An enzyme-catalyzed reaction proceeds 40 times faster at 25 °C than at 4 °C.Estimate the overall activation energy for the process responsible for thisreaction.
Question
An enzyme-catalyzed reaction proceeds 40 times faster at 25 °C than at 4 °C.Estimate the overall activation energy for the process responsible for thisreaction.
Solution
To estimate the overall activation energy for the process responsible for this reaction, we can use the Arrhenius equation, which is a simple mathematical model used to calculate the temperature dependence of reaction rates. The equation is as follows:
k = Ae^(-Ea/RT)
where:
- k is the rate constant,
- A is the pre-exponential factor,
- Ea is the activation energy,
- R is the gas constant, and
- T is the temperature (in Kelvin).
Given that the reaction is 40 times faster at 25 °C than at 4 °C, we can set up a ratio of the rate constants (k2/k1) as follows:
k2/k1 = e^((Ea/R)((1/T1) - (1/T2)))
where:
- k2 and k1 are the rate constants at T2 and T1 respectively,
- T1 and T2 are the temperatures (in Kelvin).
First, convert the temperatures from Celsius to Kelvin. T1 = 4 °C = 277.15 K and T2 = 25 °C = 298.15 K.
Then, substitute the given values into the equation:
40 = e^((Ea/8.314)((1/277.15) - (1/298.15)))
To solve for Ea, take the natural logarithm of both sides and rearrange the equation to solve for Ea:
ln(40) = (Ea/8.314)((1/277.15) - (1/298.15)) Ea = ln(40) * 8.314 / ((1/277.15) - (1/298.15))
Calculate the right side of the equation to find the activation energy. The result will be in Joules per mole, which can be converted to kilojoules per mole by dividing by 1000.
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