The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the rate constant for the reaction increases by a factor of 2.50 ✕ 103 as compared with the uncatalyzed reaction. Assuming the frequency factor A is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.
Use the Arrhenius equation.
ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
I would use k1 = 1 and k2 = 2.5E3. Solve for Ea
Alternatively, you may use the 50 as Ea for the uncatalyzed reaction and calculate k1. Then multiply k1 by 2.50E3, use that for k2 and solve for the new Ea. The first way means solving that equation just once and is a shortcut. Try it both ways to prove it to yourself.
Post your work if you get stuck.
To calculate the activation energy for the catalyzed reaction, we can use the Arrhenius equation:
k = A * e^(-Ea/RT)
Where:
k = rate constant
A = frequency factor
Ea = activation energy
R = gas constant (8.314 J/(mol·K))
T = temperature in Kelvin (37°C = 273 + 37 = 310 K)
We are given that the rate constant for the catalyzed reaction (k_catalyzed) is 2.50 * 10^3 times larger than the rate constant for the uncatalyzed reaction (k_uncatalyzed). Mathematically, we can write this as:
k_catalyzed = 2.50 * 10^3 * k_uncatalyzed
Now let's substitute the values into the Arrhenius equation:
k_catalyzed = A * e^(-Ea_catalyzed/RT)
k_uncatalyzed = A * e^(-Ea_uncatalyzed/RT)
Divide the two equations:
k_catalyzed / k_uncatalyzed = (A * e^(-Ea_catalyzed/RT)) / (A * e^(-Ea_uncatalyzed/RT))
The A value cancels out:
k_catalyzed / k_uncatalyzed = e^(-Ea_catalyzed/RT) / e^(-Ea_uncatalyzed/RT)
Substituting the given values:
2.50 * 10^3 = e^(-Ea_catalyzed/(8.314 J/(mol·K) * 310 K)) / e^(-50.0 kJ/mol / (8.314 J/(mol·K) * 310 K))
Simplifying the equation:
2.50 * 10^3 = e^(-Ea_catalyzed / (2562.94 J/mol)) / e^(-5957.68 J/mol)
We can rewrite this equation as:
2.50 * 10^3 = e^(-Ea_catalyzed / (2562.94 J/mol + 5957.68 J/mol))
Taking the natural logarithm of both sides:
ln(2.50 * 10^3) = -Ea_catalyzed / (2562.94 J/mol + 5957.68 J/mol)
Rearranging the equation:
Ea_catalyzed = -ln(2.50 * 10^3) * (2562.94 J/mol + 5957.68 J/mol)
Calculating the activation energy for the catalyzed reaction:
Ea_catalyzed ≈ -ln(2.50 * 10^3) * (8519.62 J/mol)
Therefore, the activation energy for the catalyzed reaction is approximately -ln(2.50 * 10^3) * (8519.62 J/mol).
To calculate the activation energy for the catalyzed reaction, we can use the Arrhenius equation, which relates the rate constant (k) to the activation energy (Ea) and temperature (T):
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
We are given the following information:
- Activation energy (Ea) for the uncatalyzed reaction: 50.0 kJ/mol
- Rate constant (k) for the catalyzed reaction is 2.50 * 10^3 times higher than the uncatalyzed reaction
- Temperature (T) is 37°C, which needs to be converted to Kelvin
First, let's convert the temperature to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 37 + 273.15
T(K) ≈ 310.15 K
Now, let's calculate the rate constant (k) for the uncatalyzed reaction using the Arrhenius equation. Since the frequency factor (A) is assumed to be the same for both reactions, we don't need to consider it in this calculation:
k_uncatalyzed = e^(-Ea_uncatalyzed/RT)
= e^(-50000 J/(8.314 J/(mol*K) * 310.15 K))
= e^(-19.08)
≈ 9.20 * 10^(-9) s^(-1)
Next, we can calculate the rate constant (k) for the catalyzed reaction by multiplying the rate constant for the uncatalyzed reaction by the factor given in the problem:
k_catalyzed = k_uncatalyzed * 2.50 * 10^3
≈ 2.30 * 10^(-5) s^(-1)
Now, rearranging the Arrhenius equation to solve for the activation energy (Ea_catalyzed), we have:
Ea_catalyzed = -ln(k_catalyzed / A) * RT
Plugging in the known values, we get:
Ea_catalyzed = -ln(2.30 * 10^(-5) s^(-1) / A) * (8.314 J/(mol*K)) * 310.15 K
Since we are assuming the frequency factor (A) is the same for both reactions, we can cancel it out in the equation. Therefore:
Ea_catalyzed = -ln(2.30 * 10^(-5) s^(-1)) * (8.314 J/(mol*K)) * 310.15 K
Calculating this using a natural logarithm calculator, we find:
Ea_catalyzed ≈ 62493 J/mol
Ea_catalyzed ≈ 62.5 kJ/mol
Therefore, the activation energy for the catalyzed reaction is approximately 62.5 kJ/mol.