`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) rarrr N_2O(g) +3CO_2(g)`

Options

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

Explanation

To calculate the change in enthalpy (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 substance involved in the reaction.

### Step 1: Write down the Δ_f H^@ values
- 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 2: Identify the products and reactants
**Products:**
- N_2O(g)
- CO_2(g) (3 moles)

**Reactants:**
- NO_2(g) (2 moles)
- CO(g) (3 moles)

### Step 3: Calculate the total enthalpy of the products
Using the Δ_f H^@ values:
- For N_2O(g): 1 × Δ_f H^@ of N_2O = 1 × 81 kJ/mol = 81 kJ
- For CO_2(g): 3 × Δ_f H^@ of CO_2 = 3 × (-393 kJ/mol) = -1179 kJ

Total enthalpy of products:
\[ \text{Total Products} = 81 + (-1179) = 81 - 1179 = -1098 \text{ kJ} \]

### Step 4: Calculate the total enthalpy of the reactants
Using the Δ_f H^@ values:
- For NO_2(g): 2 × Δ_f H^@ of NO_2 = 2 × 34 kJ/mol = 68 kJ
- For CO(g): 3 × Δ_f H^@ of CO = 3 × (-110 kJ/mol) = -330 kJ

Total enthalpy of reactants:
\[ \text{Total Reactants} = 68 + (-330) = 68 - 330 = -262 \text{ kJ} \]

### Step 5: Calculate ΔH for the reaction
Using the formula:
\[ \Delta H = \text{Total Products} - \text{Total Reactants} \]
\[ \Delta H = (-1098) - (-262) \]
\[ \Delta H = -1098 + 262 = -836 \text{ kJ} \]

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

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