Calculate the enthalpy of reaction `(Delta H ^(@))`when ammonia is oxidized `:`
`4NH_(3) (g) + 5O_(2)(g) rarr 6H_(2)O(g) + 4NO(g)`
Standard enthalpies of formation `(Delta_(f) H^(@))` at `25^(@)C`for`NH_(3)(g),H_(2)O(g) ` and `NO(g)` are - 46.2, -241.8 and `+ 90.4 kJ //` mole respectively.

Answer

To calculate the enthalpy of the reaction \( \Delta H^\circ \) for the oxidation of ammonia, we can use the standard enthalpies of formation of the reactants and products. The reaction is given as:

\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 6 \text{H}_2\text{O}(g) + 4 \text{NO}(g) \]

### Step 1: Write the formula for the enthalpy of reaction

The enthalpy change of the reaction can be calculated using the formula:

\[ \Delta H^\circ_{reaction} = \Sigma \Delta H^\circ_f \text{(products)} - \Sigma \Delta H^\circ_f \text{(reactants)} \]

### Step 2: Identify the standard enthalpies of formation

From the problem, we have the following standard enthalpies of formation at \( 25^\circ C \):

- \( \Delta H^\circ_f \text{(NH}_3(g)) = -46.2 \, \text{kJ/mol} \)
- \( \Delta H^\circ_f \text{(H}_2\text{O}(g)) = -241.8 \, \text{kJ/mol} \)
- \( \Delta H^\circ_f \text{(NO}(g)) = +90.4 \, \text{kJ/mol} \)
- \( \Delta H^\circ_f \text{(O}_2(g)) = 0 \, \text{kJ/mol} \) (as it is in its elemental form)

### Step 3: Calculate the total enthalpy of formation for products

For the products, we have:

- For \( 6 \text{H}_2\text{O}(g) \):
\[ 6 \times (-241.8) = -1450.8 \, \text{kJ} \]

- For \( 4 \text{NO}(g) \):
\[ 4 \times (+90.4) = +361.6 \, \text{kJ} \]

Adding these together gives:

\[ \Delta H^\circ_f \text{(products)} = -1450.8 + 361.6 = -1089.2 \, \text{kJ} \]

### Step 4: Calculate the total enthalpy of formation for reactants

For the reactants, we have:

- For \( 4 \text{NH}_3(g) \):
\[ 4 \times (-46.2) = -184.8 \, \text{kJ} \]

- For \( 5 \text{O}_2(g) \):
\[ 5 \times (0) = 0 \, \text{kJ} \]

Adding these together gives:

\[ \Delta H^\circ_f \text{(reactants)} = -184.8 + 0 = -184.8 \, \text{kJ} \]

### Step 5: Calculate the enthalpy of the reaction

Now we can substitute these values into the enthalpy change formula:

\[ \Delta H^\circ_{reaction} = (-1089.2) - (-184.8) \]

\[ \Delta H^\circ_{reaction} = -1089.2 + 184.8 = -904.4 \, \text{kJ} \]

### Final Answer

Thus, the enthalpy of the reaction \( \Delta H^\circ \) when ammonia is oxidized is:

\[ \Delta H^\circ = -904.4 \, \text{kJ} \]

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