Consider the following reactions :
(a) `H_((aq))^(+)+OH_((aq))^(-)=H_(2)O_((l)), Delta H= - X_(1)"kJ mol"^(-1)`
(b) `H_(2(g))+(1)/(2)O_(2(g))=H_(2)O_((l)), Delta H= - X_(2)"kJ mol"^(-1)`
(c ) `CO_(2(g))+H_(2(g))=CO_((g))+H_(2)O_((l))- X_(3)"kJ mol"^(-1)`
(d) `C_(2)H_(2(g))+(5)/(2)O_(2(g))=2CO_(2(g))+H_(2)O_((l))+X_(4)"kJ mol"^(-1)`
Enthalpy of formation of `H_(2)O_((l))` is :

Options

A. `+X_(1)"kJ mol"^(-1)`
B. `-X_(2)"kJ mol"^(-1)`
C. `+X_(3)"kJ mol"^(-1)`
D. `-X_(4)"kJ mol"^(-1)`

Explanation

To determine the enthalpy of formation of water (H₂O), we need to analyze the given reactions and identify which one corresponds to the formation of water from its constituent elements, hydrogen (H₂) and oxygen (O₂).

### Step-by-Step Solution:

1. **Understanding Enthalpy of Formation**:
The enthalpy of formation (ΔH_f) of a compound is defined as the heat change that results when one mole of the compound is formed from its elements in their standard states.

2. **Identify the Relevant Reaction**:
Among the provided reactions, we need to find the one that represents the formation of water from its elements:
- Reaction (a): H⁺(aq) + OH⁻(aq) → H₂O(l) (not from elements)
- Reaction (b): H₂(g) + (1/2)O₂(g) → H₂O(l) (this is from elements)
- Reaction (c): CO₂(g) + H₂(g) → CO(g) + H₂O(l) (not from elements)
- Reaction (d): C₂H₂(g) + (5/2)O₂(g) → 2CO₂(g) + H₂O(l) (not from elements)

The only reaction that shows water being formed from its constituent elements (H₂ and O₂) is reaction (b).

3. **Conclusion**:
Therefore, the enthalpy of formation of water (ΔH_f) is given by the enthalpy change of reaction (b):
\[ \Delta H_f = -X_2 \text{ kJ/mol} \]

### Final Answer:
The enthalpy of formation of H₂O(l) is \(-X_2\) kJ/mol.

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