The enthalpies of formation of `N_2O` and NO are 28 and 90 kj `mol^(-1)` respectively. The enthalpy of the reaction, `2N_2O(g)+O_2(g) rarr 4NO(g)` is equal to

Options

A. 8 kJ
B. 88 kJ
C. `-16 kJ`
D. `304 kJ`

Answer

To find the enthalpy of the reaction \(2N_2O(g) + O_2(g) \rightarrow 4NO(g)\), we will use the given enthalpies of formation for \(N_2O\) and \(NO\).

### Step-by-Step Solution:

1. **Write the given enthalpies of formation:**
- Enthalpy of formation of \(N_2O\) (\(\Delta H_f^{\circ} (N_2O)\)) = 28 kJ/mol
- Enthalpy of formation of \(NO\) (\(\Delta H_f^{\circ} (NO)\)) = 90 kJ/mol

2. **Write the formation reactions for \(N_2O\) and \(NO\):**
- For \(N_2O\):
\[ N_2(g) + \frac{1}{2}O_2(g) \rightarrow N_2O(g) \quad \Delta H = 28 \text{ kJ} \]
- For \(NO\):
\[ \frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \rightarrow NO(g) \quad \Delta H = 90 \text{ kJ} \]

3. **Multiply the formation reactions to match the coefficients in the target reaction:**
- Multiply the formation reaction of \(N_2O\) by 2:
\[ 2N_2(g) + O_2(g) \rightarrow 2N_2O(g) \quad \Delta H = 2 \times 28 = 56 \text{ kJ} \]
- Multiply the formation reaction of \(NO\) by 4:
\[ 2N_2(g) + 2O_2(g) \rightarrow 4NO(g) \quad \Delta H = 4 \times 90 = 360 \text{ kJ} \]

4. **Set up the overall reaction:**
- The overall reaction can be derived by subtracting the formation of \(N_2O\) from the formation of \(NO\):
\[ 2N_2O(g) + O_2(g) \rightarrow 4NO(g) \]

5. **Calculate the enthalpy change for the overall reaction:**
- The enthalpy change for the overall reaction is given by:
\[ \Delta H_{reaction} = \Delta H_{products} - \Delta H_{reactants} \]
- Substituting the values:
\[ \Delta H_{reaction} = 360 \text{ kJ} - 56 \text{ kJ} = 304 \text{ kJ} \]

### Final Answer:
The enthalpy of the reaction \(2N_2O(g) + O_2(g) \rightarrow 4NO(g)\) is **304 kJ**.

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