7. Consider the following graphs of In k versus 1/T for reactions A, B, C, and D. D In k In k 1 1 I (K-I) T (K-I) Which reaction of the four (A, B, C, or D) would have the largest activation energy? Explain how you know your answer.

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**Title: Understanding Activation Energy from Arrhenius Plots**

**7. Consider the following graphs of ln k versus 1/T for reactions A, B, C, and D:**

- **Graph 1:**
    - **Axes:** The x-axis represents \( \frac{1}{T} \) (in \( K^{-1} \)). The y-axis represents \( \ln k \).
    - **Lines:**
        - Line A: Steepest slope.
        - Line B: Less steep compared to A.

- **Graph 2:**
    - **Axes:** The x-axis represents \( \frac{1}{T} \) (in \( K^{-1} \)). The y-axis represents \( \ln k \).
    - **Lines:**
        - Line C: Steep slope.
        - Line D: Less steep compared to C.

**Question:** Which reaction of the four (A, B, C, or D) would have the largest activation energy? Explain how you know your answer.

**Explanation:**
The activation energy (Ea) of a reaction can be determined from the slope of the line in an Arrhenius plot (ln k versus \( \frac{1}{T} \)). The slope of the line in these graphs is given by \( -\frac{Ea}{R} \), where \( R \) is the gas constant. A steeper slope indicates a larger magnitude of \( -\frac{Ea}{R} \), and therefore a larger activation energy.

Based on the provided graphs:
- Reaction A has the steepest slope, indicating the largest negative slope.
- Therefore, Reaction A has the largest activation energy, as its slope \( -\frac{Ea}{R} \) has the greatest absolute value.

The relative steepness of the lines indicates the order of activation energies:
- **Largest activation energy:** Reaction A
- **Next largest activation energy:** Reaction B, followed by C, and finally D.
Transcribed Image Text:**Title: Understanding Activation Energy from Arrhenius Plots** **7. Consider the following graphs of ln k versus 1/T for reactions A, B, C, and D:** - **Graph 1:** - **Axes:** The x-axis represents \( \frac{1}{T} \) (in \( K^{-1} \)). The y-axis represents \( \ln k \). - **Lines:** - Line A: Steepest slope. - Line B: Less steep compared to A. - **Graph 2:** - **Axes:** The x-axis represents \( \frac{1}{T} \) (in \( K^{-1} \)). The y-axis represents \( \ln k \). - **Lines:** - Line C: Steep slope. - Line D: Less steep compared to C. **Question:** Which reaction of the four (A, B, C, or D) would have the largest activation energy? Explain how you know your answer. **Explanation:** The activation energy (Ea) of a reaction can be determined from the slope of the line in an Arrhenius plot (ln k versus \( \frac{1}{T} \)). The slope of the line in these graphs is given by \( -\frac{Ea}{R} \), where \( R \) is the gas constant. A steeper slope indicates a larger magnitude of \( -\frac{Ea}{R} \), and therefore a larger activation energy. Based on the provided graphs: - Reaction A has the steepest slope, indicating the largest negative slope. - Therefore, Reaction A has the largest activation energy, as its slope \( -\frac{Ea}{R} \) has the greatest absolute value. The relative steepness of the lines indicates the order of activation energies: - **Largest activation energy:** Reaction A - **Next largest activation energy:** Reaction B, followed by C, and finally D.
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