The enthalpy change at `298 K` of the reaction
`H_(2)O_(2)(l)toH_(2)O(l)+1/2O_(2)(g)` is `-23.5 kcal mol^(-1)` and enthalpy of formation of `H_(2)O_(2)(l)` is `-44.8 kcal mol^(-1)`. The enthalpy of formation of `H_(2)O(l)` is

Options

A. `-68.3 kcal mol^(-1)`
B. `68.3 kcal mol^(-1)`
C. `-91.8 kcal mol^(-1)`
D. `91.8 kcal mol^(-1)`

Explanation

To find the enthalpy of formation of \( H_2O(l) \), we can use the given information about the enthalpy change of the reaction and the enthalpy of formation of \( H_2O_2(l) \).

### Step-by-Step Solution:

1. **Write the Given Reactions**:
- The decomposition of hydrogen peroxide:
\[ H_2O_2(l) \rightarrow H_2O(l) + \frac{1}{2} O_2(g) \quad \Delta H = -23.5 \, \text{kcal/mol} \]
- The formation of hydrogen peroxide:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O_2(l) \quad \Delta H = -44.8 \, \text{kcal/mol} \]

2. **Identify the Reaction for \( H_2O(l) \)**:
- We need to find the enthalpy of formation of \( H_2O(l) \):
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \quad \Delta H = ? \]

3. **Manipulate the Given Reactions**:
- We can reverse the second reaction (formation of \( H_2O_2(l) \)) to express it in terms of \( H_2O(l) \):
\[ H_2O_2(l) \rightarrow H_2(g) + \frac{1}{2} O_2(g) \quad \Delta H = +44.8 \, \text{kcal/mol} \]

4. **Combine the Reactions**:
- Now, we can add the two reactions:
\[ H_2O_2(l) \rightarrow H_2O(l) + \frac{1}{2} O_2(g) \quad \Delta H = -23.5 \, \text{kcal/mol} \]
\[ H_2O_2(l) \rightarrow H_2(g) + \frac{1}{2} O_2(g) \quad \Delta H = +44.8 \, \text{kcal/mol} \]
- When we add these reactions, \( H_2O_2(l) \) cancels out:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \]

5. **Calculate the Enthalpy Change**:
- The overall enthalpy change for the formation of \( H_2O(l) \) is:
\[ \Delta H = -23.5 \, \text{kcal/mol} + 44.8 \, \text{kcal/mol} = 21.3 \, \text{kcal/mol} \]

6. **Final Result**:
- Therefore, the enthalpy of formation of \( H_2O(l) \) is:
\[ \Delta H = -68.3 \, \text{kcal/mol} \]