Determine enthalpy of formation for `H_(2)O_(2)(l)`, using the listed enthalpies of reaction :
`N_(2)H_(4)(l)+2H_(2)O_(2)(l)toN_(2)(g)+4H_(2)O(l)`,
`" "Delta_(r)H_(1)^(@)=-818kJ//mol`
`N_(2)H_(4)(l)+O_(2)(g)toN_(2)(g)+2H_(2)O(l)`
`" "Delta_(r)H_(2)^(@)=-622kJ//mol`
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)" "Delta_(r)H_(3)^(@)=-285kJ//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 will use Hess's law and the provided enthalpies of reaction. The enthalpy of formation is defined as the change in enthalpy when one mole of a compound is formed from its elements in their standard states.

### Step 1: Write the target reaction
The target reaction for the formation of hydrogen peroxide from its elements is:
\[ H_2(g) + O_2(g) \rightarrow H_2O_2(l) \]

### Step 2: Analyze the given reactions
We have the following reactions with their enthalpy changes:
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 3: Manipulate the reactions
We need to manipulate these reactions to derive the formation of \( H_2O_2(l) \).

1. **Reverse Reaction 1** and divide by 2:
\[ N_2(g) + 4H_2O(l) \rightarrow N_2H_4(l) + 2H_2O_2(l) \]
\(\Delta_rH = +\frac{818}{2} = +409 \, \text{kJ/mol}\)

2. **Keep Reaction 2** as is but divide by 2:
\[ N_2(g) + 2H_2O(l) \rightarrow N_2H_4(l) + O_2(g) \]
\(\Delta_rH = +\frac{622}{2} = +311 \, \text{kJ/mol}\)

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

### Step 4: Add the manipulated reactions
Now we will add the manipulated reactions together:
\[ \begin{align*} N_2(g) + 4H_2O(l) & \rightarrow N_2H_4(l) + 2H_2O_2(l) \quad (409 \, \text{kJ}) \\ N_2(g) + 2H_2O(l) & \rightarrow N_2H_4(l) + O_2(g) \quad (311 \, \text{kJ}) \\ H_2(g) + \frac{1}{2}O_2(g) & \rightarrow H_2O(l) \quad (-285 \, \text{kJ}) \\ \end{align*} \]

### Step 5: Cancel out the species
When we add these reactions, we can cancel out \( N_2H_4(l) \) and \( N_2(g) \):
\[ 2H_2O_2(l) + 2H_2O(l) + H_2(g) + \frac{1}{2}O_2(g) \rightarrow 2H_2O(l) + O_2(g) + H_2O(l) \]
This simplifies to:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O_2(l) \]

### Step 6: Calculate the total enthalpy change
Now, we can calculate the total enthalpy change for the formation of \( H_2O_2(l) \):
\[ \Delta H = 409 + 311 - 285 = 435 \, \text{kJ/mol} \]

### Final Answer
Thus, the enthalpy of formation for \( H_2O_2(l) \) is:
\[ \Delta H_f^\circ(H_2O_2) = 435 \, \text{kJ/mol} \]