Calculate the standard enthalpy of formation of ethylene, `C_(2)H_(4)(g)` from the following thermochemical equation :`C_(2)H_(4)(g) + 3O_(2)(g) to 2CO_(2)(g) + 2H_(2)O(g)`
`Delta_(r)H^(@) = -1323 kJ`
Given the standard enthalpy of formation of `CO_(2)(g)` and `H_(2)O(g)` are - 393.5 and -249 kJ `mol^(-1)` respectively. (Answer in KJ mol"^(-1))

Explanation

To calculate the standard enthalpy of formation of ethylene (C₂H₄), we can use the provided thermochemical equation and the standard enthalpies of formation for CO₂ and H₂O.

### Step-by-Step Solution:

1. **Write down the given thermochemical equation:**
\[ C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(g) \]
The enthalpy change for this reaction is given as:
\[ \Delta_rH^\circ = -1323 \, \text{kJ} \]

2. **Identify the standard enthalpies of formation:**
- For CO₂(g): \(\Delta_fH^\circ = -393.5 \, \text{kJ/mol}\)
- For H₂O(g): \(\Delta_fH^\circ = -249 \, \text{kJ/mol}\)
- For O₂(g): \(\Delta_fH^\circ = 0 \, \text{kJ/mol}\) (elements in their standard state have an enthalpy of formation of zero)

3. **Set up the equation for the enthalpy change of the reaction:**
The enthalpy change of the reaction can be expressed in terms of the enthalpies of formation of the products and reactants:
\[ \Delta_rH^\circ = \left( \sum \Delta_fH^\circ \text{(products)} \right) - \left( \sum \Delta_fH^\circ \text{(reactants)} \right) \]

4. **Calculate the total enthalpy of formation for the products:**
- For 2 moles of CO₂:
\[ 2 \times (-393.5) = -787 \, \text{kJ} \]
- For 2 moles of H₂O:
\[ 2 \times (-249) = -498 \, \text{kJ} \]
- Total for products:
\[ -787 + (-498) = -1285 \, \text{kJ} \]

5. **Set up the equation:**
Now, substituting the values into the enthalpy change equation:
\[ -1323 = (-1285) - \Delta_fH^\circ(C_2H_4) \]

6. **Rearranging to find the enthalpy of formation of C₂H₄:**
\[ \Delta_fH^\circ(C_2H_4) = -1285 + 1323 \]
\[ \Delta_fH^\circ(C_2H_4) = 38 \, \text{kJ/mol} \]

### Final Answer:
The standard enthalpy of formation of ethylene (C₂H₄) is:
\[ \Delta_fH^\circ(C_2H_4) = 38 \, \text{kJ/mol} \]