Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data:

C2H4(g) + 3O2(g) --> 2CO2(g) + 2H2O(l) H°rxn = –1411 kJ

Given that:

H°f[CO2(g)] = –393.5 kJ/mol

H°f[H2O(l)] = –285.8 kJ/mol

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The heat of reaction of a particular chemical reaction can be expressed as:

`H_(reaction) = sum[n Delta H^0_(f) p r oducts] -sum[n Delta H^0_(f) react a nts ] `

Where n = number of moles in the balanced chemical equation

Sum of heat of formation of the products `= [2 mol CO_2* -393.5 (kJ)/(mol)] + [2 mol H_(2)O * -285.8 (kJ)/(mol)] =-1358.6 kJ`

Sum of heat of formation of reactants `= Delta H^0_(f) _(ethyl e n e) + 0`

note: Oxygen has a `Delta H^(0)_(f)` of 0

Substituting these values to the first equation:

`-1411 kJ = -1358.6 kJ - Delta H^0_(f) _(ethyl en e) `

`Delta H^0_(f) _(ethyl en e) = -1358.6 kJ - (-1411kJ) `

`Delta H^0_(f) _(ethyl en e) = + 52.4 kJ -> answer`

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