mol.

Step by step solution

01

Choose any two data points

For example, let's choose the following two data points: 1. \(T_1\) = 320 K; \(k_1\) = 1.0 × 10\(^{-9}\) \(s^{-1}\) 2. \(T_2\) = 355 K; \(k_2\) = 2.4 × 10\(^{-7}\) \(s^{-1}\)
02

Apply the logarithmic form of Arrhenius equation

Now let's apply the logarithmic form of the Arrhenius equation for both data points: \[ \ln k_1 = \ln A - \frac{E_a}{R T_1} \] and \[ \ln k_2 = \ln A - \frac{E_a}{R T_2}. \]
03

Subtract the two equations and eliminate the ln A

By subtracting the two equations, we can eliminate the term ln A: \[ \ln k_2 - \ln k_1 = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right). \]
04

Rearrange the resulting equation for Ea

Let's rearrange the equation to make \(E_a\) the subject: \[ E_a = R ((\ln k_2 - \ln k_1) / (\frac{1}{T_1} - \frac{1}{T_2})). \]
05

Plug in the values for the chosen data points and the gas constant

Now, let's plug in the values for the chosen data points (in K) and the rate constants as well as the gas constant: \[ E_a = 8.314 \frac{J}{mol\cdot K} \left(\frac{(\ln (2.4 \times 10^{-7}) - \ln (1.0 \times 10^{-9}))}{(\frac{1}{320} - \frac{1}{355})}\right) \frac{1\,kJ}{1000\, J}. \]
06

Calculate the activation energy

Now, calculate the activation energy: \[ E_a \approx 96.844\, \frac{kJ}{mol}. \] The activation energy for the given reaction is approximately 96.8 kJ/mol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arrhenius Equation
The Arrhenius equation plays a pivotal role in chemical kinetics, as it relates the rate constant of a reaction with both temperature and activation energy. This mathematical relationship helps chemists understand how reaction rates are affected by these variables. The general form of the Arrhenius equation is:

\[\begin{equation}k = Ae^{-\frac{E_a}{RT}}\end{equation}\]
where:
    \t
  • \t\tk is the rate constant,
  • \t
  • \t\tA is the pre-exponential factor, which represents the frequency of collisions with the correct orientation,
  • \t
  • \t\tE_a is the activation energy,
  • \t
  • \t\tR is the gas constant (8.314 J/mol·K), and
  • \t
  • \t\tT is the absolute temperature in Kelvin (K).
Taking the natural logarithm of both sides of the equation allows algebraic manipulation and simplifies the process of calculating the activation energy from experimental rate constants at different temperatures, as shown in the step by step solution.
Rate Constant
In the context of chemical reactions, the rate constant, symbolized as k, is a proportionality factor that provides the relationship between the rate of a reaction and the concentrations of reactants according to the rate law. Significantly, this constant changes with temperature and the nature of the reaction, playing a key role in the kinetics of a process. The rate law can be expressed for a simple reaction as:

\[\begin{equation}rate = k[A]^{m}[B]^{n}\end{equation}\]
where A and B are reactants, and m and n are the respective reaction orders. This equation shows that the rate of reaction is directly proportional to the rate constant. By studying the rate constant across various temperatures, as given in the textbook exercise, we uncover details about the reaction's temperature dependence and activation energy.
Chemical Kinetics
Chemical kinetics is the study of the rates of chemical processes and the factors that influence these rates. It includes the examination of how different conditions, such as concentration, temperature, and presence of a catalyst, affect the speed at which a chemical reaction proceeds. Activation energy, a central concept in kinetics, is the energy barrier that must be overcome for reactants to transform into products. Calculating the activation energy, as demonstrated in the textbook solution, enables chemists to predict reaction behavior and design experiments to control reaction speeds for various applications, including material synthesis and pharmaceutical development.
Temperature Dependence
The rate of a chemical reaction is notably dependent on temperature. Generally, an increase in temperature results in an increased reaction rate. This is due to a greater number of molecules having sufficient energy to overcome the activation energy barrier when the temperature is higher. The Arrhenius equation shows the quantitative relationship between the rate constant and temperature. In solving for activation energy using the Arrhenius equation, the temperature dependence of these rates becomes clear, as even small changes in temperature can lead to significant variations in the rate constant. This sensitivity underscores the importance of controlling temperature during chemical processes and helps explain how chemical reactions tend to be faster at elevated temperatures.

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Most popular questions from this chapter

(a) What are the units usually used to express the rates of reactions occurring in solution? (b) From your everyday experience, give two examples of the effects of temperature on the rates of reactions. (c) What is the difference between average rate and instantaneous rate?

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\) is first order. Near room temperature, the rate constant equals \(7.0 \times 10^{-4} \mathrm{~s}^{-1} .\) Calculate the half-life at this temperature. (b) At \(415^{\circ} \mathrm{C},\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}\) decomposes in the gas phase, \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g) .\) If the reaction is first order with a half-life of 56.3 min at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right.\) ) and overall \(\Delta E\) are \(154 \mathrm{~kJ} / \mathrm{mol}\) and \(136 \mathrm{~kJ} / \mathrm{mol}\), respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

(a) Define the following symbols that are encountered in rate equations for the generic reaction \(\mathrm{A} \longrightarrow \mathrm{B}:[\mathrm{A}]_{0}, t_{1 / 2}[\mathrm{~A}]_{t}, k\) (b) What quantity, when graphed versus time, will yield a straight line for a first-order reaction? (c) How can you calculate the rate constant for a first- order reaction from the graph you made in part (b)?

(a) Two reactions have identical values for \(E_{a} .\) Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a larger rate constant than the other. Account for these observations.

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