The heat of combustion per mole for acetylene, C2H2(g), is -1299.5 kJ/mol. Assuming that the combustion products are CO2(g) and H2O(l), and given that the enthalpy of formation is -393.5 kJ/mol for CO2(g) and -285.8 kJ/mol for H2O(l), find the enthalpy of formation of C2H2(g).
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![\Delta H^{o}_{rxn} = \sum \Delta H_{products} - \sum \Delta H_{reactants}\\\Delta H^{o}_{rxn} = [2\Delta H^{o}_{f}(CO_{2}) + \Delta H^{o}_{f} (H_{2}O)] - [\Delta H^{o}_{f}(C_{2}H_{2}) + \frac{5}{2} \Delta H^{o}_{f} O_{2}]\\-1299.5 = 2(-393.5) + (-285.8) - \Delta H^{o}_{f} (C_{2}H_{2})\\\Delta H^{o}_{f} (C_{2}H_{2}) = 227.7 kJ/mol](/tpl/images/1323/2810/92eb3.png)
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The heat of combustion per mole for acetylene, C2H2(g), is -1299.5 kJ/mol. Assuming that the combust...
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