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.
First, we need to write the balanced chemical equation for the reaction of ethylene (C2H4) with hydrogen (H2) to form ethane (C2H6):
C2H4(g)+H2(g)→C2H6(g)
Next, we need to determine the enthalpy change for this reaction. The enthalpy change can be represented by the following equation:
ΔH = ΣnΔHf(products) - ΣmΔHf(reactants)
Where n and m represent the stoichiometric coefficients of the products and reactants, respectively, and ΔHf is the standard enthalpy of formation for each compound.
We can look up the standard enthalpy of formation values for ethylene, hydrogen, and ethane from a table. For this calculation, we need to use the following values:
ΔHf(C2H4) = 52.3 kJ/mol
ΔHf(H2) = 0 kJ/mol
ΔHf(C2H6) = -84.7 kJ/mol
Now we can plug these values into the equation to calculate the enthalpy change for the reaction:
ΔH = (1*(-84.7 kJ/mol)) - [(1*52.3 kJ/mol) + (1*0 kJ/mol)]
ΔH = -84.7 kJ/mol - 52.3 kJ/mol
ΔH = -137.0 kJ/mol
Now we need to determine the number of moles of ethylene in 52.0 g. The molar mass of ethylene (C2H4) is 28.05 g/mol, so we can calculate the number of moles:
n = (52.0 g)/(28.05 g/mol) ≈ 1.853 mol
Finally, we can calculate the energy released during the reaction of 52.0 g of ethylene by multiplying the enthalpy change by the number of moles of ethylene:
Energy released = ΔH * n
Energy released = -137.0 kJ/mol * 1.853 mol ≈ -253.7 kJ
Therefore, the energy released during the reaction of 52.0 g of ethylene is approximately -253.7 kJ.