`Delta_f H^@ of CO_2(g), CO(g) , N_2O(g) and NO_2(g)` in KJ/ mol are respectively -393 -110,81 and 34.Calculate the `Delta H` in kj of the following reaction:
`2NO_2(g)+3CO(g) rarr N_2O(g) +3CO_2(g)`

Options

A. 836
B. 1460
C. `-836`
D. `-1460`

Answer

To calculate the enthalpy change (ΔH) for the reaction:

\[ 2NO_2(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g) \]

we will use the standard enthalpy of formation (Δ_f H°) values provided for each compound:

- Δ_f H° of CO_2(g) = -393 kJ/mol
- Δ_f H° of CO(g) = -110 kJ/mol
- Δ_f H° of N_2O(g) = 81 kJ/mol
- Δ_f H° of NO_2(g) = 34 kJ/mol

### Step-by-Step Solution:

1. **Identify the reaction components**:
- Reactants: 2 moles of NO_2 and 3 moles of CO
- Products: 1 mole of N_2O and 3 moles of CO_2

2. **Write the formula for ΔH of the reaction**:
\[ \Delta H_{reaction} = \sum (\Delta_f H^\circ \text{ of products}) - \sum (\Delta_f H^\circ \text{ of reactants}) \]

3. **Calculate the sum of Δ_f H° for the products**:
- For N_2O: \( 1 \times 81 \, \text{kJ/mol} = 81 \, \text{kJ} \)
- For CO_2: \( 3 \times (-393 \, \text{kJ/mol}) = -1179 \, \text{kJ} \)
- Total for products:
\[ 81 + (-1179) = -1098 \, \text{kJ} \]

4. **Calculate the sum of Δ_f H° for the reactants**:
- For NO_2: \( 2 \times 34 \, \text{kJ/mol} = 68 \, \text{kJ} \)
- For CO: \( 3 \times (-110 \, \text{kJ/mol}) = -330 \, \text{kJ} \)
- Total for reactants:
\[ 68 + (-330) = -262 \, \text{kJ} \]

5. **Combine the totals to find ΔH for the reaction**:
\[ \Delta H_{reaction} = (-1098) - (-262) = -1098 + 262 = -836 \, \text{kJ} \]

### Final Answer:
The enthalpy change (ΔH) for the reaction is:
\[ \Delta H = -836 \, \text{kJ} \]