Question 10.9: Given the following information, calculate the standard enth...
Given the following information, calculate the standard enthalpy of formation of acetylene (\text{C}_2\text{H}_2) from its constituent elements:
\text{C(graphite)} + \text{O}_2(\text{g}) → \text{CO}_2(\text{g}) ∆H^\circ_\text{rxn} = –393.5 \text{kJ/mol} (1)
\text{H}_2(\text{g}) +\frac{1}{2}\text{O}_2(\text{g}) → \text{H}_2\text{O}(l) ∆H^\circ_\text{rxn} = –285.8 \text{kJ/mol} (2)
2\text{C}_2\text{H}_2(\text{g}) + 5\text{O}_2(\text{g}) → 4\text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(l) ∆H^\circ_\text{rxn} = –2598.8 \text{kJ/mol} (3)
Strategy Arrange the equations that are provided so that they will sum to the desired equation. This may require reversing or multiplying one or more of the equations. For any such change, the corresponding change must also be made to the ∆H^\circ_\text{rxn} value.
Setup The desired equation, corresponding to the standard enthalpy of formation of acetylene, is
2\text{C(graphite)} + \text{H}_2(\text{g}) → \text{C}_2\text{H}_2(\text{g})
We multiply Equation (1) and its ∆H^\circ_\text{rxn} value by 2:
2\text{C(graphite)} + 2\text{O}_2(\text{g}) → 2\text{CO}_2(\text{g}) ∆H^\circ_\text{rxn} = –787.0 \text{kJ/mol}We include Equation (2) and its ∆H^\circ_\text{rxn} value as is:
\text{H}_2(\text{g}) +\frac{1}{2} \text{O}_2(\text{g}) → \text{H}_2\text{O}(l) ∆H^\circ_\text{rxn} = –285.8 \text{kJ/mol}We reverse Equation (3) and divide it by 2 (i.e., multiply through by 1/2):
2\text{CO}_2(\text{g}) + \text{H}_2\text{O}(l) → \text{C}_2\text{H}_2(\text{g}) +\frac{5}{2} \text{CO}_2(\text{g}) ∆H^\circ_\text{rxn} = +1299.4 \text{kJ/mol}^*Summing the resulting equations and the corresponding ∆H^\circ_\text{rxn} values:
2\text{C(graphite)} + \cancel{2\text{O}_2 (\text{g})} → \cancel{2\text{CO}_2 (\text{g})} ∆H^\circ_\text{rxn} = –787.0 \text{kJ/mol} \text{H}_2(\text{g}) +\cancel{\frac{1}{2} \text{O}_2(\text{g})} → \cancel{\text{H}_2\text{O}(l)} ∆H^\circ_\text{rxn} = –285.8 \text{kJ/mol} \underline{\cancel{2\text{CO}_2(\text{g})} + \cancel{\text{H}_2\text{O}(l)} → \text{C}_2\text{H}_2(\text{g}) +\cancel{\frac{5}{2} \text{CO}_2(\text{g})} ∆H^\circ_\text{rxn} = +1299.4 \text{kJ/mol}} 2\text{C(graphite)} + \text{H}_2(\text{g}) → \text{C}_2\text{H}_2(\text{g}) ∆H^\circ_\text{f} = +226.6 \text{kJ/mol}^*Student Annotation: The original ∆H^\circ_\text{rxn} value of Equation (3) has its sign reversed and it is divided by 2.