For the reaction
`3N_(2)O_((g))+2NH_(3(g)) rarr 4N_(2(g))+3H_(2)O_((g))Delta H^(@) = -879.6 KJ` if standard heat of formation of ammonia and water are -45.9KJ/mol and -241.8KJ/mol. Then standard heat of formation of nitrous oxide in KJ/mol is

Options

A. `-102`
B. `-82`
C. 82
D. 98

Explanation

To find the standard heat of formation of nitrous oxide (N₂O) from the given reaction and the standard heats of formation of ammonia (NH₃) and water (H₂O), we can follow these steps:

### Step 1: Write down the given reaction and data
The reaction is:
\[ 3N_{2}O_{(g)} + 2NH_{3(g)} \rightarrow 4N_{2(g)} + 3H_{2}O_{(g)} \]
Given:
- \(\Delta H^{\circ} = -879.6 \, \text{kJ}\)
- Standard heat of formation of ammonia, \(\Delta H_f^{\circ}(NH_3) = -45.9 \, \text{kJ/mol}\)
- Standard heat of formation of water, \(\Delta H_f^{\circ}(H_2O) = -241.8 \, \text{kJ/mol}\)

### Step 2: Write the formula for the enthalpy change
The enthalpy change for the reaction can be expressed as:
\[ \Delta H^{\circ} = \sum (\Delta H_f^{\circ} \text{ of products}) - \sum (\Delta H_f^{\circ} \text{ of reactants}) \]

### Step 3: Calculate the enthalpy of the products
The products of the reaction are \(4N_2\) and \(3H_2O\). The standard heat of formation for \(N_2\) is zero because it is in its elemental form. Therefore:
\[ \Delta H_f^{\circ}(products) = 0 + 3 \times (-241.8 \, \text{kJ/mol}) = -725.4 \, \text{kJ} \]

### Step 4: Calculate the enthalpy of the reactants
The reactants are \(3N_2O\) and \(2NH_3\):
\[ \Delta H_f^{\circ}(reactants) = 3 \times \Delta H_f^{\circ}(N_2O) + 2 \times (-45.9 \, \text{kJ/mol}) \]
\[ = 3 \times \Delta H_f^{\circ}(N_2O) - 91.8 \, \text{kJ} \]

### Step 5: Substitute into the enthalpy change equation
Now we can substitute into the enthalpy change equation:
\[ -879.6 \, \text{kJ} = (-725.4 \, \text{kJ}) - (3 \times \Delta H_f^{\circ}(N_2O) - 91.8 \, \text{kJ}) \]

### Step 6: Simplify the equation
Rearranging gives:
\[ -879.6 \, \text{kJ} = -725.4 \, \text{kJ} - 3 \Delta H_f^{\circ}(N_2O) + 91.8 \, \text{kJ} \]
Combine like terms:
\[ -879.6 \, \text{kJ} = -633.6 \, \text{kJ} - 3 \Delta H_f^{\circ}(N_2O) \]

### Step 7: Isolate \(\Delta H_f^{\circ}(N_2O)\)
Now, isolate \(\Delta H_f^{\circ}(N_2O)\):
\[ -879.6 + 633.6 = -3 \Delta H_f^{\circ}(N_2O) \]
\[ -246 = -3 \Delta H_f^{\circ}(N_2O) \]
\[ \Delta H_f^{\circ}(N_2O) = \frac{246}{3} = 82 \, \text{kJ/mol} \]

### Final Answer
The standard heat of formation of nitrous oxide (N₂O) is:
\[ \Delta H_f^{\circ}(N_2O) = 82 \, \text{kJ/mol} \]