Acetylene burns in air according to the following

QUESTION:

Acetylene burns in air according to the following equation:

\(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H} 2 \mathrm{O}(\mathrm{g}) \quad \Delta \mathrm{H}^{\circ}{ }_{\mathrm{r} \times \mathrm{n}}=-1255.8 \mathrm{~kJ}\)

Given that \(\Delta H_{f}^{\circ} \text { of } \mathrm{CO}_{2}(\mathrm{~g})=-393.5 \mathrm{~kJ} / \mathrm{mol}\) and that \(\Delta H_{f}^{\circ} \text { of } \mathrm{H}_{2} \mathrm{O}(\mathrm{g})=-241.8 \mathrm{~kJ} / \mathrm{mol}\), what is \(\Delta H_{f}^{\circ} \text { of } \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})\)

Equation Transcription:

C2H2(g) + (g) 2CO2(g) + H2O(g)                H°rxn  = -1255.8 kJ

 of CO2(g) = -393.5 kJ/mol

 of H2O(g) = -241.8 kJ/mol

 of C2H2(g)

Text Transcription:

C_2H_2(g) + 5/2O_2(g) rightarrow 2CO_2(g) + H2O(g)                DeltaH°_rxn  = -1255.8 kJ

Delta H_f° of CO_2(g) = -393.5 kJ/mol

Delta H_f° of H_2O(g) = -241.8 kJ/mol

Delta H_f° of C_2H_2(g)