The enthalpy of the reaction,
`H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)O(g) " is " Delta H_(1)` and that of `H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)O(l)` is `Delta H_(2)`. Then

Options

A. `Delta H_(1) lt Delta H_(2)`
B. `Delta H_(1) + Delta H_(2)= 0`
C. `Delta H_(1) gt Delta H_(2)`
D. `Delta H_(1)= Delta H_(2)`

Explanation

To find the relationship between the enthalpy changes of the two reactions given, we need to analyze the two reactions and apply the concept of the change in the number of moles of gas (ΔNg) and the enthalpy change equation.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- Reaction 1: \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \) with enthalpy change \( \Delta H_1 \)
- Reaction 2: \( H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \) with enthalpy change \( \Delta H_2 \)

2. **Calculate ΔNg for Reaction 1**:
- For Reaction 1, the total moles of gaseous reactants = 1 (from \( H_2 \)) + 0.5 (from \( O_2 \)) = 1.5 moles.
- The total moles of gaseous products = 1 (from \( H_2O(g) \)).
- Therefore, \( \Delta N_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 1 - 1.5 = -0.5 \).

3. **Calculate ΔNg for Reaction 2**:
- For Reaction 2, the total moles of gaseous reactants remain the same = 1.5 moles.
- The total moles of products = 0 (since \( H_2O \) is in the liquid state and does not count as a gas).
- Therefore, \( \Delta N_g = 0 - 1.5 = -1.5 \).

4. **Apply the Enthalpy Change Equation**:
- The general equation for the enthalpy change is given by:
\[ \Delta H = \Delta U + \Delta N_g RT \]
- For Reaction 1:
\[ \Delta H_1 = \Delta U + (-0.5)RT \]
- For Reaction 2:
\[ \Delta H_2 = \Delta U + (-1.5)RT \]

5. **Compare ΔH1 and ΔH2**:
- Since both reactions have the same internal energy change (\( \Delta U \)), we can compare the two:
\[ \Delta H_2 = \Delta U - 1.5RT \]
\[ \Delta H_1 = \Delta U - 0.5RT \]
- This implies:
\[ \Delta H_2 < \Delta H_1 \]
- Therefore, \( \Delta H_2 \) is more negative than \( \Delta H_1 \), indicating that the enthalpy change for the formation of liquid water is lower (more exothermic) than that for the formation of gaseous water.

### Conclusion:
The relationship between the enthalpy changes is:
\[ \Delta H_2 < \Delta H_1 \]