Question

The enthalpy change for the reaction \( \text{C}_2\text{H}_4 (\text{g}) + 3\text{O}_2 (\text{g}) \rightarrow 2\text{CO}_2 (\text{g}) + 2\text{H}_2\text{O} (\text{l}) \) is \( -1410 \, \text{kJ/mol} \). If the standard enthalpies of formation of \( \text{CO}_2 (\text{g}) \) and \( \text{H}_2\text{O} (\text{l}) \) are \( -393.5 \, \text{kJ/mol} \) and \( -286 \, \text{kJ/mol} \) respectively, what is the standard enthalpy of formation of \( \text{C}_2\text{H}_4 (\text{g}) \)?

• A. \( +52 \, \text{kJ/mol} \)
• B. \( -52 \, \text{kJ/mol} \)
• C. \( +104 \, \text{kJ/mol} \)
• D. \( -104 \, \text{kJ/mol} \)

Hint:

When calculating enthalpy of formation using Hess’s law, ensure all stoichiometric coefficients from the balanced equation are accounted for. Elements in their standard state have \( \Delta H_f^\circ = 0 \).

Answer 1

The Correct Option is A

The enthalpy change for a reaction is related to the standard enthalpies of formation (\( \Delta H_f^\circ \)) of the reactants and products by Hess’s law:
\[ \Delta H_{\text{reaction}}^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \] Given:
- Enthalpy change of the reaction: \( \Delta H_{\text{reaction}}^\circ = -1410 \, \text{kJ/mol} \),
- Standard enthalpy of formation of \( \text{CO}_2 (\text{g}) \): \( \Delta H_f^\circ (\text{CO}_2) = -393.5 \, \text{kJ/mol} \),
- Standard enthalpy of formation of \( \text{H}_2\text{O} (\text{l}) \): \( \Delta H_f^\circ (\text{H}_2\text{O}) = -286 \, \text{kJ/mol} \),
- Standard enthalpy of formation of \( \text{O}_2 (\text{g}) \): \( \Delta H_f^\circ (\text{O}_2) = 0 \, \text{kJ/mol} \) (as it is an element in its standard state).
We need to find the standard enthalpy of formation of \( \text{C}_2\text{H}_4 (\text{g}) \), denoted as \( \Delta H_f^\circ (\text{C}_2\text{H}_4) \).
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 is:
\[ \Delta H_{\text{reaction}}^\circ = [2 \Delta H_f^\circ (\text{CO}_2) + 2 \Delta H_f^\circ (\text{H}_2\text{O})] - [\Delta H_f^\circ (\text{C}_2\text{H}_4) + 3 \Delta H_f^\circ (\text{O}_2)] \] Substitute the known values:
\[ -1410 = [2 \times (-393.5) + 2 \times (-286)] - [\Delta H_f^\circ (\text{C}_2\text{H}_4) + 3 \times 0] \] Calculate the products’ enthalpies:
\[ 2 \times (-393.5) = -787 \, \text{kJ/mol}, \quad 2 \times (-286) = -572 \, \text{kJ/mol} \] \[ -787 + (-572) = -1359 \, \text{kJ/mol} \] So:
\[ -1410 = -1359 - \Delta H_f^\circ (\text{C}_2\text{H}_4) \] Solve for \( \Delta H_f^\circ (\text{C}_2\text{H}_4) \):
\[ \Delta H_f^\circ (\text{C}_2\text{H}_4) = -1359 + 1410 = +51 \, \text{kJ/mol} \approx +52 \, \text{kJ/mol} \] Thus, the standard enthalpy of formation of \( \text{C}_2\text{H}_4 (\text{g}) \) is \( +52 \, \text{kJ/mol} \).