Given #DeltaH_"combustion"^@# for butane is #-2657kJmol^-1#, what is #"heat of combustion"# for butane in #J*g^-1#?

Answer 1

#"Heat of combustion of butane"# #-=# #45.7xx10^3*J*g^-1#

We have data for the stoichiometric equation:

#C_4H_10(g)+ 13/2O_2(g)rarr 4CO_2 + 5H_2O# #DeltaH_"rxn"=-2657*kJ*mol^-1#

Note that I halved the equation because it makes the arithmetic a little bit easier. I also had to halve the enthalpy change accordingly.

And thus #"Heat of combustion of butane"# #-=# #(-2657xx10^3*kJ*cancel(mol^-1))/(58.12*g*cancel(mol^-1))=45.7*kJ*g^-1#.

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Answer 2

Owen Ambrose

The heat of combustion for butane is -2657 kJ/mol, which is equivalent to -2657000 J/mol. To convert this to J/g, divide by the molar mass of butane (58.12 g/mol). The heat of combustion for butane in J/g is approximately -45700 J/g.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Given #DeltaH_"combustion"^@# for butane is #-2657*kJ*mol^-1#, what is #"heat of combustion"# for butane in #J*g^-1#?

Given #DeltaH_"combustion"^@# for butane is #-2657kJmol^-1#, what is #"heat of combustion"# for butane in #J*g^-1#?