Consider the following two reactions :
(i) `"Propene "+H_(2) rarr "Propane, "DeltaH_(1)`
(ii) `"Cyclopropane"+H_(2)rarr"Propane, "DeltaH_(2)`
Then, `DeltaH_(2)-DeltaH_(1)` will be

Options

A. (a) 0
B. (b) `2BE_(C-C)-BE_(C=C)`
C. (c) BE_(C=C)`
D. (d) `2BE_(C=C)-BE_(C-C)`

Explanation

To find the value of \( \Delta H_2 - \Delta H_1 \) for the given reactions, we will analyze the bond energies involved in both reactions.

### Step 1: Write the reactions and identify the bonds
1. **Reaction (i)**:
\[ \text{Propene} + H_2 \rightarrow \text{Propane} \]
- **Reactants**: Propene (C3H6) has:
- 1 C=C bond
- 2 C-C bonds
- 6 C-H bonds
- 1 H-H bond
- **Products**: Propane (C3H8) has:
- 2 C-C bonds
- 8 C-H bonds

2. **Reaction (ii)**:
\[ \text{Cyclopropane} + H_2 \rightarrow \text{Propane} \]
- **Reactants**: Cyclopropane (C3H6) has:
- 3 C-C bonds
- 6 C-H bonds
- 1 H-H bond
- **Products**: Propane (C3H8) has:
- 2 C-C bonds
- 8 C-H bonds

### Step 2: Calculate \( \Delta H_1 \) for Reaction (i)
Using the bond energy formula:
\[ \Delta H_1 = \text{(Bonds in reactants)} - \text{(Bonds in products)} \]
- **Bonds in Reactants**:
- C=C bond: \( \text{BE}_{C=C} \)
- C-C bonds: \( 2 \times \text{BE}_{C-C} \)
- C-H bonds: \( 6 \times \text{BE}_{C-H} \)
- H-H bond: \( \text{BE}_{H-H} \)

- **Bonds in Products**:
- C-C bonds: \( 2 \times \text{BE}_{C-C} \)
- C-H bonds: \( 8 \times \text{BE}_{C-H} \)

Thus,
\[ \Delta H_1 = \left( \text{BE}_{C=C} + 2 \text{BE}_{C-C} + 6 \text{BE}_{C-H} + \text{BE}_{H-H} \right) - \left( 2 \text{BE}_{C-C} + 8 \text{BE}_{C-H} \right) \]
Simplifying this gives:
\[ \Delta H_1 = \text{BE}_{C=C} - 2 \text{BE}_{C-H} + \text{BE}_{H-H} \]

### Step 3: Calculate \( \Delta H_2 \) for Reaction (ii)
Using the same bond energy formula:
\[ \Delta H_2 = \text{(Bonds in reactants)} - \text{(Bonds in products)} \]
- **Bonds in Reactants**:
- C-C bonds: \( 3 \times \text{BE}_{C-C} \)
- C-H bonds: \( 6 \times \text{BE}_{C-H} \)
- H-H bond: \( \text{BE}_{H-H} \)

- **Bonds in Products**:
- C-C bonds: \( 2 \times \text{BE}_{C-C} \)
- C-H bonds: \( 8 \times \text{BE}_{C-H} \)

Thus,
\[ \Delta H_2 = \left( 3 \text{BE}_{C-C} + 6 \text{BE}_{C-H} + \text{BE}_{H-H} \right) - \left( 2 \text{BE}_{C-C} + 8 \text{BE}_{C-H} \right) \]
Simplifying this gives:
\[ \Delta H_2 = \text{BE}_{C-C} - 2 \text{BE}_{C-H} + \text{BE}_{H-H} \]

### Step 4: Calculate \( \Delta H_2 - \Delta H_1 \)
Now, we find \( \Delta H_2 - \Delta H_1 \):
\[ \Delta H_2 - \Delta H_1 = \left( \text{BE}_{C-C} - 2 \text{BE}_{C-H} + \text{BE}_{H-H} \right) - \left( \text{BE}_{C=C} - 2 \text{BE}_{C-H} + \text{BE}_{H-H} \right) \]
This simplifies to:
\[ \Delta H_2 - \Delta H_1 = \text{BE}_{C-C} - \text{BE}_{C=C} \]

### Conclusion
Thus, the final result is:
\[ \Delta H_2 - \Delta H_1 = \text{2 BE}_{C-C} - \text{BE}_{C=C} \]