The activation energy for the reaction in which CO2 decomposes to CO and a free radical oxygen atom, O, has activation energy of 460 kJ mol^-2. The frequency factor is 2 x 1011 s^-1. What is the rate constant for this reaction at 298 K?
Answer
The activation energy for the reaction in which CO2 decomposes to CO and a free radical oxygen atom, O, has activation energy of 460 kJ mol^-2. The frequency factor is 2 x 1011 s^-1. What is the rate constant for this reaction at 298 K?
To find the rate constant for the reaction at 298 K, we can use the Arrhenius equation:
k = A * e^(-Ea / (RT))
Where:
k is the rate constant
A is the frequency factor
Ea is the activation energy
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
First, we need to convert the activation energy to J/mol by multiplying it by 1000:
Ea = 460 kJ/mol * 1000 J/kJ = 460,000 J/mol
Next, we need to convert the temperature from Celsius to Kelvin by adding 273:
T = 298 K
Now, we can substitute these values into the equation to calculate the rate constant:
k = (2 x 10^11 s^-1) * e^(-460,000 J/mol / (8.314 J/(mol·K) * 298 K))
Calculating the expression inside the exponential:
Ea / (RT) = 460,000 J/mol / (8.314 J/(mol·K) * 298 K) = 195.14 mol^-1
Now we can solve for k:
k = (2 x 10^11 s^-1) * e^(-195.14 mol^-1)
Using the exponential function, we get:
k ≈ (2 x 10^11 s^-1) * e^(-195.14) ≈ 9.010 x 10^(-23) s^-1
Therefore, the rate constant for this reaction at 298 K is approximately 9.010 x 10^(-23) s^-1.
You can read about the Arrhenius equation here.
http://en.wikipedia.org/wiki/Arrhenius_equation
k = Ae-Ea/RT