**Title**: Calculating Activation Energy from a Plot**Text**: Using the data in the plot below, calculate the value of the activation energy (in kJ/mol) for this reaction.**Graph Description**:- The graph is a plot of ln(k) (natural logarithm of the rate constant) on the y-axis versus 1/T (temperature in Kelvin) on the x-axis.- The x-axis is labeled as "1/T (1/K)" and ranges from 0.002 to 0.009.- The y-axis is labeled as "ln k" and ranges from 0.5 to 3.5.- Data points are plotted and a linear trend line is fitted to these points.- The equation of the line is given as $y = -5073.6x + 1.866$ with an R² value of 0.9687, indicating a good fit.**Instruction**: Use the equation of the line to determine the activation energy. The slope of the line is related to the activation energy by the formula:\[ \text{Slope} = -\frac{E_a}{R} \]where $E_a$ is the activation energy and $R$ is the universal gas constant (8.314 J/mol·K). Convert the energy to kJ/mol and solve for $E_a$.

Answered: Image /qna-images/answer/dc793686-6b6a-4b09-8ddd-75e9d1259cfc.jpg

Using the data in the plot below, calculate the value of the activation energy (in kJ/mol) for this reaction: 3.5 y-507.36x 3.866 R'=0.9687 25 2 1.5 0.5 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 1/T (1/K)

Using the data in the plot below, calculate the value of the activation energy (in kJ/mol) for this reaction: 3.5 y-507.36x 3.866 R'=0.9687 25 2 1.5 0.5 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 1/T (1/K)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

**Title**: Calculating Activation Energy from a Plot

**Text**:
Using the data in the plot below, calculate the value of the activation energy (in kJ/mol) for this reaction.

**Graph Description**:
- The graph is a plot of ln(k) (natural logarithm of the rate constant) on the y-axis versus 1/T (temperature in Kelvin) on the x-axis.
- The x-axis is labeled as "1/T (1/K)" and ranges from 0.002 to 0.009.
- The y-axis is labeled as "ln k" and ranges from 0.5 to 3.5.
- Data points are plotted and a linear trend line is fitted to these points.
- The equation of the line is given as $y = -5073.6x + 1.866$ with an R² value of 0.9687, indicating a good fit.

**Instruction**:
Use the equation of the line to determine the activation energy. The slope of the line is related to the activation energy by the formula:
\[ \text{Slope} = -\frac{E_a}{R} \]
where $E_a$ is the activation energy and $R$ is the universal gas constant (8.314 J/mol·K). Convert the energy to kJ/mol and solve for $E_a$.

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