To solve the problem of how the change in the reaction energy upon adding an enzyme affects the rate of a biochemical reaction, we can follow these steps:
### Step 1: Understand the Rate Increase
The problem states that the reaction occurs \(10^6\) times faster with the enzyme than without it. This means that the rate constant \(k\) with the enzyme (\(k_{enzyme}\)) is \(10^6\) times the rate constant without the enzyme (\(k_{no \, enzyme}\)):
\[
k_{enzyme} = 10^6 \cdot k_{no \, enzyme}
\]
### Step 2: Apply the Arrhenius Equation
The Arrhenius equation relates the rate constant \(k\) to the activation energy \(E_a\) and temperature \(T\):
\[
k = A e^{-\frac{E_a}{RT}}
\]
Where:
- \(A\) = pre-exponential factor
- \(E_a\) = activation energy
- \(R\) = universal gas constant
- \(T\) = temperature in Kelvin
### Step 3: Write the Arrhenius Equation for Both Cases
For the reaction without the enzyme:
\[
k_{no \, enzyme} = A e^{-\frac{E_{a, no}}{RT}}
\]
For the reaction with the enzyme:
\[
k_{enzyme} = A e^{-\frac{E_{a, enzyme}}{RT}}
\]
### Step 4: Set Up the Equation
From the relationship established in Step 1, we can write:
\[
10^6 \cdot k_{no \, enzyme} = k_{enzyme}
\]
Substituting the Arrhenius equations, we get:
\[
10^6 \cdot A e^{-\frac{E_{a, no}}{RT}} = A e^{-\frac{E_{a, enzyme}}{RT}}
\]
### Step 5: Simplify the Equation
We can cancel \(A\) from both sides (assuming \(A\) is not zero):
\[
10^6 \cdot e^{-\frac{E_{a, no}}{RT}} = e^{-\frac{E_{a, enzyme}}{RT}}
\]
Taking the natural logarithm of both sides gives:
\[
\ln(10^6) - \frac{E_{a, no}}{RT} = -\frac{E_{a, enzyme}}{RT}
\]
### Step 6: Rearranging the Equation
Rearranging the equation to isolate the activation energy terms:
\[
\frac{E_{a, enzyme}}{RT} = \frac{E_{a, no}}{RT} - \ln(10^6)
\]
This simplifies to:
\[
E_{a, enzyme} = E_{a, no} - RT \cdot \ln(10^6)
\]
### Step 7: Calculate the Change in Activation Energy
The change in activation energy (\(\Delta E_a\)) when adding the enzyme is:
\[
\Delta E_a = E_{a, enzyme} - E_{a, no} = -RT \cdot \ln(10^6)
\]
### Step 8: Final Expression
Since \(\ln(10^6) = 6 \ln(10)\):
\[
\Delta E_a = -6RT \cdot \ln(10)
\]
### Conclusion
The change in the reaction energy upon adding the enzyme is:
\[
\Delta E_a = -6RT \cdot \ln(10)
\]
