Given the following data:
C2H4 (g) + 3O2 (g) -> 2CO2 (g) + 2H2O (l) H = –1411.0 kJ
2C2H6 (g) + 7O2 (g) -> 4CO2 (g) + 6H2O (l) H = –3119.8 kJ
2H2 (g) + O2 (g) -> 2H2O (l) H = –571.7 kJ
Calculate H for the reaction
C2H4 (g) + H2 -> C2H6 (g)
First, we need to write out the balanced chemical equation for the reaction:
C2H4 (g) + H2 (g) -> C2H6 (g)
Now, we can use the following formula to calculate the enthalpy change for the reaction:
ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants)
where ΔHf is the standard enthalpy of formation for each compound. We can find these values in a table of standard enthalpies of formation.
ΔHf(C2H4) = 52.3 kJ/mol
ΔHf(H2) = 0 kJ/mol
ΔHf(C2H6) = -84.7 kJ/mol
Plugging these values into the formula, we get:
ΔHrxn = [ΔHf(C2H6) - ΔHf(C2H4) - ΔHf(H2)]
ΔHrxn = [-84.7 - 52.3 - 0] kJ/mol
ΔHrxn = -137.0 kJ/mol
Therefore, the enthalpy change for the reaction C2H4 (g) + H2 (g) -> C2H6 (g) is -137.0 kJ/mol.