3. A reaction has a rate constant of 0.000122 s¹ at 27.0°C and 0.228 s'at 77.0°C. a. Determine the activation energy for this reaction. b. What is the value of the rate constant at 17.0°C?

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**Title: Understanding Chemical Reaction Rates and Activation Energy**

**1. Measuring Reaction Rates as a Function of Temperature**

The rate constant (\(k\)) for a reaction was assessed based on temperature changes. A plot of \(\ln k\) versus \(1/T\) (in Kelvin) is linear with a slope of \(-1.01 \times 10^4\) K. To discover the activation energy for this reaction, the equation:

\[ E_a = -\text{slope} \times R \]

is used, where \(R\) is the gas constant (\(8.314 \text{ J/mol K}\)):

\[ E_a = -(-1.01 \times 10^4 \, \text{K}) \times 8.314 \, \text{J/mol K} \]

\[ E_a \approx 8.39 \times 10^4 \, \text{J/mol} \]

**2. Rate Constant Calculation at Different Temperatures**

For a reaction at 32.0°C with a rate constant of \(0.0550 \, \text{s}^{-1}\) and a frequency factor of \(1.20 \times 10^{13} \, \text{s}^{-1}\), the activation energy is calculated using:

\[ E_a = RT(\ln A - \ln k) \]

Substituting values and calculating energy:

\[ E_a \approx 89372.257 \, \text{J/mol} \]

**3. Calculating Activation Energy and Rate Constants at Various Conditions**

- **Given Conditions:**
  - Rate constant at 27.0°C: \(0.000122 \, \text{s}^{-1}\)
  - Rate constant at 77.0°C: \(0.228 \, \text{s}^{-1}\)

  **a. Determine Activation Energy:**
  - Utilize the Arrhenius equation.

  **b. Rate Constant at 17.0°C:**
  - Once activation energy is known, use it to find \(k\) at 17.0°C.

**4. Activation Energy in Biological Reactions**

An apple rots extensively in about 4.00 days at room temperature (25.0°C). If refrigerated at 5.0°C, rotting requires roughly 16.0 days. To determine activation energy for the rotting process,
Transcribed Image Text:**Title: Understanding Chemical Reaction Rates and Activation Energy** **1. Measuring Reaction Rates as a Function of Temperature** The rate constant (\(k\)) for a reaction was assessed based on temperature changes. A plot of \(\ln k\) versus \(1/T\) (in Kelvin) is linear with a slope of \(-1.01 \times 10^4\) K. To discover the activation energy for this reaction, the equation: \[ E_a = -\text{slope} \times R \] is used, where \(R\) is the gas constant (\(8.314 \text{ J/mol K}\)): \[ E_a = -(-1.01 \times 10^4 \, \text{K}) \times 8.314 \, \text{J/mol K} \] \[ E_a \approx 8.39 \times 10^4 \, \text{J/mol} \] **2. Rate Constant Calculation at Different Temperatures** For a reaction at 32.0°C with a rate constant of \(0.0550 \, \text{s}^{-1}\) and a frequency factor of \(1.20 \times 10^{13} \, \text{s}^{-1}\), the activation energy is calculated using: \[ E_a = RT(\ln A - \ln k) \] Substituting values and calculating energy: \[ E_a \approx 89372.257 \, \text{J/mol} \] **3. Calculating Activation Energy and Rate Constants at Various Conditions** - **Given Conditions:** - Rate constant at 27.0°C: \(0.000122 \, \text{s}^{-1}\) - Rate constant at 77.0°C: \(0.228 \, \text{s}^{-1}\) **a. Determine Activation Energy:** - Utilize the Arrhenius equation. **b. Rate Constant at 17.0°C:** - Once activation energy is known, use it to find \(k\) at 17.0°C. **4. Activation Energy in Biological Reactions** An apple rots extensively in about 4.00 days at room temperature (25.0°C). If refrigerated at 5.0°C, rotting requires roughly 16.0 days. To determine activation energy for the rotting process,
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