When a biochemical reaction is carried out in laboratory from outside of human body in the absence of enzyme, the rate of reaction obtained is `10^(-6)` times, then activation energy of the reaction in the presence of enzyme is

Options

A. `(6)/(RT)`
B. `P` is required
C. different from `E_(a)` obtained in laboratory
D. can't say anything

Explanation

To solve the problem, we need to understand the relationship between the rate of reaction and activation energy, particularly in the context of enzyme-catalyzed reactions.

### Step-by-Step Solution:

1. **Understanding the Rate of Reaction**:
- In the absence of an enzyme, the rate of the biochemical reaction is given as \(10^{-6}\) times the rate of the reaction in the presence of an enzyme. This indicates that the enzyme significantly increases the rate of the reaction.

2. **Relationship Between Rate and Activation Energy**:
- The rate of a reaction can be described by the Arrhenius equation:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \(k\) is the rate constant,
- \(A\) is the pre-exponential factor,
- \(E_a\) is the activation energy,
- \(R\) is the gas constant,
- \(T\) is the temperature in Kelvin.
- When an enzyme is present, it lowers the activation energy (\(E_a\)) for the reaction.

3. **Comparing Activation Energies**:
- Let \(E_{a, \text{lab}}\) be the activation energy of the reaction in the laboratory (without enzyme) and \(E_{a, \text{enzyme}}\) be the activation energy in the presence of the enzyme.
- Since the rate in the presence of the enzyme is significantly higher, we can infer that:
\[ k_{\text{enzyme}} \gg k_{\text{lab}} \]
- This implies that:
\[ e^{-\frac{E_{a, \text{enzyme}}}{RT}} \gg e^{-\frac{E_{a, \text{lab}}}{RT}} \]

4. **Calculating Activation Energy**:
- Given that the rate in the absence of the enzyme is \(10^{-6}\) times that with the enzyme, we can express this as:
\[ \frac{k_{\text{enzyme}}}{k_{\text{lab}}} = 10^6 \]
- From the Arrhenius equation, we can derive the relationship:
\[ \frac{A e^{-\frac{E_{a, \text{enzyme}}}{RT}}}{A e^{-\frac{E_{a, \text{lab}}}{RT}}} = 10^6 \]
- Simplifying gives:
\[ e^{-\frac{E_{a, \text{enzyme}}}{RT}} = 10^6 e^{-\frac{E_{a, \text{lab}}}{RT}} \]

5. **Taking Natural Logarithm**:
- Taking the natural logarithm of both sides:
\[ -\frac{E_{a, \text{enzyme}}}{RT} = \ln(10^6) - \frac{E_{a, \text{lab}}}{RT} \]
- Rearranging gives:
\[ E_{a, \text{enzyme}} = E_{a, \text{lab}} - RT \ln(10^6) \]

6. **Conclusion**:
- Since the presence of the enzyme lowers the activation energy, we can conclude that \(E_{a, \text{enzyme}}\) is significantly lower than \(E_{a, \text{lab}}\). The exact value will depend on the specific values of \(R\) and \(T\) used in the calculation.