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.
here is the reactant C2H4(g)+H2(g)→C2H6(g)
To find the energy released during the reaction of 52.0 g of ethylene (C2H4), we first need to balance the chemical equation:
C2H4(g) + H2(g) -> C2H6(g)
Next, we need to calculate the molar mass of each compound:
C2H4: 2 * (12.01 g/mol) + 4 * (1.01 g/mol) = 28.05 g/mol
H2: 2 * (1.01 g/mol) = 2.02 g/mol
C2H6: 2 * (12.01 g/mol) + 6 * (1.01 g/mol) = 30.07 g/mol
Now, we can calculate the molar enthalpy change for this reaction using the enthalpies of formation for each compound:
ΔH° = ΣnΔH°f(products) - ΣmΔH°f(reactants)
Using the values for enthalpies of formation from a table, we can calculate ΔH° for the reaction:
ΔH° = [2 * (-84.68 kJ/mol) + 6 * (-20.43 kJ/mol)] - [1 * (52.26 kJ/mol) + 1 * 0 kJ/mol]
ΔH° = [-169.36 kJ/mol + (-122.58 kJ/mol)] - [52.26 kJ/mol]
ΔH° = -169.36 kJ/mol - 122.58 kJ/mol - 52.26 kJ/mol
ΔH° = -344.20 kJ/mol
So, the enthalpy change for the reaction of 52.0 g of ethylene is -344.20 kJ/mol. To find the total energy released, we need to calculate the moles of ethylene in 52.0 g:
52.0 g / 28.05 g/mol = 1.852 mol
Finally, we can calculate the total energy released:
1.852 mol * -344.20 kJ/mol = -637.35 kJ
Therefore, the energy released during the reaction of 52.0 g of ethylene is -637.35 kJ.