An organic compound dissociates into n parallel first order reactions simultaneously and produces n different products `P_(1), P_(2), P_(3), ......., P_(n)` having rate constants `k, 2k, 3k, ......, nk` and activation energies E, 2E, 3E ..............., nE respectively. Calculate the overall energy of activation of the compound P.

Options

A. 2E/3(2n+1)
B. E/3(2n+1)
C. E/3(n+1)
D. E/3(n+2)

Answer

To solve the problem of calculating the overall activation energy of the compound P that dissociates into n parallel first-order reactions, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have n parallel first-order reactions producing n different products \( P_1, P_2, \ldots, P_n \).
- Each product has a rate constant \( k, 2k, 3k, \ldots, nk \) and activation energies \( E, 2E, 3E, \ldots, nE \).

2. **Using the Formula for Overall Activation Energy**:
- The overall activation energy \( E_{\text{Net}} \) can be calculated using the formula:
\[ E_{\text{Net}} = \frac{(E_1 \cdot k_1 + E_2 \cdot k_2 + \ldots + E_n \cdot k_n)}{(k_1 + k_2 + \ldots + k_n)} \]
- Here, \( E_i \) are the activation energies and \( k_i \) are the rate constants.

3. **Substituting the Values**:
- Substitute \( E_i \) and \( k_i \):
\[ E_{\text{Net}} = \frac{(E \cdot k + 2E \cdot 2k + 3E \cdot 3k + \ldots + nE \cdot nk)}{(k + 2k + 3k + \ldots + nk)} \]

4. **Simplifying the Numerator**:
- The numerator becomes:
\[ E(k + 4k + 9k + \ldots + n^2k) = kE(1^2 + 2^2 + 3^2 + \ldots + n^2) \]

5. **Simplifying the Denominator**:
- The denominator becomes:
\[ k(1 + 2 + 3 + \ldots + n) \]

6. **Using the Formulas for Sums**:
- The sum of the first n squares is given by:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6} \]
- The sum of the first n natural numbers is given by:
\[ \sum_{i=1}^{n} i = \frac{n(n + 1)}{2} \]

7. **Substituting the Sums into the Formula**:
- Substitute these sums into the equation for \( E_{\text{Net}} \):
\[ E_{\text{Net}} = \frac{kE \cdot \frac{n(n + 1)(2n + 1)}{6}}{k \cdot \frac{n(n + 1)}{2}} \]

8. **Canceling Out Common Terms**:
- The \( k \) and \( n(n + 1) \) terms cancel out:
\[ E_{\text{Net}} = \frac{E(2n + 1)}{3} \]

9. **Final Result**:
- Therefore, the overall activation energy of the compound P is:
\[ E_{\text{Net}} = \frac{(2n + 1)E}{3} \]