Problem: Three common hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation:HydrocarbonFormulaΔH˚f(kJ/mol)1,3-ButadieneC4H6(g)111.91-ButeneC4H8(g)1.2n-ButaneC4H10(g)–124.7For 1,3-Butadiene, calculate the fuel value in kJ/g.

FREE Expert Solution

We are asked to calculate the fuel value in kJ/g for 1,3-Butadiene


ΔH°rxn=ΔH°f, prod-ΔH°f, react


Note that we need to multiply each ΔH˚f by the stoichiometric coefficient since ΔH˚f is in kJ/mol. 

Also, note that ΔH˚f for elements in their standard state is 0.


Balanced reaction: 

C4H6(g) + 11/2 O2(g)→ 4 CO2(g) + 3 H2O(g) 


ΔH°rxn=ΔH°f, prod-ΔH°f, reactΔH°rxn=[4ΔH°f, CO2+3ΔH°f, H2O ]-[ΔH°f, C4H6+112ΔH°f,O2 ]ΔH°rxn=[4(393.5 kJ)+3(241.8 kJ )]-[(111.9 kJ)+112(0 kJ) ]

ΔH°rxn = -2411.3 kJ 


Molar mass C4H6(g) = 54.10 g/mol


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Problem Details
Three common hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation:
HydrocarbonFormulaΔH˚f(kJ/mol)
1,3-ButadieneC4H6(g)111.9
1-ButeneC4H8(g)1.2
n-ButaneC4H10(g)–124.7

For 1,3-Butadiene, calculate the fuel value in kJ/g.

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