It was mentioned that the rates of many reactions approximately double for each 10 °C rise in temperature. Assuming a starting temperature of 25 °C, the activation energy would be_____________ in kJ mol-1, if the rate of a reaction were to be twice as large at 35 °C?
Use the Clausius-Clapeyron equation. Use P1 = any number of your choosing and P2 is twice that or plug in for P2 = 2P1
To determine the activation energy, we can use the Arrhenius equation:
k = A * e^(-Ea/RT)
Where:
- k is the rate constant for the reaction
- A is the pre-exponential factor (related to the frequency of successful collisions between the reactant molecules)
- Ea is the activation energy (the energy barrier that must be overcome for the reaction to occur)
- R is the ideal gas constant (8.314 J/(mol*K))
- T is the temperature in Kelvin
In this case, we are given the starting temperature (25 °C) and the corresponding reaction rate. Let's convert the temperatures to Kelvin:
T1 = 25 °C + 273.15 = 298.15 K
T2 = 35 °C + 273.15 = 308.15 K
We know that the rate of the reaction approximately doubles for each 10 °C rise in temperature. Therefore, if the rate at T2 (35 °C) is twice as large as the rate at T1 (25 °C), we can write:
k2 = 2 * k1
Now, let's substitute the values into the Arrhenius equation:
A * e^(-Ea/RT2) = 2 * A * e^(-Ea/RT1)
We can simplify the equation by canceling out the pre-exponential factor:
e^(-Ea/RT2) = 2 * e^(-Ea/RT1)
Now, we can take the natural logarithm (ln) of both sides to eliminate the exponential term:
-Ea/RT2 = ln(2) - Ea/RT1
Let's rearrange the equation to solve for the activation energy Ea:
Ea/RT2 - Ea/RT1 = ln(2)
Ea(RT1 - RT2)/(RT1 * RT2) = ln(2)
Substituting the known values:
Ea(298.15 K * 308.15 K)/(298.15 K * 308.15 K) = ln(2)
Ea = ln(2) * (298.15 K * 308.15 K)
Calculating this expression gives us the activation energy in J/mol. To convert it to kJ/mol, we divide by 1000:
Ea = ln(2) * (298.15 K * 308.15 K) / 1000
Evaluating this expression will give us the activation energy Ea in kJ/mol.