The enthalpy of reaction for thr equation
`2H_2(g) +O_2(g) rarr 2H_2O(l)` is `DeltaH^@=-572kj/mol`
what will be the standard enthalpy for the formation of `H_2O`(l)?

Options

A. (-572 kj/mol)
B. (+57 kj/mol)
C. (+286 kj/mol)
D. (-286 kj/mol0

Answer

To determine the standard enthalpy of formation for water (H₂O) in its liquid state from the given reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Reaction**:
The reaction provided is:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]
The enthalpy change for this reaction is given as:
\[ \Delta H^\circ = -572 \text{ kJ/mol} \]

2. **Identify the Formation Reaction**:
The standard enthalpy of formation (\(\Delta H_f^\circ\)) for a compound is defined as the enthalpy change when one mole of the compound is formed from its elements in their standard states. For water, the formation reaction is:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \]

3. **Relate the Two Reactions**:
The reaction for the formation of 2 moles of water is:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]
The enthalpy change for this reaction is \(-572 \text{ kJ}\) for 2 moles of water.

4. **Calculate the Enthalpy Change for One Mole**:
To find the standard enthalpy of formation for one mole of water, we need to divide the enthalpy change of the reaction by 2:
\[ \Delta H_f^\circ(H_2O) = \frac{-572 \text{ kJ}}{2} = -286 \text{ kJ/mol} \]

5. **Final Answer**:
Therefore, the standard enthalpy of formation for liquid water is:
\[ \Delta H_f^\circ(H_2O) = -286 \text{ kJ/mol} \]