STEM Problem Detail

Problem Statement

Prompt

Provide step by step solution of Problem 3.2

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## Detailed Solution for Potential Energy Surface

### 1. Morse Potential Calculation for $$H_2$$

The Morse equation is defined as:
\[ E = D \left( e^{-2\beta(r-r_0)} - 2e^{-\beta(r-r_0)} \right) \]

Given constants for $$H_2$$:

- $$D = 438.9 \text{ kJ mol}^{-1}$$
- $$r_0 = 74 \text{ pm}$$
- $$\beta = 19.54 \text{ nm}^{-1} = 0.01954 \text{ pm}^{-1}$$

#### Example: Calculation for $$r = 50$$ pm

First, calculate the displacement:
\[ \Delta r = r - r_0 = 50 - 74 = -24 \text{ pm} \]
Now, calculate the exponent term $$x = \beta(r-r_0)$$:
\[ x = 0.01954 \times (-24) = -0.46896 \]
Substitute into the Morse equation:
\[ E = 438.9 \left( e^{-2(-0.46896)} - 2e^{-(-0.46896)} \right) \]
\[ E = 438.9 \left( 2.5547 - 2(1.5983) \right) \]
\[ E = 438.9 \left( 2.5547 - 3.1966 \right) = \mathbf{-281.7 \text{ kJ mol}^{-1}} \]

### 2. Coulombic and Exchange Energies

Based on the assumption that Coulombic energy ($$Q$$) is 14% of the total Morse energy:

- $$Q = 0.14 \times E$$
- $$J = 0.86 \times E$$ (Exchange Energy)

For $$r = 50$$ pm:
\[ Q = 0.14 \times (-281.7) = -39.44 \text{ kJ mol}^{-1} \]
\[ J = 0.86 \times (-281.7) = -242.26 \text{ kJ mol}^{-1} \]

### 3. Linear $$H \cdots H \cdots H$$ (London Equation)

The total energy $$E$$ for the triatomic system is calculated using:
\[ E = A + B + C - \left\{ \frac{1}{2} \left[ (\alpha-\beta)^2 + (\beta-\gamma)^2 + (\gamma-\alpha)^2 \right] \right\}^{1/2} \]
Where:

- $$A, B, C$$ are Coulombic energies for pairs $$r_1, r_2,$$ and $$(r_1+r_2)$$.
- $$\alpha, \beta, \gamma$$ are Exchange energies for pairs $$r_1, r_2,$$ and $$(r_1+r_2)$$.

### 4. Results from Surface Analysis

From the constructed potential energy surface (PES):

- **Basin depth ($$E_d$$):** $$\approx 13 \text{ kJ mol}^{-1}$$
- **Classical activation energy ($$E_c$$):** $$\approx 50 \text{ kJ mol}^{-1}$$

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Problem #56