For the reaction
`3N_(2)O(g)+2NH_(3)(g)to4N_(2)(g)+3H_(2)O(g),DeltaH^(@)=-879.6kJ`
If `DeltaH_(f)^(@)[NH_(3)(g)]=-45.9kJ" "mol^(-1)`,
`DeltaH_(f)^(@)[H_(2)O(g)]=-241.8kJ" "mol^(-1)`
Then `DeltaH_(f)^(@)[N_(2)O(g)]` will be:

Options

A. `+246` kJ
B. `+82` kJ
C. `-82kJ`
D. `-246kJ`

Explanation

To find the standard enthalpy of formation (ΔH_f^(@)) for N2O(g), we can use the given reaction and the enthalpies of formation for the other compounds involved. The reaction is:

\[ 3N_{2}O(g) + 2NH_{3}(g) \rightarrow 4N_{2}(g) + 3H_{2}O(g) \]

Given:
- ΔH^(@) = -879.6 kJ (for the reaction)
- ΔH_f^(@)[NH3(g)] = -45.9 kJ/mol
- ΔH_f^(@)[H2O(g)] = -241.8 kJ/mol

We can use the following equation for the enthalpy change of the reaction:

\[ \Delta H_{reaction} = \sum \Delta H_f^(@) \text{(products)} - \sum \Delta H_f^(@) \text{(reactants)} \]

### Step 1: Calculate the total enthalpy of formation for the products
The products of the reaction are \(4N_{2}(g)\) and \(3H_{2}O(g)\).

- The enthalpy of formation for \(N_{2}(g)\) is 0 kJ/mol (elements in their standard state).
- For \(H_{2}O(g)\), we have:
\[ \Delta H_f^(@)[H_{2}O(g)] = -241.8 \text{ kJ/mol} \]

Thus, the total enthalpy of formation for the products is:
\[ \Delta H_f^(@) \text{(products)} = 4 \times 0 + 3 \times (-241.8) = 0 - 725.4 = -725.4 \text{ kJ} \]

### Step 2: Calculate the total enthalpy of formation for the reactants
The reactants are \(3N_{2}O(g)\) and \(2NH_{3}(g)\).

- For \(NH_{3}(g)\), we have:
\[ \Delta H_f^(@)[NH_{3}(g)] = -45.9 \text{ kJ/mol} \]

Thus, the total enthalpy of formation for the reactants is:
\[ \Delta H_f^(@) \text{(reactants)} = 3 \times \Delta H_f^(@)[N_{2}O(g)] + 2 \times (-45.9) \]
\[ = 3 \times \Delta H_f^(@)[N_{2}O(g)] - 91.8 \]

### Step 3: Set up the equation using the reaction enthalpy
Now we can set up the equation using the enthalpy change of the reaction:
\[ -879.6 = (-725.4) - (3 \times \Delta H_f^(@)[N_{2}O(g)] - 91.8) \]

### Step 4: Simplify the equation
Rearranging gives:
\[ -879.6 = -725.4 - 3 \times \Delta H_f^(@)[N_{2}O(g)] + 91.8 \]
\[ -879.6 = -633.6 - 3 \times \Delta H_f^(@)[N_{2}O(g)] \]
\[ -879.6 + 633.6 = -3 \times \Delta H_f^(@)[N_{2}O(g)] \]
\[ -246 = -3 \times \Delta H_f^(@)[N_{2}O(g)] \]

### Step 5: Solve for ΔH_f^(@)[N2O(g)]
Dividing both sides by -3 gives:
\[ \Delta H_f^(@)[N_{2}O(g)] = \frac{246}{3} = 82 \text{ kJ/mol} \]

### Final Answer:
\[ \Delta H_f^(@)[N_{2}O(g)] = 82 \text{ kJ/mol} \]