I have two questions that I really don't understand.
The first one is: A reaction proceeds with delta H=-10kJ/mol. The energy of activation of the uncatalyzed reaction is 150kJ/mol, whereas it is 100kJ/mol for the uncatalyzed reaction. How many times faster is the catalyzed reaction than the uncatalyzed reaction at 25 degrees C.
The second question is: At 30 degrees C, a reaction produces product at the rate of 0.613 mol/Lxh. If the activation energy is 88 kJ/mol, what will the reaction rate be at 45 degrees C.
I appreciate any help, thanks!
Use the Arrhenius equation for both problems.
ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
I realize this, but I don't understand how to get k1 and k2 in the first question or the second.
Also it only gives one temperature in the first question so would t1 and t2 be the same?
No the second one gives two T values, one at 30 the other at 45.
i figured out question 2, i don't understand question 1 at all still
Sure! Let's break down each question and calculate the answers step by step.
Question 1:
To find the rate of the catalyzed reaction compared to the uncatalyzed reaction, we need to use the Arrhenius equation, which relates the rate constant (k) of a reaction to the activation energy (Ea) and temperature (T).
The Arrhenius equation is given by:
k = Ae^(-Ea/RT)
Where:
k: rate constant
A: pre-exponential factor (constant)
Ea: activation energy
R: gas constant (= 8.314 J/mol·K)
T: temperature in Kelvin (K)
For the uncatalyzed reaction, the activation energy is 150 kJ/mol, and for the catalyzed reaction, it is 100 kJ/mol.
Let's assume a constant value for the pre-exponential factor (A) for both reactions, as it is not given in the question.
At 25 degrees C, the temperature in Kelvin is:
T1 = 298 K
Now, let's calculate the rate constant (k1) for the uncatalyzed reaction using the Arrhenius equation:
k1 = Ae^(-Ea1/RT1)
Since the activation energy (Ea1) for the uncatalyzed reaction is 150 kJ/mol:
k1 = Ae^(-150,000 J/mol / (8.314 J/mol·K * 298 K))
Similarly, calculate the rate constant (k2) for the catalyzed reaction using the Arrhenius equation:
k2 = Ae^(-Ea2/RT1)
Since the activation energy (Ea2) for the catalyzed reaction is 100 kJ/mol:
k2 = Ae^(-100,000 J/mol / (8.314 J/mol·K * 298 K))
Now, the ratio of catalyzed reaction rate to uncatalyzed reaction rate can be found by dividing k2 by k1:
ratio = k2 / k1
That will give you the multiple times faster the catalyzed reaction is compared to the uncatalyzed reaction at 25 degrees Celsius.
Question 2:
To find the reaction rate at 45 degrees Celsius, we can use the Arrhenius equation again.
Given:
Rate at 30 degrees C (R1) = 0.613 mol/L·h
Activation energy (Ea) = 88 kJ/mol
We need to find the rate at 45 degrees C (R2).
First, convert the given temperature values to Kelvin:
T1 = 303 K (30 degrees C)
T2 = 318 K (45 degrees C)
Now, we can use the Arrhenius equation to find the rate constant (k1) at 30 degrees C:
k1 = Ae^(-Ea/RT1)
Since the activation energy (Ea) is given as 88 kJ/mol:
k1 = Ae^(-88,000 J/mol / (8.314 J/mol·K * 303 K))
Similarly, calculate the rate constant (k2) at 45 degrees C:
k2 = Ae^(-Ea/RT2)
Now, the reaction rate (R2) at 45 degrees C can be calculated by multiplying the rate constant (k2) with the same concentration:
R2 = k2 * [concentration]
That will give you the reaction rate at 45 degrees Celsius.
Remember to provide the concentration of the reactant(s) to get an accurate reaction rate calculation.
I hope this helps!