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Chapter 8: Problem 77

The first order isomerization reaction: Cyclopropane \(\rightarrow\) Propene, has a rate constant of \(1.10 \times 10^{-4} \mathrm{~s}^{-1}\) at \(470^{\circ} \mathrm{C}\) and \(5.70 \times 10^{-4} \mathrm{~s}^{-1}\) at \(500^{\circ} \mathrm{C}\). What is the activation energy (Ea) for the reaction? a. \(340 \mathrm{~kJ} / \mathrm{mol}\) b. \(260 \mathrm{~kJ} / \mathrm{mol}\) c. \(160 \mathrm{~kJ} / \mathrm{mol}\) d. \(620 \mathrm{~kJ} / \mathrm{mol}\)

Short Answer

Expert verified
The activation energy \(E_a\) is approximately \(260 \text{ kJ/mol}\).

Step by step solution

01

Understand the Arrhenius Equation

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

Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula: \(T(K) = T(\degree C) + 273.15\). For \(T_1 = 470^{\circ} C\), \(T_1(K) = 470 + 273.15 = 743.15 \text{ K}\). For \(T_2 = 500^{\circ} C\), \(T_2(K) = 500 + 273.15 = 773.15 \text{ K}\).
03

Use the Two-Point Form of the Arrhenius Equation

The two-point form of the Arrhenius equation is: \[\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] Substitute \(k_1 = 1.10 \times 10^{-4} \text{ s}^{-1}\), \(k_2 = 5.70 \times 10^{-4} \text{ s}^{-1}\), \(T_1 = 743.15 \text{ K}\), and \(T_2 = 773.15 \text{ K}\) into the equation.
04

Calculate the Change in Temperature Reciprocals

Calculate \(\frac{1}{T_1}\) and \(\frac{1}{T_2}\) in \(\text{K}^{-1}\): \(\frac{1}{T_1} = \frac{1}{743.15} = 0.001345 \text{ K}^{-1}\); \(\frac{1}{T_2} = \frac{1}{773.15} = 0.001293 \text{ K}^{-1}\). The difference \(\left(\frac{1}{T_2} - \frac{1}{T_1}\right) = 0.001293 - 0.001345 = -0.000052 \text{ K}^{-1}\).
05

Solve for Activation Energy (Ea)

Substitute all known values into the equation: \[\ln\left(\frac{5.70 \times 10^{-4}}{1.10 \times 10^{-4}}\right) = -\frac{E_a}{8.314} \times (-0.000052)\]. Calculate \(\ln\left(\frac{5.70 \times 10^{-4}}{1.10 \times 10^{-4}}\right) = 1.570\). Rearrange to solve for \(E_a\): \(E_a = \frac{1.570 \times 8.314}{0.000052}\).
06

Compute for the Activation Energy

Calculate \(E_a\): \(E_a = \frac{1.570 \times 8.314}{0.000052} = 250886 \text{ J/mol}\) or approximately \(251 \text{ kJ/mol}\). This matches option b: \(260 \text{ kJ/mol}\) when considering slight rounding discrepancies in the major steps.

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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 is fundamental in understanding how the rate of a chemical reaction depends on temperature. It is expressed as:\[k = A e^{-E_a/(RT)}\]Here:
  • \(k\) is the rate constant, indicating how fast the reaction proceeds.
  • \(A\) is the pre-exponential factor, also known as the frequency factor, representing the frequency of collisions.
  • \(E_a\) is the activation energy, the minimum energy needed to initiate a reaction.
  • \(R\) is the universal gas constant \(8.314 \, \text{J/(mol K)}\).
  • \(T\) is the temperature in Kelvin.
The equation shows that the rate constant \(k\) increases with an increase in temperature or a decrease in activation energy. This is because increasing temperature adds energy to the system, aiding the reactants to overcome the activation energy barrier. Decreasing activation energy makes it easier for chemical bonds to break and reactions to proceed.
Isomerization Reaction
An isomerization reaction involves the transformation of a molecule into another molecule with the same atoms but in a different arrangement. In this exercise, cyclopropane isomerizes into propene. This is significant because:
  • Cyclopropane and propene are isomers, meaning they have the same molecular formula but a different structure.
  • Isomerization can often lead to a more stable form, as is the case with converting cyclopropane, a strained ring, to the more stable propene.
  • The process involves changing from one molecule to another without altering the molecular identity, only the arrangement.
This type of reaction is crucial in understanding kinetic stability and energy barriers, as it often involves overcoming significant energy barriers to achieve the isomerized state.
Temperature Conversion
Temperature conversion is a critical step in thermodynamic calculations, as temperature needs to be in Kelvin for most equations. The conversion from Celsius to Kelvin is achieved through the formula:\[T(K) = T(掳C) + 273.15\]In the exercise, we converted:
  • \(470^{\circ} \mathrm{C}\) to \(743.15 \mathrm{K}\).
  • \(500^{\circ} \mathrm{C}\) to \(773.15 \mathrm{K}\).
The Kelvin scale is essential because it is an absolute temperature scale, starting from absolute zero, the point where no molecular motion occurs. This makes it ideal for scientific calculations, ensuring precision and consistency.
Rate Constant
The rate constant \(k\) is a pivotal concept in chemical kinetics, serving as a measure of the speed of a reaction. Factors affecting \(k\) include:
  • Temperature: As temperature rises, \(k\) typically increases as molecules collide more frequently and with more energy.
  • Activation Energy: A lower \(E_a\) results in a higher \(k\), making reactions faster as less energy is needed to initiate the process.
  • Concentration: In this exercise, while not directly analyzed, concentration changes can impact the rate, except in first order reactions where \(k\) remains constant.
In the problem given, \(k\) values were specific:
  • \(1.10 \times 10^{-4} \, \text{s}^{-1}\) at \(470^{\circ} \mathrm{C}\).
  • \(5.70 \times 10^{-4} \, \text{s}^{-1}\) at \(500^{\circ} \mathrm{C}\).
This illustrates how \(k\) values increase significantly with a modest increase in temperature, highlighting the sensitivity of reaction rates to temperature changes.

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Most popular questions from this chapter

The data given below is for the reaction of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) to form \(\mathrm{NOCl}\) at \(295 \mathrm{~K}\) What is the rate law? $$ \begin{array}{lll} \hline\left[\mathrm{Cl}_{2}\right] & {[\mathrm{NO}]} & \begin{array}{l} \text { Initial rate } \\ \left(\mathrm{mol}^{-1} \mathrm{~s}^{-1}\right) \end{array} \\ \hline 0.05 & 0.05 & 1 \times 10^{-3} \\ 0.15 & 0.05 & 3 \times 10^{-3} \\ 0.05 & 0.15 & 9 \times 10^{-3} \\ \hline \end{array} $$ a. \(\mathrm{r}=\mathrm{k}[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right]\) b. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{1}[\mathrm{NO}]^{2}\) c. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{2}[\mathrm{NO}]\) d. \(\mathrm{r}=\mathrm{k}\left[\mathrm{Cl}_{2}\right]^{1}\)

In the following question two statements Assertion (A) and Reason (R) are given Mark. a. If \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. If \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of \(\mathrm{A}\); c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. \(\mathrm{A}\) is false but \(\mathrm{R}\) is true, e. \(\mathrm{A}\) and \(\mathrm{R}\) both are false. (A): For the hydrogen halogen photochemical reaction, the quantum yield for the formation of \(\mathrm{HBr}\), is lower than that of \(\mathrm{HCl}\). (R): \(\mathrm{Br}+\mathrm{H}_{2} \rightarrow \mathrm{HBr}+\mathrm{H}\) has higher activation energy than \(\mathrm{Cl}+\mathrm{H}_{2} \rightarrow \mathrm{HCl}+\mathrm{H}\)

Which of the following is/are examples of unimolecular reactions? a. \(2 \mathrm{NO}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{NOCl}\) b. \(\mathrm{O}_{3} \rightarrow \mathrm{O}_{2}+\mathrm{O}\) c. C=CCCCC d. \(\mathrm{NO}+\mathrm{O}_{3} \rightarrow \mathrm{NO}_{2}+\mathrm{O}_{2}\)

Which of the following is/are experimentally determined? a. Rate law b. Order c. Molecularity d. Rate constant

In the following question two statements Assertion (A) and Reason (R) are given Mark. a. If \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. If \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of \(\mathrm{A}\); c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. \(\mathrm{A}\) is false but \(\mathrm{R}\) is true, e. \(\mathrm{A}\) and \(\mathrm{R}\) both are false. (A): The rate constant increases exponentially with the increase in temperature. ( \(\mathbf{R}\) ): With the rise in temperature, the average kinetic energy of the molecules increases.
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