Rate of a reaction increases by `10^6` times when a reaction is carried out in presence of enzyme catalyst at same temperature. Determine change in activation energy.

Options

A. (a) `-6(2.303)RT`
B. (b) `+6(2.303)RT`
C. (c) `+6RT`
D. (d) `-6RT`

Answer

To determine the change in activation energy when the rate of a reaction increases by \(10^6\) times in the presence of an enzyme catalyst, we can use the Arrhenius equation and the relationship between rate constants and activation energy.

### Step-by-Step Solution:

1. **Understand the Arrhenius Equation**:
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \(k\) = rate constant
- \(A\) = pre-exponential factor
- \(E_a\) = activation energy
- \(R\) = universal gas constant
- \(T\) = temperature in Kelvin

2. **Set Up the Problem**:
Let:
- \(E\) = activation energy in the absence of the catalyst
- \(E_C\) = activation energy in the presence of the catalyst
- The rate constant in the absence of the catalyst is \(k\).
- The rate constant in the presence of the catalyst is \(10^6 k\).

3. **Write the Rate Constant Equations**:
For the reaction without the catalyst:
\[ k = A e^{-\frac{E}{RT}} \]
For the reaction with the catalyst:
\[ 10^6 k = A e^{-\frac{E_C}{RT}} \]

4. **Divide the Two Equations**:
Dividing the second equation by the first gives:
\[ 10^6 = \frac{A e^{-\frac{E_C}{RT}}}{A e^{-\frac{E}{RT}}} \]
This simplifies to:
\[ 10^6 = e^{-\frac{E_C - E}{RT}} \]

5. **Take the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln(10^6) = -\frac{E_C - E}{RT} \]
We know that \(\ln(10^6) = 6 \ln(10)\) and \(\ln(10) \approx 2.303\), so:
\[ 6 \cdot 2.303 = -\frac{E_C - E}{RT} \]

6. **Rearranging for Activation Energy Change**:
Rearranging gives:
\[ E_C - E = -6 \cdot 2.303 \cdot RT \]
Thus, the change in activation energy (\(\Delta E_a\)) is:
\[ \Delta E_a = E_C - E = -6 \cdot 2.303 \cdot RT \]

### Final Answer:
The change in activation energy when the reaction is carried out in the presence of an enzyme catalyst is:
\[ \Delta E_a = -6 \cdot 2.303 \cdot RT \]