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.

Answer 1

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.