The ehthalpy change taking place during the reaction `H_2O(l) rarr H_2O(g) is ,` [ Given `Delta H_f=H_2O(g) =- 57 kcal, Delta H_f = H_2 O(l) =- 68.3 kcal `

Options

A. `+11.3 " kcal " `
B. `-11.3 " kcal " `
C. `115 .3 " kcal " `
D. `+115.3` kcal

Explanation

To find the enthalpy change (ΔH) for the reaction \( H_2O(l) \rightarrow H_2O(g) \), we will use the enthalpy of formation values provided for both states of water.

### Step-by-Step Solution:

1. **Write the Enthalpy of Formation Reactions**:
- The enthalpy of formation for \( H_2O(g) \):
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g) \quad \Delta H_f = -57 \text{ kcal} \]
- The enthalpy of formation for \( H_2O(l) \):
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H_f = -68.3 \text{ kcal} \]

2. **Identify the Target Reaction**:
- The target reaction we want to find the enthalpy change for is:
\[ H_2O(l) \rightarrow H_2O(g) \quad \Delta H = ? \]

3. **Use Hess's Law**:
- According to Hess's Law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We can manipulate the above equations to find ΔH for the target reaction.

4. **Subtract the Enthalpy of Formation of Liquid Water from that of Gaseous Water**:
- We can express the target reaction as:
\[ H_2O(l) \rightarrow H_2O(g) \]
- To find ΔH for this reaction, we can subtract the enthalpy of formation of \( H_2O(l) \) from that of \( H_2O(g) \):
\[ \Delta H = \Delta H_f(H_2O(g)) - \Delta H_f(H_2O(l)) \]
- Substituting the values:
\[ \Delta H = (-57 \text{ kcal}) - (-68.3 \text{ kcal}) \]
\[ \Delta H = -57 + 68.3 = 11.3 \text{ kcal} \]

5. **Conclusion**:
- The enthalpy change for the reaction \( H_2O(l) \rightarrow H_2O(g) \) is \( +11.3 \text{ kcal} \).

### Final Answer:
\[ \Delta H = +11.3 \text{ kcal} \]