By what factor does the fraction of collisions with energy equal to or greater than activation energy at 100 kJ/mol. change if the temperature increases from 32 degrees celsius to 63 degrees celsius. Give your answer in SCIENTIFIC NOTATION
where is the answer
To determine the factor by which the fraction of collisions with energy equal to or greater than the activation energy changes, we can use the Arrhenius equation. The Arrhenius equation relates the rate constant of a reaction to the temperature and the activation energy.
The equation is given as:
k = Ae^(-Ea/RT)
Where:
k is the rate constant
A is the pre-exponential factor or frequency factor
Ea is the activation energy
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
In this case, we are interested in the fraction of collisions with energy equal to or greater than the activation energy. This fraction depends on the rate constant and the total number of collisions.
Since the total number of collisions is not provided, we can assume it remains constant. Therefore, the fraction of collisions with energy equal to or greater than the activation energy is directly proportional to the rate constant.
To compare the effect of temperature on the rate constant, we can take the ratio of the rate constants at the two temperatures (T1 and T2):
k2/k1 = (Ae^(-Ea/RT2))/(Ae^(-Ea/RT1))
Simplifying the equation by canceling out the A and rearranging, we get:
k2/k1 = e^((Ea/R)*((1/T1) - (1/T2)))
Now, let's calculate the factor by substituting the given temperatures into the equation:
T1 = 32 degrees Celsius = 32 + 273.15 = 305.15 K
T2 = 63 degrees Celsius = 63 + 273.15 = 336.15 K
k2/k1 = e^((Ea/R)*(1/305.15 - 1/336.15))
Calculating the value using the given value for the activation energy (100 kJ/mol):
k2/k1 = e^((100000 J/mol)/(8.314 J/(mol·K))*(1/305.15 - 1/336.15))
k2/k1 = e^(40789.5*(0.0032744 - 0.002977))
k2/k1 = e^(40789.5*(0.00029744))
k2/k1 = e^(12.108)
Using the scientific notation for e^(12.108), the factor can be expressed as:
k2/k1 ≈ 1.38 x 10^5
Therefore, the fraction of collisions with energy equal to or greater than the activation energy increases by a factor of approximately 1.38 x 10^5 when the temperature increases from 32 degrees Celsius to 63 degrees Celsius.