In the absence of a catalyst, a particular reaction occurs slowly at 23°C. You are planning to design a catalyst to increase the rate of the reaction by a factor of 1.370 ✕ 103 at the same temperature. If the frequency factor is unchanged by the addition of the catalyst, calculate the amount by which the catalyst should lower the activation energy of the reaction. (The gas constant R = 8.314472 J/mol · K, approximately.)
To increase the reaction rate by a factor of 1.370 ✕ 103, the catalyst should lower the activation energy by WebAssign will check your answer for the correct number of significant figures. 
 kJ/mol.

Useful Information:
1.013 bar = 760 torr = 1 atm = 760 mm Hg
h = 6.62607 ✕ 10−34 J.s
c = 2.99792 ✕ 108 m/s
R = 8.314472 J/mol.K
NA = 6.02214 ✕ 1023 1/mol
R = 0.08205746 L.atm/mol.K
F = 9.648534 ✕ 104 C/mol


To calculate the decrease in activation energy required for the reaction rate to increase by a factor of 1.370 × 10³, we can use the **Arrhenius equation** in the form:

\[
k = A e^{-\frac{E_a}{RT}}
\]

Where:
- \(k\) is the rate constant,
- \(A\) is the frequency factor (assumed unchanged),
- \(E_a\) is the activation energy,
- \(R\) is the gas constant,
- \(T\) is the temperature in Kelvin.

### Step-by-Step Approach

1. **Arrhenius Equation Ratios**: 
   Since the rate constant increases by a factor of 1.370 × 10³, we can set up the ratio of the rate constants before (\(k_{\text{before}}\)) and after (\(k_{\text{after}}\)) adding the catalyst:
   \[
   \frac{k_{\text{after}}}{k_{\text{before}}} = 1.370 \times 10^3
   \]

   Using the Arrhenius equation for both situations:
   \[
   \frac{A e^{-\frac{E_a^{\text{after}}}{RT}}}{A e^{-\frac{E_a^{\text{before}}}{RT}}} = 1.370 \times 10^3
   \]
   The frequency factor \(A\) cancels out, and the equation simplifies to:
   \[
   e^{-\frac{E_a^{\text{after}}}{RT}} / e^{-\frac{E_a^{\text{before}}}{RT}} = 1.370 \times 10^3
   \]
   Taking the natural logarithm of both sides:
   \[
   -\frac{E_a^{\text{after}}}{RT} + \frac{E_a^{\text{before}}}{RT} = \ln(1.370 \times 10^3)
   \]
   \[
   \frac{E_a^{\text{before}} - E_a^{\text{after}}}{RT} = \ln(1.370 \times 10^3)
   \]
   
2. **Solve for the Energy Difference**: 
   Rearranging for the difference in activation energy (\(\Delta E_a = E_a^{\text{before}} - E_a^{\text{after}}\)):
   \[
   \Delta E_a = R T \ln(1.370 \times 10^3)
   \]

3. **Convert Temperature to Kelvin**: 
   The temperature is given as 23°C, so:
   \[
   T = 23 + 273.15 = 296.15 \, \text{K}
   \]

4. **Calculate the Natural Logarithm**:
   \[
   \ln(1.370 \times 10^3) \approx \ln(1370) \approx 7.224
   \]

5. **Substitute Values**:
   Now substitute the values into the equation:
   \[
   \Delta E_a = (8.314472 \, \text{J/mol.K}) (296.15 \, \text{K}) (7.224)
   \]
   \[
   \Delta E_a \approx 17783.9 \, \text{J/mol}
   \]
   Convert to kJ/mol:
   \[
   \Delta E_a \approx 17.78 \, \text{kJ/mol}
   \]

Thus, the catalyst should lower the activation energy by **17.78 kJ/mol**.

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Let me know if you'd like further details or clarifications! Here are some follow-up questions:

1. What effect does activation energy have on the rate of a reaction?
2. How does temperature influence the rate constant in the Arrhenius equation?
3. Why does the frequency factor remain unchanged when adding a catalyst?
4. What would happen if the activation energy wasn't reduced as much as calculated?
5. How does a catalyst generally affect the mechanism of a chemical reaction?

**Tip**: A lower activation energy means more molecules will have enough energy to react, thus speeding up the reaction!

In the absence of a catalyst, a particular reaction occurs slowly at 23°C. You are planning to design a catalyst to increase the rate of the reaction by a factor of 1.370 ✕ 103 at the same temperature. If the frequency factor is unchanged by the addition of the catalyst, calculate the amount by which the catalyst should lower the activation energy of the reaction. (The gas constant R = 8.314472 J/mol · K, approximately.)

Learn how to calculate the effect of a catalyst on activation energy using the Arrhenius equation. This problem involves determining the decrease in activation energy required to increase the reaction rate by a factor of 1.370 ✕ 103 at 23°C. Suitable for undergraduate chemistry students.

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