Question

Given the following data, what is the heat of formation of methane (CH4) gas?

I came up
with the chemical equation or methane gas
C + 2H2 -> CH4
though not sure it was needed. I know Hesses law applies here right?

Answer 1

I think you got it mate...

Explanation:

The heat of formation of a substance is the enthalpy associated with ONE mole of substance from its elements in their standard states under standard conditions...

And for...

`C(s) + 2H_2(g) rarr CH_4(g) + Delta_"rxn"`

`Delta_"rxn"="enthalpy of formation of methane"`...

Answer 2

Warning! Long Answer. `Δ_text(f)H = "-75 kJ"`

Explanation:

You gave us four equations, but we need only the first three:

`bb"(1)"color(white)(m) "CH"_4"(g)" + "2O"_2"(g)" → "CO"_2"(g)" + 2"H"_2"O"; DeltaH = "-803 kJ"`

`bb"(2)"color(white)(m)"H"_2"(g)" + "½O"_2"(g)" → "H"_2"O(g)"; ΔH = "-242 kJ"`

`bb"(3)"color(white)(m)"C(s)" + "O"_2"(g)" → "CO"_2"(g)"; ΔH = "-394 kJ"`

From these, you must devise the target equation:

`bb"(4)"color(white)(m)color(blue)("C(s)" + "2H"_2"(g)" rarr "CH"_4"(g)"; Δ_text(f)H^@ = ?)`

The target equation has `"C(s)"` on the left, so you rewrite equation (3).

`bb"(5)"color(white)(m)"C(s)" + "O"_2"(g)" → "CO"_2"(g)"; ΔH = "-394 kJ"`

Equation (5) has `"CO"_2"(g)"` on the right, and that is not in the target equation.

You need an equation with `"CO"_2"(g)"` on the left.

Reverse Equation (1).

When you reverse an equation, you reverse the sign of its `ΔH`.

`bb"(6)"color(white)(m) "CO"_2"(g)" +2"H"_2"O(g)"→ "CH"_4"(g)" + "2O"_2"(g)"; DeltaH = "+803 kJ"`

Equation (6) has `"2H"_2"O(g)"` on the left, and that is not in the target equation.

You need an equation with `2"H"_2"O"` on the right.

Double Equation (2).

When you double an equation, you double its `ΔH`.

`bb"(7)"color(white)(m)"2H"_2"(g)" + "O"_2"(g)" → "2H"_2"O(g)"; DeltaH = "-484 kJ"`

Now, you add equations (5), (6) and (7), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their `ΔH` values.

This gives us the target equation (8):
`color(white)(mmmmmmmmmmmmmmmmmmmmmmml)ul(ΔH^@"/kJ")`
`bb"(5)"color(white)(m)"C(s)" + color(red)(cancel(color(black)("O"_2"(g)"))) → color(red)(cancel(color(black)("CO"_2"(g)")));color(white)(mmmmmmmmll)"-394"`
`bb"(6)"color(white)(m) color(red)(cancel(color(black)("CO"_2"(g)"))) + 2"H"_2"O(g)" → "CH"_4"(g)" + color(red)(cancel(color(black)("2O"_2"(g)"))); +803"`

`bb"(7)"color(white)(m) ul("2H"_2"(g)" + color(red)(cancel(color(black)("O"_2"(g)"))) → color(red)(cancel(color(black)("2H"_2"O(g)")));color(white)(mmmmm))color(white)(mm)ul("-484")`
`bb"(8)"color(white)(m)color(blue)("C(s)" + "2H"_2"(g)" rarr "CH"_4"(g)"; color(white)(mmmmmmmmmml)"-75")`

`Δ_text(f)H = "-75 kJ·mol"^"-1"`