# 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 kJGiven that:H°f[CO2(g)] = –393.5 kJ/molH°f[H2O(l)] = –285.8 kJ/mol

The standard enthalpy of formation of ethylene, C2H4(g), given the above data, is H°f(C2H4(g) =52.4 kJ/mol.

The heat of reaction is the result of the breaking of bonds within the reactant species and the reforming of bonds to make the product species. The heat of formation (`Delta` H°f) is the energy required to form a compound in kilojoules per mole of substance. Heats of formation can be used to calculate the heat of reaction (`Delta` H°rxn) by subtracting the heats of formation of the reactants from the heats of formation of the products:

Heat of reaction = Heats of formation of products - Heats of formation of reactants

Heats of formation can be found in tables such as the one found here. Elements have a heat of formation of zero (they are the building blocks for compounds). The heats of reaction are reported per mole of substance, so we'll need to multiply the heat of formation by the number of moles of each substance in the reaction.

So, for our reaction, C2H4(g) + 3O2(g)  -->  2CO2(g) + 2H2O(l), we have:

H°rxn = 2 mol CO2(g) H°f(CO2(g)) + 2 mol H2O(g) H°f(H2O(g))

- [H°f(C2H4(g)) - 3 mol O2(g) H°f(O2(g))]

Substituting the values for H°rxn and H°f (remember H°f for O2 is zero):

-1411 kJ = 2 mol (-393.5 kJ/mol) + 2 mol (-285.8 kJ/mol) - [H°f(C2H4(g) + 0]

Rearranging to solve for H°f(C2H4(g):

H°f(C2H4(g) = 2 mol (-393.5 kJ/mol) + 2 mol (-285.8 kJ/mol) + 1141 kJ

H°f(C2H4(g) =52.4 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`