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Chapter 9: Problem 21

Calculate the activation energy for a reaction in which the rate constant is \(\left.5 \text { times faster at } 50^{\circ} \mathrm{C} \text { than at } 20^{\circ} \mathrm{C} \text { . (Section } 9.7\right)\)

Short Answer

Expert verified
The activation energy is approximately 43.2 kJ/mol.

Step by step solution

01

Identify Known Values

First, we identify the known values given in the problem. We know two temperatures: \(T_1 = 20^{\circ}C\) and \(T_2 = 50^{\circ}C\). We convert these to Kelvin by adding 273.15. Thus, \(T_1 = 293.15\, K\) and \(T_2 = 323.15\, K\). The rate constant at \(50^{\circ}C\) (\(k_2\)) is 5 times the rate constant at \(20^{\circ}C\) (\(k_1\)), so \(k_2 = 5k_1\).
02

Determine the Arrhenius Equation

The Arrhenius equation relates the rate constant \(k\) to the activation energy \(E_a\) and temperature \(T\): \[ k = A e^{-\frac{E_a}{RT}} \]where \(A\) is the frequency factor, \(R\) is the universal gas constant \(8.314\, J/mol\cdot K\), and \(T\) is the absolute temperature in Kelvin.
03

Use the Arrhenius Equation Ratio

We take the natural logarithm of the ratio of rate constants \(\frac{k_2}{k_1} = 5\).\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]Plug in \(\ln(5)\), \(T_1 = 293.15\, K\), \(T_2 = 323.15\, K\), and \(R = 8.314\, J/mol\cdot K\).
04

Calculate the Activation Energy

Rearrange the equation to solve for the activation energy \(E_a\):\[ E_a = \frac{\ln(5) \times R}{\left(\frac{1}{T_1} - \frac{1}{T_2}\right)} \]Substitute \(\ln(5) \approx 1.6094\) and calculate:\[ E_a = \frac{1.6094 \times 8.314}{\left(\frac{1}{293.15} - \frac{1}{323.15}\right)} \]Calculate the numerical value for \(E_a\).
05

Perform Final Calculation

Perform the calculation:\[ \frac{1}{293.15} \approx 0.00341\]\[ \frac{1}{323.15} \approx 0.00310\]\[ E_a = \frac{1.6094 \times 8.314}{0.00341 - 0.00310} = \frac{13.377}{0.00031} \approx 43152\, J/mol \]Convert \(E_a\) to \(kJ/mol\) by dividing by 1000: \(E_a \approx 43.2\, kJ/mol\).
06

Verify Units and Calculation

Ensure the units are correct and the calculation aligns with standard thermodynamics. \(E_a\) is typically expressed in \(kJ/mol\), which matches our improved result of \(43.2 \ kJ/mol\).

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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 is a crucial concept in chemical kinetics, providing insight into how temperature affects reaction rates. This equation links the rate constant, activation energy, and temperature. It states:
\[ k = A e^{-\frac{E_a}{RT}} \]
  • \(k\): the rate constant, determining how quickly a reaction proceeds.
  • \(A\): the frequency factor, indicative of the number of successful collisions between reactant molecules.
  • \(E_a\): activation energy, the minimum energy required to initiate the reaction.
  • \(R\): the universal gas constant, playing a role in converting energy units.
  • \(T\): temperature in Kelvin, crucial for capturing the effect of temperature on reaction rates.
At the core, the Arrhenius Equation explains that as temperature increases, the rate constant also increases, because more molecules have the required activation energy to react.
This explains why reactions generally proceed faster at higher temperatures.
Rate Constants
Rate constants (
\(k\)) in chemical reactions are essential for understanding how fast a reaction occurs. They are influenced by factors such as temperature and presence of catalysts. One important point is that they change with temperature, as seen in the Arrhenius Equation.
  • At a higher temperature, molecules possess more energy, thus increasing the frequency of effective collisions, resulting in a higher rate constant.
  • In our example: at \(50^{\circ}C\), the rate constant (\(k_2\)) is 5 times higher than at \(20^{\circ}C\) (\(k_1\)), i.e., the reaction occurs 5 times faster.
Understanding rate constants helps in accurately predicting how quickly a chemical process will happen, which is crucial in both industrial and research settings.
Temperature Conversion
Temperature conversion between Celsius and Kelvin is pivotal in chemical equations since the Arrhenius Equation requires temperature in Kelvin. The formula is simple:
\[ T(K) = T(\text{in Celsius}) + 273.15 \]
  • This conversion standardizes the temperature, ensuring it reflects absolute temperatures needed for thermodynamic equations.
  • In the provided solution, the conversions were: \(T_1 = 20^{\circ}C = 293.15 \, K\) and \(T_2 = 50^{\circ}C = 323.15 \, K\).
Using Kelvin ensures that calculations related to energy transformations, such as activation energy, are consistent and accurate, reflecting true thermodynamic properties.
Universal Gas Constant
The Universal Gas Constant (\(R\)) is a fundamental constant that appears in various equations relating to gases and reactions. It facilitates calculations across different scientific areas, ensuring consistency in units. For our purpose:
\[ R = 8.314 \, \text{J/mol鈰匥} \]
  • \(R\) ensures that energy calculations such as those involving \(E_a\) (activation energy) are accurate.
  • It also bridges the energy factors from the microscopic world (molecular) to the macroscopic world (observable reaction rates).
In the calculation of activation energy, \(R\) is used to balance the units across the equation for consistent and correct results. It plays a vital role in converting energy quantities to values that can be compared and applied in practical scenarios.

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Most popular questions from this chapter

Rate constants at a series of temperatures were obtained for the decomposition of azomethane \\[ \mathrm{CH}_{3} \mathrm{N}_{2} \mathrm{CH}_{3} \rightarrow 2 \mathrm{CH}_{3}^{*}+\mathrm{N}_{2} \\] $$\begin{array}{llllll} \hline T / \mathrm{K} & 523 & 541 & 560 & 576 & 593 \\ k / 10^{-6} \mathrm{s}^{-1} & 1.8 & 15 & 60 & 160 & 950 \end{array}$$ Use the data in the table to find the activation energy, \(E_{a}\), for the reaction. (Section \(9.7)\)

For a particular first order reaction, half of the reactant is used up after 15 s. What fraction of the reactant will remain after 1 \(\min ?(\text { Section } 9.4)\)

From the data provided in the table, deduce the rate equation and the value of the rate constant for the following reaction. (Section \(9.4)\) \\[ \begin{aligned} \mathrm{CH}_{3} \mathrm{COCH}_{3}(\mathrm{aq})+\mathrm{Br}_{2}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \rightarrow & \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{Br}(\mathrm{aq}) \\ &+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Br}(\mathrm{aq}) \end{aligned} \\] $$\begin{array}{llll} \hline & \text { Initial } & \text { Initial } & \text { Initial } & \text { Initial } \\ & \text { concentration } & \text { concentration } & \text { concentration } & \text { rate of } \\ & \begin{array}{l} \text { of } \mathrm{CH}_{3} \mathrm{COCH}_{3} \\ \text { /moldm }^{-3} \end{array} & \begin{array}{l} \text { of } \mathrm{Br}_{2} \\ \text { /moldm }^{-3} \end{array} & \begin{array}{l} \text { of } \mathrm{H}^{+} \\ \text {/moldm }^{-3} \end{array} & \begin{array}{l} \text { reaction } \\ \text { /moldm }^{-3} \mathrm{s}^{-1} \end{array} \\ \hline 1 & 1.00 & 1.00 & 1.00 & 4.0 \times 10^{-3} \\ 2 & 2.00 & 1.00 & 1.00 & 8.0 \times 10^{3} \\ 3 & 2.00 & 2.00 & 1.00 & 8.0 \times 10^{3} \\ 4 & 1.00 & 1.00 & 2.00 & 8.0 \times 10^{-3} \end{array}$$

Write the rate equation for the following elementary reactions and give the molecularity for each reaction. (Sections 9.4 and 9.8) (a) \(\mathrm{Cr}+\mathrm{O}_{3} \rightarrow \mathrm{ClO}^{*}+\mathrm{O}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{N}_{2} \mathrm{CH}_{3} \rightarrow 2 \mathrm{CH}_{3}^{*}+\mathrm{N}_{2}\) (c) \(2 \mathrm{Cl}^{\prime} \rightarrow \mathrm{Cl}_{2}\) (d) \(\mathrm{NO}_{2}^{*}+\mathrm{F}_{2} \rightarrow \mathrm{NO}_{2} \mathrm{F}+\mathrm{F}\)

The decomposition of ammonia on a platinum surface at \(856^{\circ} \mathrm{C}\) \\[ 2 \mathrm{NH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \\] shows the following dependence of the concentration of ammonia gas on time. $$\begin{array}{llllllll} t / s & 0 & 200 & 400 & 600 & 800 & 1000 & 1200 \\ {\left[\mathrm{NH}_{3} / 110^{-3} \mathrm{moldm}^{-3}\right.} & 2.10 & 1.85 & 1.47 & 1.23 & 0.86 & 0.57 & 0.34 \end{array}$$ Find the order of the reaction and a value for the rate constant at \(856^{\circ} \mathrm{C}\). Suggest an explanation for the order you obtain. (Section \(9.5)\)
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