The enthalpy of the reaction
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g)` is `DeltaH_(1)` and that of `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)` is `DeltaH_(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 solve the problem, we need to analyze the two reactions given and their corresponding enthalpy changes, \(\Delta H_1\) and \(\Delta H_2\).

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The first reaction is:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g) \quad \text{(Enthalpy change = } \Delta H_1\text{)} \]
- The second reaction is:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \text{(Enthalpy change = } \Delta H_2\text{)} \]

2. **Understanding the Nature of the Reactions**:
- The first reaction produces water in the gaseous state (steam), while the second reaction produces water in the liquid state.
- The transition from gas to liquid (condensation) is an exothermic process, meaning it releases heat.

3. **Comparing the Enthalpy Changes**:
- In the first reaction, energy is absorbed to form gaseous water. Thus, \(\Delta H_1\) is positive.
- In the second reaction, energy is released when gaseous water condenses into liquid water. Thus, \(\Delta H_2\) is negative.

4. **Relating \(\Delta H_1\) and \(\Delta H_2\)**:
- Since condensation releases energy, the enthalpy change for the condensation process (\(\Delta H_2\)) will be less than that for the formation of gaseous water (\(\Delta H_1\)).
- Therefore, we can conclude:
\[ \Delta H_2 < \Delta H_1 \]
- This implies that \(\Delta H_2\) is more negative (or less positive) than \(\Delta H_1\).

5. **Final Conclusion**:
- The correct relationship between the two enthalpy changes is:
\[ \Delta H_2 < \Delta H_1 \]