Question

The activation energy for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. If an enzyme increases the rate of the hydrolysis reaction by a factor of 5.00x10^8 at T=298.15 K, what is the activation energy when the sucrose is in the active site of the un-catalyzed reaction are the same.

Answer 1

The activation energy for the enzyme-catalyzed reaction is calculated by comparing the rate constants of the catalyzed and uncatalyzed reactions using the Arrhenius equation. The enzyme's influence on the activation energy is determined through a mathematical derivation based on the rate enhancement factor, providing insight into the efficiency of enzymatic catalysis.

The activation energy (Ea) for a reaction is a measure of the energy barrier that must be overcome for the reaction to proceed. Enzymes play a crucial role in biological systems by lowering the activation energy required for a reaction. The relationship between the rate constant (k) of a reaction, the activation energy, and temperature is described by the Arrhenius equation:

`\[ k = Ae^{-(E_a)/(RT)}, \]`

where:

• K is the rate constant,

• A is the pre-exponential factor,

• 
  `\( E_a \)` is the activation energy,

• R is the ideal gas constant (8.314 J/(mol·K)),

• T is the temperature in Kelvin.

Given that the enzyme increases the rate of the hydrolysis reaction by a factor of
`\(5.00 * 10^8\)`, we can express this as:

`\[ k_{\text{enzyme}} = 5.00 * 10^8 * k_{\text{uncatalyzed}}. \]`

Now, comparing the Arrhenius equations for the catalyzed and uncatalyzed reactions, we get:

`\[ A_{\text{enzyme}}e^{-\frac{E_{a, \text{enzyme}}}{RT}} = 5.00 * 10^8 * A_{\text{uncatalyzed}}e^{-\frac{E_{a, \text{uncatalyzed}}}{RT}}. \]`

Since both reactions occur at the same temperature
`(\(T=298.15\) K)`, the temperature terms cancel out, leaving us with:

`\[ A_{\text{enzyme}}e^{-\frac{E_{a, \text{enzyme}}}{RT}} = 5.00 * 10^8 * A_{\text{uncatalyzed}}e^{-\frac{E_{a, \text{uncatalyzed}}}{RT}}. \]`

Now, we know that the pre-exponential factors
`(\(A_{\text{enzyme}}\)` and
`\(A_{\text{uncatalyzed}}\))` are the same since they represent the frequency of collisions and the nature of the reactants, and the temperature is constant. Therefore, the ratio of the exponential terms gives us the factor by which the activation energy is reduced:

`\[ e^{-\frac{E_{a, \text{enzyme}}}{RT}} = 5.00 * 10^8 * e^{-\frac{E_{a, \text{uncatalyzed}}}{RT}}. \]`

Taking the natural logarithm (ln) of both sides:

`\[ -\frac{E_{a, \text{enzyme}}}{RT} = \ln(5.00 * 10^8) - \frac{E_{a, \text{uncatalyzed}}}{RT}. \]`

Solving for
`\(E_{a, \text{enzyme}}\)`:

`\[ E_{a, \text{enzyme}} = -RT \ln(5.00 * 10^8) + E_{a, \text{uncatalyzed}}. \]`

Now, substitute the values:

`\[ E_{a, \text{enzyme}} = -(8.314 \, \text{J/(mol·K)}) * (298.15 \, \text{K}) * \ln(5.00 * 10^8) + 108 \, \text{kJ/mol}. \]`

After evaluating this expression, you can determine the activation energy for the enzyme-catalyzed reaction.