`DeltaH` for the reaction, `OF_(2)+H_(2)OrarrO_(2)+2HF`
(B.E. of `O-F, O-H, H-F and O=O` are 44, 111, 135 and `"119 kcal mol"^(-1)` respectively)

Options

A. `-222" kcal"`
B. `-"88 kcal"`
C. `-"111 kcal"`
D. `-"79 kcal"`

Explanation

To calculate the enthalpy change (ΔH) for the reaction:

\[ \text{OF}_2 + \text{H}_2\text{O} \rightarrow \text{O}_2 + 2\text{HF} \]

we will use the bond energies provided for the bonds involved in the reaction. The bond energies are as follows:

- O-F: 44 kcal/mol
- O-H: 111 kcal/mol
- H-F: 135 kcal/mol
- O=O: 119 kcal/mol

### Step-by-Step Solution:

1. **Identify the Bonds in Reactants:**
- In OF2, there are 2 O-F bonds.
- In H2O, there are 2 O-H bonds.

Therefore, the total bond energy for the reactants can be calculated as:
\[ \text{Total bond energy of reactants} = 2 \times \text{(O-F bond energy)} + 2 \times \text{(O-H bond energy)} \]
\[ = 2 \times 44 \, \text{kcal/mol} + 2 \times 111 \, \text{kcal/mol} \]
\[ = 88 \, \text{kcal/mol} + 222 \, \text{kcal/mol} = 310 \, \text{kcal/mol} \]

2. **Identify the Bonds in Products:**
- In O2, there is 1 O=O bond.
- In 2HF, there are 2 H-F bonds.

Therefore, the total bond energy for the products can be calculated as:
\[ \text{Total bond energy of products} = 1 \times \text{(O=O bond energy)} + 2 \times \text{(H-F bond energy)} \]
\[ = 1 \times 119 \, \text{kcal/mol} + 2 \times 135 \, \text{kcal/mol} \]
\[ = 119 \, \text{kcal/mol} + 270 \, \text{kcal/mol} = 389 \, \text{kcal/mol} \]

3. **Calculate ΔH using Hess's Law:**
According to Hess's Law:
\[ \Delta H = \text{Total bond energy of reactants} - \text{Total bond energy of products} \]
\[ \Delta H = 310 \, \text{kcal/mol} - 389 \, \text{kcal/mol} \]
\[ \Delta H = -79 \, \text{kcal/mol} \]

### Final Answer:
Thus, the enthalpy change (ΔH) for the reaction is:
\[ \Delta H = -79 \, \text{kcal/mol} \]