Question

Question #01551

Answer 1

Reference states

Explanation:

The main difference is that if the water formed is in gaseous state then we lose the heat required to evaporate the water also from the combustion. This is the difference between Gross calorific value and Net calorific value. Net calorific value is the GCV-heat required to evaporate water from liquid state to gas state

Answer 2

The difference is 132.0 kJ/mol (3 times the molar enthalpy of vaporization of water).

Explanation:

You can calculate the enthalpy change of a reaction by using the
enthalpies of formation of reactants and products.

The formula is

`color(blue)(bar(ul(|color(white)(a/a)Δ_text(rxn)H° = Δ_text(f)H^@("products") - Δ_text(f)H^@("reactants")color(white)(a/a)|)))" "`

Forming liquid water

We have

`color(white)(mmmmmmmmm)"C"_2"H"_6"(g)" + "3.5O"_2"(g)" → "2CO"_2"(g)" + "3H"_2"O(l)"`
`Δ_text(f)H^°"/kJ·mol"^"-1": color(white)(m) "-84.667"color(white)(mmml)0color(white)(mmmm)"-393.513"color(white)(ml)"-285.830"`

`Δ_text(r)H^° = "[2×(-393.513) + 3×(-285.830)) - (-84.667)] kJ" = "-1559.849 kJ"`

Forming gaseous water

`color(white)(mmmmmmmmm)"C"_2"H"_6"(g)" + "3.5O"_2"(g)" → "2CO"_2"(g)" + "3H"_2"O(g)"`
`Δ_text(f)H^°"/kJ·mol"^"-1": color(white)(m) "-84.667"color(white)(mmml)0color(white)(mmmm)"-393.513"color(white)(ml)"-241.826"`

`Δ_text(r)H^° = "[2×(-393.513) + 3×(-241.826)) - (-84.667)] kJ" = "-1427.837 kJ"`

Calculate the enthalpy difference

`ΔΔH = "-1427.837 - (-1559.849) kJ" = "132.012 kJ"`

The only difference is that in one reaction we are forming 3 mol of liquid water, while in the other reaction we are forming 3 mol of gaseous water.

Hence, this difference represents the enthalpy of vaporization of 3 mol of water.

`color(white)(mmmmmmmmm)3"H"_2"O(l)" → 3 "H"_2"O(g)"`
`Δ_text(f)H^°"/kJ·mol"^"-1": color(white)(m) "-285.84"color(white)(mm)"-241.83"`

`Δ_text(rxn)H =[ "3×(-241.83) - 3×(-285.84)] kJ" = "132.012 kJ"`

`Δ_text(vap)H = "132.012 kJ"/"3 mol" = "44.00 kJ/mol"`