By what factor is the rate of a reaction changed if an enzyme lowers the Ea by 9.0 kJ/mol at 37°C?

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Waste of time

To determine the factor by which the rate of a reaction is changed when an enzyme lowers the activation energy (Ea), we need to use the Arrhenius equation.

The Arrhenius equation is given as:

k = Ae^(-Ea/RT)

Where:
k is the rate constant of the reaction,
A is the pre-exponential factor,
Ea is the activation energy,
R is the gas constant (8.314 J/mol⋅K),
T is the temperature in Kelvin.

In this case, we are given that the enzyme lowers the Ea by 9.0 kJ/mol at a temperature of 37°C (which is 310 K).

To find out the factor by which the rate changes, we need to compare the rate constant before and after the enzyme is introduced. Let's call the rate constant before the enzyme k1, and after the enzyme k2.

To simplify the calculations, we can assume that the value of A remains constant.

Taking the ratio of the rate constants:

k2/k1 = (Ae^(-Ea2/RT))/(Ae^(-Ea1/RT))

The A values cancel out, and we're left with:

k2/k1 = e^((-Ea2 + Ea1)/RT)

Substituting the known values:

k2/k1 = e^((-Ea2 + Ea1)/(R * 310))

k2/k1 = e^((-9000 + 0)/(8.314 * 310))

k2/k1 = e^(-2.984)

k2/k1 ≈ 0.0507

Therefore, the factor by which the rate of the reaction changes when the enzyme lowers the Ea by 9.0 kJ/mol at 37°C is approximately 0.0507.

Use the Arrhenius equation.

Pick a number, for example choose 50,000 J/mol for Ea, the redo the equation using 50,000-9000 = ? J/mol and see the ratio of the two k values you calculated. That will be the ratio (factor) you want.