Questions: The activation energy for a decomposition reaction is 189 kJ/mol. The rate constant at 563 K is 3.52 x 10^-7 L/mol·s. What is the rate constant at 643 K? HOW DO WE GET THERE? ln (k2/k1)=5.02 Now solve for k2/k1

Solution Steps

The Arrhenius equation relates the rate constants at two different temperatures:

\[ \ln \left(\frac{k_{2}}{k_{1}}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

where:

• \( E_a \) is the activation energy,
• \( R \) is the gas constant (\(8.314 \, \text{J/mol} \cdot \text{K}\)),
• \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin,
• \( k_1 \) and \( k_2 \) are the rate constants at \( T_1 \) and \( T_2 \), respectively.

We are given:

• \( E_a = 189 \, \text{kJ/mol} = 189,000 \, \text{J/mol} \)
• \( T_1 = 563 \, \text{K} \)
• \( T_2 = 643 \, \text{K} \)
• \( k_1 = 3.52 \times 10^{-7} \, \text{L/mol} \cdot \text{s} \)
• \(\ln \left(\frac{k_{2}}{k_{1}}\right) = 5.02\)

Given the equation:

\[ \ln \left(\frac{k_{2}}{k_{1}}\right) = 5.02 \]

We solve for \(\frac{k_{2}}{k_{1}}\) by exponentiating both sides:

\[ \frac{k_{2}}{k_{1}} = e^{5.02} \]

Calculate the value:

\[ \frac{k_{2}}{k_{1}} = e^{5.02} \approx 151.9911 \]

Final Answer