Find the standard enthalpy of formation of ethylene, C₂H₄(g), given the following data: heat of combustion of C₂H₄(g)= -1411 kJ/mol; $\Delta$H^$\circ$_f[CO₂(g)] = -393.5 kJ/mol; $\Delta$H^$\circ$_f[H₂O(l)] = -285.8 kJ/mol. A) 52 kJ/mol B) 87 kJ/mol C) 731 kJ/mol D) 1.41 * 10³ kJ/mol E) 2.77 * 10³ kJ/mol

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The standard enthalpy of formation of ethylene can be calculated using the given heat of combustion and the standard enthalpies of formation for CO2(g) and H2O(l). The combustion reaction for ethylene (C2H4) is: C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l). Using the given data: ΔH_combustion = -1411 kJ/mol, ΔH_f[CO2(g)] = -393.5 kJ/mol, ΔH_f[H2O(l)] = -285.8 kJ/mol. The total enthalpy of products = 2(-393.5) + 2(-285.8) = -1358.6 kJ. The enthalpy of reactants includes the enthalpy of formation of ethylene and O2 (which is 0 because it's an element in its standard state). Setting up the equation: -1411 = ΔH_f[C2H4] + (-1358.6), solving for ΔH_f[C2H4] gives approximately 52 kJ/mol.

Find the Standard Enthalpy of Formation of Ethylene, C2H4(g), Given Δ\DeltaΔ

Find the Standard Enthalpy of Formation of Ethylene, C₂H₄(g), Given $\Delta$