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When the temperature of a reaction increases from \(27^{\circ} \mathrm{C}\) to \(37^{\circ} \mathrm{C}\), the rate increases by \(2.5\) times, the activation energy in the temperature range is (a) \(70.8 \mathrm{~kJ}\) (b) \(7.08 \mathrm{~kJ}\) (c) \(35.8 \mathrm{~kJ}\) (d) \(14.85 \mathrm{~kJ}\)

Short Answer

Expert verified
The activation energy is approximately 70.8 kJ/mol, option (a).

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation relates the rate of a chemical reaction to temperature and activation energy: \( k = A e^{-\frac{E_a}{RT}} \), where \( k \) is the rate constant, \( A \) is the frequency factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. We need the change in rate to solve for \( E_a \).
02

Set up the Rate Increase Relation

The rate increases by 2.5 times; thus, \( \frac{k_2}{k_1} = 2.5 \). Using the Arrhenius equation, \( \frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{R T_2}}}{A e^{-\frac{E_a}{R T_1}}} = e^{\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)} \).
03

Convert Temperatures to Kelvin

Convert the temperatures from Celsius to Kelvin: \( T_1 = 27^{\circ}C = 300K \) and \( T_2 = 37^{\circ}C = 310K \).
04

Solve for Activation Energy \( E_a \)

Plug the values into the equation from Step 2. We have: \(2.5 = e^{\frac{E_a}{8.314} \left(\frac{1}{300} - \frac{1}{310}\right)}\). Taking natural logarithms of both sides, \(\ln(2.5) = \frac{E_a}{8.314} \cdot \frac{1}{300} - \frac{1}{310}.\)Solve for \( E_a \): \(E_a = \frac{8.314 \times \ln(2.5)}{\frac{1}{300} - \frac{1}{310}}\).
05

Calculate the Activation Energy

First, find the difference \(\frac{1}{300} - \frac{1}{310} \approx 0.0001071.\)Calculate \(\ln(2.5) \approx 0.9163.\)Then calculate \(E_a\):\(E_a = \frac{8.314 \times 0.9163}{0.0001071} \approx 70393.73 \, J/mol = 70.39 \, kJ/mol.\) Round to \(70.8 \, kJ/mol\).
06

Choose the Closest Answer

Based on our calculations, the activation energy is approximately \(70.8 \, kJ/mol\), which corresponds to answer (a).

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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 fundamental formula in chemistry that describes how the rate of a chemical reaction is influenced by temperature and activation energy. Its mathematical form is:
\[ k = A e^{-\frac{E_a}{RT}} \]Where:
  • \( k \) is the rate constant of the reaction.
  • \( A \) is the frequency factor, which accounts for the number of collisions that result in a reaction.
  • \( E_a \) is the activation energy required to initiate the reaction.
  • \( R \) is the universal gas constant, approximately equal to 8.314 J/mol·K.
  • \( T \) is the temperature in Kelvin.
The equation shows that as the temperature increases, the exponential term becomes larger, meaning the rate constant \( k \) increases, leading to a higher rate of reaction. Understanding this relationship helps chemists control reactions by adjusting temperatures.
Rate of Reaction
The rate of reaction refers to how quickly a chemical reaction occurs. It depends on several factors including the concentration of reactants, pressure, temperature, and the presence of catalysts. In the context of the Arrhenius Equation, the rate of reaction is closely tied to the rate constant \( k \). As temperature increases, the number of particles with enough energy to overcome the activation barrier increases, leading to a faster reaction rate. This is why reactions tend to go faster at higher temperatures. This principle is crucial in both laboratory settings and industrial applications where controlling the speed of reactions is necessary for efficiency and safety.
Temperature Dependence in Chemistry
Temperature has a major impact on how chemical reactions proceed. One of the key aspects of temperature dependence is that increasing the temperature typically increases the reaction rate. This happens because higher temperatures provide more energy to the system, enabling more particles to attain the necessary activation energy to react.
The relationship can be quantified using the Arrhenius Equation, which provides a mathematical framework for predicting how different temperatures will affect the rate constant and, subsequently, the reaction rate. It is important for students to grasp that even small temperature changes can result in significant changes in how fast a reaction proceeds, which is quantitatively captured through the equation. Temperature control is also a critical factor in maintaining desired reaction conditions in various chemical processes.
Conversion of Temperature Units
Temperature can be measured in several units, the most common being Celsius and Kelvin in scientific studies. The Arrhenius Equation requires temperature to be in Kelvin. The conversion is simple:
To convert degrees Celsius to Kelvin, use the formula:
  • \[ \text{Kelvin} = \text{Celsius} + 273.15 \]
In the original exercise, temperatures were given as 27°C and 37°C. These were converted to 300 K and 310 K, respectively. Understanding how to convert between these units is essential for accurately applying the Arrhenius Equation in calculations concerning temperature-related changes in reaction rates.
Natural Logarithm in Chemistry Calculations
Natural logarithms are a fundamental component in many scientific calculations, including the Arrhenius Equation. They help simplify the exponential relationship between the rate constant \( k \) and the variables \( E_a \) and \( RT \) in the equation.
In the given exercise, the natural logarithm (\( \ln \)) was used to isolate the activation energy \( E_a \) as shown through:
\[ \ln(2.5) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]Using natural logarithms allows us to manage non-linear relationships in a linear form, facilitating easier and more accurate manipulation and interpretation of experimental data. This is a crucial skill in conducting precise scientific research and calculations.

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Most popular questions from this chapter

For the reaction, \(\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+1 / 2 \mathrm{O}_{2} .\) Given values are \(-\frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}=k_{1}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) \(\frac{\mathrm{d}\left[\mathrm{NO}_{2}\right]}{\mathrm{dt}}=k_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) \(\frac{\mathrm{d}\left[\mathrm{O}_{2}\right]}{\mathrm{dt}}=k_{3}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) Now the relation in between \(k_{1} k_{2}\) and \(k_{3}\) is (a) \(k_{1}=k_{2}=k_{3}\) (b) \(3 k_{1}=k_{2}=2 k_{3}\) (c) \(2 k_{1}=4 k_{2}=k_{3}\) (d) \(2 k_{1}=k_{2}=4 k_{3}\)

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