If for `H_(2)(g)+1/2 O_(2)(g) to H_(2)O (g), triangleH_(1)` is enthalpy of reaction and for `H_(2) (g)+1/2 O_(2)(g) to H_(2)O (l), triangleH_(2)` is the enthalpy of reaction, then the magnitude of

Options

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

Answer

To solve the problem, we need to analyze the enthalpy changes for the two reactions given:

1. The reaction for the formation of water vapor (H2O(g)) from hydrogen gas (H2(g)) and oxygen gas (O2(g)):
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(g) \quad \Delta H_1 \]

2. The reaction for the formation of liquid water (H2O(l)) from hydrogen gas and oxygen gas:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \quad \Delta H_2 \]

### Step 1: Understanding the Enthalpy Changes
- **ΔH1** is the enthalpy change for the formation of water vapor (H2O(g)).
- **ΔH2** is the enthalpy change for the formation of liquid water (H2O(l)).

### Step 2: Breaking Down the Reactions
We can break down the formation of water vapor into two steps:
1. Formation of liquid water from hydrogen and oxygen (exothermic process).
2. Vaporization of liquid water to water vapor (endothermic process).

The overall reaction can be represented as:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \quad \Delta H_2 \quad \text{(exothermic)} \]
\[ H_2O(l) \rightarrow H_2O(g) \quad \Delta H_{\text{vap}} \quad \text{(endothermic)} \]

### Step 3: Relating ΔH1 and ΔH2
From the above steps, we can express ΔH1 in terms of ΔH2 and the heat of vaporization (ΔHvap):
\[ \Delta H_1 = \Delta H_2 + \Delta H_{\text{vap}} \]

### Step 4: Analyzing the Signs of Enthalpy Changes
- ΔH2 is negative (exothermic reaction).
- ΔHvap is positive (endothermic process).
- Therefore, ΔH1 will be less negative than ΔH2 because we are adding a positive value (ΔHvap) to a negative value (ΔH2).

### Step 5: Conclusion
Since ΔH1 is less negative than ΔH2, we can conclude:
\[ |\Delta H_1| < |\Delta H_2| \]

Thus, the magnitude of ΔH1 is less than the magnitude of ΔH2.

### Final Answer
\[ |\Delta H_1| < |\Delta H_2| \]