For the reaction of the peroxide decomposition that has an activation energy of 17.4 kJ/mol and a rate constant of 0.027 s⁻¹ (25 °C), we have:
a) The rate constant at 65 °C is 0.062 s⁻¹.
b) The rate constant is 25% greater than the rate constant at 25 °C when the temperature is 35.1 °C.
a) The rate constant at 65°C can be calculated with the Arrhenius equation:
[tex]k_{(T)} = Ae^{-\frac{E}{RT}}[/tex]
Where:
[tex]k_{(T)}[/tex]: is the rate constant =?
A: is the pre-exponential constant
E: is the activation energy = 17.4 kJ/mol
R: is the gas constant = 8.314 J/K*mol
T: is the temperature = 65 °C = 338 K
First, we need to find the constant A. We know values of E and [tex] k_(T)[/tex] for the decomposition of peroxide when T = 25 °C = 298 K, so we can find A:
[tex]A = \frac{k_{(25)}}{e^{-\frac{E}{RT}}} = \frac{0.027 s^{-1}}{e^{-\frac{17.4 \cdot 10^{3} J/mol}{8.314 J/K*mol*298 K}}} = 30.3 s^{-1}[/tex]
Now, the rate constant at 65 °C is:
[tex]k_{(65)} = Ae^{-\frac{E}{RT}} = 30.3 s^{-1}*e^{-\frac{17.4 \cdot 10^{3} J/mol}{8.314 J/K*mol*338 K}} = 0.062 s^{-1}[/tex]
Therefore, the rate constant at 65 °C is 0.062 s⁻¹.
b) The value of the rate constant when is 25% greater than the rate constant at 25 °C is:
[tex]k_{(T)} = \frac{25}{100}*k_{(25)} + k_{(25)} = 0.027 s^{-1}(0.25 + 1) = 0.034 s^{-1}[/tex]
Now, the temperature at which we get the above rate constant is:
[tex]T = -\frac{E}{R*ln(\frac{k_{(T)}}{A})} = -\frac{17.4 \cdot 10^{3} J/mol}{8.314 J/K*mol*ln(\frac{0.034 s^{-1}}{30.3 s^{-1}})} = 308.1 K = 35.1 ^{\circ} C[/tex]
Therefore, the rate constant is 25% greater than the rate constant at 25 °C when the temperature is 35.1 °C.
You can learn more about activation energy here: https://brainly.com/question/9967920?referrer=searchResults
I hope it helps you!
