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QUESTION

How do you calculate enthalpy change of combustion?

You usually calculate the change of combustion from enthalpies of formation.

The standard enthalpy of combustion is ΔH_"c"^°.

It is the heat evolved when 1 mol of a substance burns completely in oxygen at standard conditions. For example,

"C"_2"H"_2"(g)" + 5/2"O"_2"(g)" → "2CO"_2"(g)" + "H"_2"O(l)"

You calculate ΔH_"c"^° from standard enthalpies of formation:

ΔH_"c"^o = ∑ΔH_"f"^°"(p)" - ∑ΔH_"f"^°"(r)"

where "p" stands for "products" and "r" stands for "reactants".

For each product, you multiply its ΔH_"f"^° by its coefficient in the balanced equation and add them together.

Do the same for the reactants. Subtract the reactant sum from the product sum.

EXAMPLE:

Use the following enthalpies of formation to calculate the standard enthalpy of combustion of acetylene, "C"_2"H"_2.

"C"_2"H"_2"(g)" + 5/2"O"_2"(g)" → "2CO"_2"(g)" + "H"_2"O(l)"

DeltaH_("C"_2"H"_2"(g)")^o = "226.73 kJ/mol"; DeltaH_("CO"_2"(g)")^o = "-393.5 kJ/mol";

DeltaH_("H"_2"O(l)")^o = "-285.8 kJ/mol"

Solution:

"C"_2"H"_2"(g)" + 5/2"O"_2"(g)" → "2CO"_2"(g)" + "H"_2"O(l)"

ΔH_"c"^o = ∑ΔH_"f"^°"(p)" - ∑ΔH_"f"^°"(r)"

"[2 × (-393.5) + (-295.8)] – [226.7 + 0] kJ" = "-1082.8 - 226.7" =

"-1309.5 kJ"

The heat of combustion of acetylene is -1309.5 kJ/mol.

Here is a video that discusses how to calculate the enthalpy change when 0.13 g of butane is burned.

Video from: Noel Pauller