A reaction rate increases by 60.3% when the temperature is increased from 21.5 to 35.6°C. Calculate the activation energy (in kJ/mol) for the reaction.
Give an arbitrary rate to the reaction at the 21.5C
Lets say it is 100 moles/sec
Then the rate at 35.6 is 160.3 M/s
Now, take the natural log of each of those rates.
Plot ln rate vs 1/tempinKelvins
The slope of that line is Ea/R where R is the universal gas constant.
To calculate the activation energy for the reaction, you can use the Arrhenius equation:
k = A * e^(-Ea/RT)
In this equation:
- k is the reaction rate constant
- A is the pre-exponential factor (a constant specific to the reaction)
- Ea is the activation energy
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
First, let's convert the temperatures from Celsius to Kelvin:
T1 = 21.5 + 273.15 = 294.65 K
T2 = 35.6 + 273.15 = 308.75 K
Next, we'll use the information given that the reaction rate increases by 60.3%. The rate constant (k2) at the higher temperature (T2) is 60.3% greater than the rate constant (k1) at the lower temperature (T1). Mathematically, this can be represented as:
k2 = k1 + 0.603 * k1
k2 = (1 + 0.603) * k1
k2 = 1.603 * k1
Now, let's rewrite the Arrhenius equation for both temperatures:
k1 = A * e^(-Ea/(R*T1))
k2 = A * e^(-Ea/(R*T2))
Divide the second equation by the first equation:
k2/k1 = e^(-Ea/(R*T2))/e^(-Ea/(R*T1))
Since the exponential terms have the same base (e), we can simplify the equation:
k2/k1 = e^(-Ea/(R*T2) + Ea/(R*T1))
k2/k1 = e^((Ea/R)*(1/T1 - 1/T2))
Substitute the values we have:
1.603 * k1 / k1 = e^((Ea/R)*(1/T1 - 1/T2))
1.603 = e^((Ea/8.314)*(1/T1 - 1/T2))
Take the natural logarithm of both sides:
ln(1.603) = (Ea/8.314)*(1/T1 - 1/T2)
Solve for Ea:
Ea = ln(1.603) * 8.314 * ((1/T1) - (1/T2))
Plug in the temperature values to calculate Ea:
Ea = ln(1.603) * 8.314 * ((1/294.65) - (1/308.75))
Ea ≈ 54.1 kJ/mol
Therefore, the activation energy for the reaction is approximately 54.1 kJ/mol.