Hess's law relates the enthalpy change of a reaction to the standard enthalpies of formation (\( \Delta H_f^\circ \)) of reactants and products:
\[\Delta H_{\text{reaction}}^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})\]
Given:
- \( \Delta H_{\text{reaction}}^\circ = -1410 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ (\text{CO}_2) = -393.5 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ (\text{H}_2\text{O}) = -286 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ (\text{O}_2) = 0 \, \text{kJ/mol} \) (element in standard state).
We seek \( \Delta H_f^\circ (\text{C}_2\text{H}_4) \).
For the reaction:
\[\text{C}_2\text{H}_4 (\text{g}) + 3\text{O}_2 (\text{g}) \rightarrow 2\text{CO}_2 (\text{g}) + 2\text{H}_2\text{O} (\text{l})\]
The enthalpy change equation is:
\[\Delta H_{\text{reaction}}^\circ = [2 \Delta H_f^\circ (\text{CO}_2) + 2 \Delta H_f^\circ (\text{H}_2\text{O})] - [\Delta H_f^\circ (\text{C}_2\text{H}_4) + 3 \Delta H_f^\circ (\text{O}_2)]\]
Substituting known values:
\[-1410 = [2 \times (-393.5) + 2 \times (-286)] - [\Delta H_f^\circ (\text{C}_2\text{H}_4) + 3 \times 0]\]
Calculating products' enthalpies:
\[2 \times (-393.5) = -787 \, \text{kJ/mol}\]
\[2 \times (-286) = -572 \, \text{kJ/mol}\]
\[-787 + (-572) = -1359 \, \text{kJ/mol}\]
Thus:
\[-1410 = -1359 - \Delta H_f^\circ (\text{C}_2\text{H}_4)\]
Solving for \( \Delta H_f^\circ (\text{C}_2\text{H}_4) \):
\[\Delta H_f^\circ (\text{C}_2\text{H}_4) = -1359 + 1410 = +51 \, \text{kJ/mol} \approx +52 \, \text{kJ/mol}\]
The standard enthalpy of formation of \( \text{C}_2\text{H}_4 (\text{g}) \) is \( +52 \, \text{kJ/mol} \).