Math Problem Statement

The Arrhenius equation for a reaction rate constant k can be written in the form \(\ln k = \ln A - \frac{E}{R} \times \frac{1}{T}\), where \(T\) is the temperature, \(E\) is the activation energy, and \(A\) is the pre-exponential factor. Given that \(\ln k\) is 13.9 at 300 K and 15 at 310 K, calculate \(E\) in kJ and \(\ln A\), rounding final answers to one decimal place.

Solution

To solve for EE (activation energy) and lnA\ln A using the given Arrhenius equation data:

  1. The Arrhenius equation is: lnk=lnAER×1T\ln k = \ln A - \frac{E}{R} \times \frac{1}{T}

  2. We have two data points for lnk\ln k at different temperatures:

    • lnk1=13.9\ln k_1 = 13.9 at T1=300KT_1 = 300 \, \text{K}
    • lnk2=15\ln k_2 = 15 at T2=310KT_2 = 310 \, \text{K}
  3. Using the equation: lnk2lnk1=ER(1T21T1)\ln k_2 - \ln k_1 = -\frac{E}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) Substitute the known values:

    • R=8.314Jmol1K1R = 8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1}

    Calculating the terms: lnk2lnk1=1513.9=1.1\ln k_2 - \ln k_1 = 15 - 13.9 = 1.1 1T21T1=131013000.000107K1\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{310} - \frac{1}{300} \approx -0.000107 \, \text{K}^{-1} Therefore: 1.1=E8.314×(0.000107)1.1 = -\frac{E}{8.314} \times (-0.000107) Rearranging to solve for EE: E=1.1×8.3140.000107J85,987JE = \frac{1.1 \times 8.314}{0.000107} \, \text{J} \approx 85,987 \, \text{J} Converting to kJ: E86.0kJE \approx 86.0 \, \text{kJ}

  4. To find lnA\ln A, we use: lnk=lnAER×1T\ln k = \ln A - \frac{E}{R} \times \frac{1}{T} Using the first data point: 13.9=lnA860008.314×130013.9 = \ln A - \frac{86000}{8.314} \times \frac{1}{300} Calculating the term: 860008.314×30034.5\frac{86000}{8.314 \times 300} \approx 34.5 So: lnA=13.9+34.5=48.4\ln A = 13.9 + 34.5 = 48.4

Final Answers:

  • E=86.0kJE = 86.0 \, \text{kJ}
  • lnA=48.4\ln A = 48.4

Would you like further details or explanations?


Relative Questions:

  1. What is the significance of the activation energy EE in a chemical reaction?
  2. How does temperature influence the rate of a reaction according to the Arrhenius equation?
  3. What other methods can be used to determine activation energy experimentally?
  4. How does the pre-exponential factor AA relate to molecular collisions and reaction orientation?
  5. Can you derive the Arrhenius equation from first principles of kinetics?

Tip:

When calculating activation energy, small changes in temperature can have a significant impact on reaction rates; precise measurements and calculations are key!

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Math Problem Analysis

Mathematical Concepts

Logarithmic Equations
Linear Equations
Chemical Kinetics

Formulas

Arrhenius equation: \(\ln k = \ln A - \frac{E}{R} \times \frac{1}{T}\)
Two-point form of Arrhenius equation: \(\ln k_2 - \ln k_1 = -\frac{E}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)

Theorems

Arrhenius Equation

Suitable Grade Level

Undergraduate Chemistry