To solve the problem, we need to analyze the given reaction:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]
with the enthalpy change (\(\Delta H\)) given as \(-x \, \text{kJ}\).
### Step-by-Step Solution:
1. **Understanding the Reaction**:
- The reaction shows the formation of liquid water from hydrogen and oxygen gases. The negative sign in \(\Delta H\) indicates that the reaction is exothermic, meaning heat is released during the reaction.
2. **Defining Heat of Formation**:
- The heat of formation (\(\Delta H_f\)) is defined as the heat change when one mole of a compound is formed from its elements in their standard states. For water (\(H_2O\)), the reaction for the formation of one mole is:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \]
3. **Relating the Given Reaction to Heat of Formation**:
- The given reaction produces 2 moles of \(H_2O\). Therefore, if we want to find the heat of formation for one mole of \(H_2O\), we need to divide the entire reaction by 2:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \]
- The enthalpy change for this reaction will be:
\[ \Delta H_f = \frac{-x}{2} \, \text{kJ} \]
4. **Conclusion on Heat of Formation**:
- Thus, the heat of formation of \(H_2O\) is \(-\frac{x}{2} \, \text{kJ}\). This indicates that for the formation of one mole of water, \(\frac{x}{2} \, \text{kJ}\) of heat is released.
5. **Analyzing the Options**:
- Option A: Incorrect, as it states \(x \, \text{kJ}\) is the heat of formation.
- Option B: Correct, as it states \(x \, \text{kJ}\) is the heat of the reaction (since the reaction releases \(x \, \text{kJ}\)).
- Option C: Incorrect, as it states \(x \, \text{kJ}\) is the heat of combustion of \(H_2\) (which would be \(-\frac{x}{2} \, \text{kJ}\)).
- Option D: Correct, as it states the heat of formation of \(H_2O\) is \(-\frac{x}{2} \, \text{kJ}\).
### Final Answer:
The correct options are B and D.
