Determine enthalpy of formation for `H_(2)O_(2)(l)` , using the listed enthalpies of reactions:
`N_(2)H_(4)(l) + 2H_(2)O_(2)(l) rarr N_(2)(g) + 4H_(2)O(l), Delta_(r)H_(1)^(@) =-818 KJ//"mol"`
`N_(2)H_(4)(l)+O_(2)(g) rarr N_(2)(g) + 2H_(2)O(l), Delta_(r)H_(2)^(@)=-622 KJ//"mol"`
`H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)O(l), Delta_(r)H_(3)^(@)=-285 KJ//"mol"`

Options

A. `-383 KJ//"mol"`
B. `-187 KJ//"mol"`
C. `-498 KJ//"mol"`
D. None of these

Explanation

To determine the enthalpy of formation for \( H_2O_2(l) \), we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction. We will manipulate the given reactions to derive the formation reaction for \( H_2O_2(l) \).

### Given Reactions:
1. \( N_2H_4(l) + 2H_2O_2(l) \rightarrow N_2(g) + 4H_2O(l) \)
\( \Delta_rH_1^\circ = -818 \, \text{kJ/mol} \)

2. \( N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(l) \)
\( \Delta_rH_2^\circ = -622 \, \text{kJ/mol} \)

3. \( H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \)
\( \Delta_rH_3^\circ = -285 \, \text{kJ/mol} \)

### Step 1: Write the Formation Reaction for \( H_2O_2 \)
The formation reaction for \( H_2O_2(l) \) can be written as:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O_2(l) \]

### Step 2: Manipulate the Given Reactions
To find the enthalpy of formation for \( H_2O_2 \), we need to manipulate the given reactions to isolate \( H_2O_2 \).

1. **Reverse Reaction 1**:
\[ N_2(g) + 4H_2O(l) \rightarrow N_2H_4(l) + 2H_2O_2(l) \]
This will change the sign of \( \Delta_rH_1 \):
\( \Delta_rH_1^\circ = +818 \, \text{kJ/mol} \)

2. **Use Reaction 2 as is**:
\[ N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(l) \]
\( \Delta_rH_2^\circ = -622 \, \text{kJ/mol} \)

3. **Use Reaction 3 as is**:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \]
\( \Delta_rH_3^\circ = -285 \, \text{kJ/mol} \)

### Step 3: Combine the Reactions
Now we will combine the manipulated reactions:

1. From the reversed Reaction 1:
\[ N_2(g) + 4H_2O(l) \rightarrow N_2H_4(l) + 2H_2O_2(l) \quad (+818 \, \text{kJ/mol}) \]

2. From Reaction 2:
\[ N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(l) \quad (-622 \, \text{kJ/mol}) \]

3. From Reaction 3 (we will need it for the formation of \( H_2O \)):
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad (-285 \, \text{kJ/mol}) \]

### Step 4: Add the Reactions
When we add these reactions, we need to ensure that the products and reactants cancel appropriately to yield the desired formation reaction for \( H_2O_2 \).

After combining and simplifying, we will find:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O_2(l) \]

### Step 5: Calculate the Total Enthalpy Change
The total enthalpy change \( \Delta H_f^\circ \) for the formation of \( H_2O_2 \) will be:
\[ \Delta H_f^\circ = +818 - 622 - 285 \]
\[ \Delta H_f^\circ = -89 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of formation for \( H_2O_2(l) \) is:
\[ \Delta H_f^\circ = -89 \, \text{kJ/mol} \]