The question asks about the energy released during the reaction of 52.0 g of ethylene. To find the energy released, we’ll need to calculate the enthalpy change for this reaction.
To calculate the energy released during the reaction of 52.0 g of ethylene (C2H4), we first need to write the balanced chemical equation for the combustion of ethylene:
C2H4 + 3O2 -> 2CO2 + 2H2O
Next, we need to calculate the molar mass of C2H4:
C: 12.01 g/mol
H: 1.01 g/mol
Molar mass of C2H4 = 2(12.01 g/mol) + 4(1.01 g/mol) = 28.05 g/mol
Now, we need to calculate the moles of ethylene in 52.0 g:
Moles = 52.0 g / 28.05 g/mol = 1.85 mol
Next, we need to use the enthalpy of formation values to determine the change in enthalpy for the reaction. The enthalpy of formation values are:
ΔHf(C2H4) = 52.5 kJ/mol
ΔHf(CO2) = -393.5 kJ/mol
ΔHf(H2O) = -285.8 kJ/mol
Using these values, we can calculate the change in enthalpy for the reaction:
ΔH = ΣΔHf(products) - ΣΔHf(reactants)
ΔH = [2(-393.5 kJ/mol) + 2(-285.8 kJ/mol)] - [1(52.5 kJ/mol) + 3(0 kJ/mol)]
ΔH = -1278.2 kJ/mol
Now, we can calculate the energy released during the reaction of 52.0 g of ethylene:
Energy released = moles of C2H4 * ΔH
Energy released = 1.85 mol * -1278.2 kJ/mol = -2364.27 kJ
Therefore, the energy released during the reaction of 52.0 g of ethylene is -2364.27 kJ.