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!