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By what factor is the rate of a reaction changed if an enzyme lowers the ea by 2.0 kj/mol at 37°c?

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Answer:

Step-by-step explanation:

The relationship between the rate constant of a chemical reaction (k) and the activation energy (Ea) is described by the Arrhenius equation:

k = A * exp(-Ea / (R * T))

where A is the pre-exponential factor, R is the gas constant, and T is the temperature in Kelvin.

If an enzyme lowers the activation energy (Ea) by 2.0 kJ/mol at 37°C (which is 310 Kelvin), we can calculate the new rate constant (k') using the modified activation energy (Ea') as:

k' = A * exp(-Ea' / (R * T))

where Ea' = Ea - 2.0 kJ/mol.

Substituting the values into the equation, we get:

k' = A * exp(-((Ea - 2.0) / (R * T)))

k' = A * exp(-Ea / (R * T)) * exp(2.0 / (R * T))

The ratio of the new rate constant (k') to the original rate constant (k) is:

k' / k = (A * exp(-Ea / (R * T)) * exp(2.0 / (R * T))) / (A * exp(-Ea / (R * T)))

k' / k = exp(2.0 / (R * T))

Plugging in the values of R and T, we get:

k' / k = exp(2.0 / (8.3145 J/(mol*K) * 310 K))

k' / k = exp(0.965)

k' / k = 2.62

Therefore, the rate of the reaction is increased by a factor of approximately 2.62 if an enzyme lowers the activation energy by 2.0 kJ/mol at 37°C.

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