The activation energy for the isomerization ol cyclopropane to propene is 274 kJ/mol. By what factor does the rate of this reaction increase as the temperature rises from 256 to 297 oC? (The factor is the ratio of the rates.)
A catalyst decreases the activation energy of a particular exothermic reaction by 60 kJ/mol, to 27 kJ/mol. Assuming that the mechanism has only one step, and that the products are 46 kJ lower in energy than the reactants, sketch approximate energy-level diagrams for the catalyzed and uncatalyzed reactions.
What is the activation energy for the uncatalyzed reverse reaction?
So I'm not sure how to do either of the questions. I've looked through my notes and the book, but I still can't get it. Thanks in advance!
You are going to have to search graphs that depict the energy of activation with and without a catalyst for you to understand how to sketch the graphs.
Is it necessary to graph it?
To understand how to sketch it, yes.
To do the first part, I believe ou have to use Arrhenius equation
ln(k2/k1) = Ea/R(1/T1 - 1/T2)
To calculate the factor by which the rate of the reaction increases as the temperature rises, we can use the Arrhenius equation:
k = A * e^(-Ea/RT)
where:
k is the rate constant
A is the pre-exponential factor
Ea is the activation energy
R is the ideal gas constant (8.314 J/(mol*K))
T is the temperature in kelvin
First, we need to convert the temperatures given from Celsius to Kelvin. To do this, we add 273 to the Celsius temperature. So, 256 oC is equivalent to (256 + 273) K = 529 K, and 297 oC is equivalent to (297 + 273) K = 570 K.
Next, we can calculate the rate constant (k) at each temperature using the Arrhenius equation. Since we are comparing the rates, we can ignore the pre-exponential factor (A) since it will cancel out in the ratio.
k1 = e^(-Ea/(R * T1))
k2 = e^(-Ea/(R * T2))
where:
T1 = 529 K
T2 = 570 K
Now, we can calculate the ratio of the rate constants:
factor = k2 / k1
Substituting the expressions for k1 and k2, we get:
factor = (e^(-Ea/(R * T2))) / (e^(-Ea/(R * T1)))
To simplify this expression, we can use the property that e^(a - b) = (e^a) / (e^b):
factor = e^((-Ea/(R * T2)) + (Ea/(R * T1)))
Simplifying further:
factor = e^((Ea/R) * ((1/T1) - (1/T2)))
Now, we can substitute the values of Ea, R, T1, and T2 into the equation, calculate the factor by raising e to the power of the resulting expression, and obtain the answer.
Similarly, for the second question regarding the uncatalyzed reverse reaction, we can use the same principles. By assuming that the mechanism has only one step and products are 46 kJ lower in energy, we can deduce that the activation energy for the uncatalyzed reverse reaction will be equal to the activation energy of the catalyzed forward reaction (27 kJ/mol) plus the energy difference of the products (46 kJ/mol). Therefore, the activation energy for the uncatalyzed reverse reaction would be 27 kJ/mol + 46 kJ/mol = 73 kJ/mol.
I hope this explanation helps you understand how to approach these questions! Feel free to ask if you have any further doubts or need additional clarification.