The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy E = 50.0 kJ/mol. If the rate constant reaction is 4.6 × 10² M-¹-1 "S at 83.0 °C, what will the rate constant be at 42.0 °C? Round your answer to 2 significant digits. -1 -1 k = MS S X

Chemistry: The Molecular Science
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Chapter11: Chemical Kinetics: Rates Of Reactions
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Problem 92QRT
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**Title: Calculating the Rate Constant of a Reaction Using the Arrhenius Equation**

**Introduction:**
The rate constant \((k)\) of a chemical reaction is a crucial factor in predicting how fast a reaction occurs. Understanding its dependence on temperature and activation energy allows chemists to control reaction rates effectively.

**Problem Statement:**
The rate constant of a certain reaction is known to obey the Arrhenius equation and has an activation energy \(E_a = 50.0\) kJ/mol. If the rate constant of this reaction is \(4.6 \times 10^2\) M\(^{-1}\)s\(^{-1}\) at \(83.0\)°C, what will the rate constant be at \(42.0\)°C?

**Solution Steps:**
1. **Restate Given Data:**
   - Activation Energy (\(E_a\)): 50.0 kJ/mol
   - Initial Rate Constant (\(k_1\)): \(4.6 \times 10^2\) M\(^{-1}\)s\(^{-1}\)
   - Initial Temperature (\(T_1\)): 83.0°C
   - Final Temperature (\(T_2\)): 42.0°C

2. **Convert Temperatures to Kelvin:**
   \[ T_1 = 83.0 + 273.15 = 356.15 \text{ K} \]
   \[ T_2 = 42.0 + 273.15 = 315.15 \text{ K} \]

3. **Apply the Arrhenius Equation:**
   The Arrhenius equation links the rate constant \(k\) with temperature \(T\) and activation energy \(E_a\):
   \[ k = A \exp\left(-\frac{E_a}{RT}\right) \]
   Where:
   - \(A\) is the pre-exponential factor
   - \(R\) is the universal gas constant (8.314 J/mol·K)
   - \(E_a\) is the activation energy (J/mol)

4. **Use the Two-Point Form of the Arrhenius Equation:**
   Relate the rate constants at two temperatures:
   \[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(
Transcribed Image Text:**Title: Calculating the Rate Constant of a Reaction Using the Arrhenius Equation** **Introduction:** The rate constant \((k)\) of a chemical reaction is a crucial factor in predicting how fast a reaction occurs. Understanding its dependence on temperature and activation energy allows chemists to control reaction rates effectively. **Problem Statement:** The rate constant of a certain reaction is known to obey the Arrhenius equation and has an activation energy \(E_a = 50.0\) kJ/mol. If the rate constant of this reaction is \(4.6 \times 10^2\) M\(^{-1}\)s\(^{-1}\) at \(83.0\)°C, what will the rate constant be at \(42.0\)°C? **Solution Steps:** 1. **Restate Given Data:** - Activation Energy (\(E_a\)): 50.0 kJ/mol - Initial Rate Constant (\(k_1\)): \(4.6 \times 10^2\) M\(^{-1}\)s\(^{-1}\) - Initial Temperature (\(T_1\)): 83.0°C - Final Temperature (\(T_2\)): 42.0°C 2. **Convert Temperatures to Kelvin:** \[ T_1 = 83.0 + 273.15 = 356.15 \text{ K} \] \[ T_2 = 42.0 + 273.15 = 315.15 \text{ K} \] 3. **Apply the Arrhenius Equation:** The Arrhenius equation links the rate constant \(k\) with temperature \(T\) and activation energy \(E_a\): \[ k = A \exp\left(-\frac{E_a}{RT}\right) \] Where: - \(A\) is the pre-exponential factor - \(R\) is the universal gas constant (8.314 J/mol·K) - \(E_a\) is the activation energy (J/mol) 4. **Use the Two-Point Form of the Arrhenius Equation:** Relate the rate constants at two temperatures: \[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(
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