Question

How do you calculate enthalpy change of combustion?

Answer 1

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

Explanation:

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