At body temperature (37°C), the rate constant of an enzyme-catalyzed decomposition is 3.0 × 108 times that of the uncatalyzed reaction. If the frequency factor, A, is the same for both processes, by how much does the enzyme lower the Ea?
I would choose something like 50,000 J/mol for Ea, use the Arrhenius equation, and solve for k(rate). Then multiply k by 3D8 and calculate the new Ea. Compare the two.
To determine by how much the enzyme lowers the activation energy (Ea), we can use the rate constant ratio and the Arrhenius equation.
The Arrhenius equation relates the rate constant (k) to the activation energy (Ea) and the 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
Given that the rate constant of the enzyme-catalyzed decomposition is 3.0 × 10^8 times that of the uncatalyzed reaction, we can express this as:
k_enzyme / k_uncatalyzed = 3.0 × 10^8
Since the frequency factor is the same for both processes, we can cancel it out in the equation.
Now, let's assign the rate constant of the uncatalyzed reaction as k_uncatalyzed and the rate constant of the enzyme-catalyzed reaction as k_enzyme.
So, we have:
k_enzyme / k_uncatalyzed = 3.0 × 10^8
Now, we can rewrite this equation as:
k_enzyme = 3.0 × 10^8 * k_uncatalyzed
We know that the rate constant depends on the activation energy and temperature, so we'll use the Arrhenius equation for each case:
For the uncatalyzed reaction:
k_uncatalyzed = A * e^(-Ea_uncatalyzed/RT)
And for the enzyme-catalyzed reaction:
k_enzyme = A * e^(-Ea_enzyme/RT)
Since we are interested in the Ea difference between the two reactions, we can divide the two equations:
(k_enzyme / k_uncatalyzed) = (A * e^(-Ea_enzyme/RT)) / (A * e^(-Ea_uncatalyzed/RT))
Canceling out the A's and rearranging the equation:
(k_enzyme / k_uncatalyzed) = e^(-Ea_enzyme/RT + Ea_uncatalyzed/RT)
Substituting the given rate constant ratio:
3.0 × 10^8 = e^(-Ea_enzyme/RT + Ea_uncatalyzed/RT)
Taking natural logarithm (ln) on both sides of the equation:
ln(3.0 × 10^8) = -Ea_enzyme/RT + Ea_uncatalyzed/RT
Now, solving for the difference in activation energy (ΔEa = Ea_uncatalyzed - Ea_enzyme):
ΔEa = -RT * (ln(3.0 × 10^8) - Ea_uncatalyzed/RT)
Given that the temperature is 37°C, we need to convert it to Kelvin (T = 37 + 273.15 = 310.15 K).
Substituting the values:
ΔEa = - (8.314 J/(mol*K)) * (310.15 K) * (ln(3.0 × 10^8) - Ea_uncatalyzed/(8.314 J/(mol*K) * (310.15 K)))
Simplifying the equation further will give you the answer for the amount by which the enzyme lowers the activation energy.