Questions: A certain first-order reaction has a rate constant of 2.75 × 10-2 s^-1 at 20.0°C. What is the value of k at 60.0°C if Ea=75.7 kJ / mol?

Solution Steps

The Arrhenius equation relates the rate constant \( k \) of a reaction to the temperature \( T \) and the activation energy \( \mathrm{Ea} \):

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

where:

• \( A \) is the pre-exponential factor,
• \( \mathrm{Ea} \) is the activation energy,
• \( R \) is the universal gas constant (\(8.314 \, \mathrm{J/mol \cdot K}\)),
• \( T \) is the temperature in Kelvin.

To find the rate constant at a different temperature, we use the two-point form of the Arrhenius equation:

\[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{\mathrm{Ea}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

where:

• \( k_1 = 2.75 \times 10^{-2} \, \mathrm{s}^{-1} \) is the rate constant at \( T_1 = 20.0^\circ \mathrm{C} \),
• \( k_2 \) is the rate constant at \( T_2 = 60.0^\circ \mathrm{C} \),
• \( \mathrm{Ea} = 75.7 \, \mathrm{kJ/mol} = 75700 \, \mathrm{J/mol} \).

Convert the temperatures from Celsius to Kelvin:

\[ T_1 = 20.0 + 273.15 = 293.15 \, \mathrm{K} \] \[ T_2 = 60.0 + 273.15 = 333.15 \, \mathrm{K} \]

Substitute the known values into the two-point Arrhenius equation:

\[ \ln\left(\frac{k_2}{2.75 \times 10^{-2}}\right) = -\frac{75700}{8.314} \left(\frac{1}{333.15} - \frac{1}{293.15}\right) \]

Calculate the right-hand side:

\[ \frac{1}{333.15} - \frac{1}{293.15} = -0.0001201 \]

\[ -\frac{75700}{8.314} \times -0.0001201 = 1.094 \]

Thus:

\[ \ln\left(\frac{k_2}{2.75 \times 10^{-2}}\right) = 1.094 \]

Solve for \( k_2 \):

\[ \frac{k_2}{2.75 \times 10^{-2}} = e^{1.094} \]

\[ k_2 = 2.75 \times 10^{-2} \times e^{1.094} \]

Calculate \( e^{1.094} \):

\[ e^{1.094} \approx 2.986 \]

Finally, calculate \( k_2 \):

\[ k_2 = 2.75 \times 10^{-2} \times 2.986 \approx 0.0821 \, \mathrm{s}^{-1} \]

Final Answer