At body temperature (37C), K of an enzyme-catalyzed reaction is 2.3x10^14 times greater than k of the uncatalyzed reaction. Assuming that the frequency factor A is the same for both reaction, by how much does the enzyme the E^a?
ln 2.3E14 = -Ea/RT
Solve for Ea
To determine by how much the enzyme reduces the activation energy (E^a) of a reaction, we need to use the equation for the rate constant (k):
k = A * e^(-E^a / RT)
Where:
- k is the rate constant
- A is the frequency factor
- E^a is the activation energy
- R is the gas constant (8.314 J/(mol⋅K))
- T is the temperature in Kelvin
Given that the frequency factor A is the same for both the catalyzed and uncatalyzed reactions, we can compare the rate constants to find the difference in activation energy provided by the enzyme.
We are told that at body temperature (37°C or 310 K), the catalyzed reaction has a rate constant (k_enzyme) that is 2.3x10^14 times greater than the rate constant (k_uncatalyzed) of the uncatalyzed reaction.
Using this information, we can set up the following equation:
k_enzyme = 2.3x10^14 * k_uncatalyzed
Since the frequency factor A is the same, and both reactions are at the same temperature, we can cancel out these variables:
e^(-E^a_enzyme / RT) = 2.3x10^14 * e^(-E^a_uncatalyzed / RT)
Taking the natural logarithm (ln) of both sides of the equation, we get:
ln(e^(-E^a_enzyme / RT)) = ln(2.3x10^14 * e^(-E^a_uncatalyzed / RT))
Simplifying further:
-E^a_enzyme / RT = ln(2.3x10^14) + (-E^a_uncatalyzed / RT)
Now, we can isolate the difference in activation energy caused by the enzyme:
E^a_enzyme - E^a_uncatalyzed = - RT * (ln(2.3x10^14))
Since we are given the temperature (310 K), the gas constant (8.314 J/(mol⋅K)), and the value of the natural logarithm, we can calculate the difference in activation energy.
Plugging in the numbers:
E^a_enzyme - E^a_uncatalyzed = - (8.314 J/(mol⋅K)) * (310 K) * ln(2.3x10^14)
Calculating this expression will give you the difference in activation energy provided by the enzyme.
To determine the effect of the enzyme on the activation energy (E^a), we can use the equation:
K (enzyme) / K (uncatalyzed) = e^(-E^a (enzyme) / RT) / e^(-E^a (uncatalyzed) / RT)
Since the frequency factor A is assumed to be the same for both reactions, we can cancel it out in the equation:
K (enzyme) / K (uncatalyzed) = e^(-E^a (enzyme) / RT) / e^(-E^a (uncatalyzed) / RT)
Given that K (enzyme) / K (uncatalyzed) = 2.3x10^14, we can express this as:
2.3x10^14 = e^(-E^a (enzyme) / RT) / e^(-E^a (uncatalyzed) / RT)
Taking the natural logarithm (ln) of both sides of the equation:
ln(2.3x10^14) = ln(e^(-E^a (enzyme) / RT) / e^(-E^a (uncatalyzed) / RT))
Using the property of logarithms (ln(a/b) = ln(a) - ln(b)):
ln(2.3x10^14) = -E^a (enzyme) / RT + E^a (uncatalyzed) / RT
Rearranging the equation to solve for E^a (enzyme):
E^a (enzyme) = -RT * (ln(2.3x10^14) - E^a (uncatalyzed))
Where:
- E^a (enzyme) is the activation energy of the enzyme-catalyzed reaction,
- E^a (uncatalyzed) is the activation energy of the uncatalyzed reaction,
- R is the gas constant, and
- T is the temperature in Kelvin.
Please note that the specific values for R and T need to be specified in order to calculate the exact value of E^a (enzyme).