Enthalpy formation from enthalpy reaction

Question

A scientist measures the standard enthalpy change for the following reaction to be -208.5 kJ :

CO(g) + 3 H₂(g)CH₄(g) + H₂O(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CH₄(g) is   kJ/mol.

Answer, please. in kj/mol

Answer 1

You do not need extra data handed to you for this one. The other two enthalpies of formation are standard table values that appear in the appendix of every general chemistry text, so here is the problem worked end to end with them stated openly.

The rule: ΔH°_rxn = Σ ΔH°_f(products) - Σ ΔH°_f(reactants)

Reaction: CO(g) + 3 H₂(g) → CH₄(g) + H₂O(g),   ΔH°_rxn = -208.5 kJ

Table values: ΔH°_f[CO(g)] = -110.5 kJ/mol, ΔH°_f[H₂O(g)] = -241.8 kJ/mol, and ΔH°_f[H₂(g)] = 0.

That last one is not a lookup, it is a definition: the standard enthalpy of formation of any element in its standard state is exactly zero. H₂(g) is elemental hydrogen in its standard state, so the entire 3 H₂ term drops out no matter what coefficient sits in front of it.

Substituting, with x = ΔH°_f[CH₄(g)]:

-208.5 = [x + (-241.8)] - [(-110.5) + 3(0)]

-208.5 = x - 241.8 + 110.5

-208.5 = x - 131.3

x = -208.5 + 131.3 = -77.2 kJ/mol

Watch the sign on that bracket. Subtracting the reactant sum means subtracting a negative number, so -110.5 comes back in as +110.5. That single sign is the most common way this calculation goes wrong.

Two checks worth making. The value is negative, which it should be: methane is a stable molecule, so forming it from carbon and hydrogen releases heat. And the magnitude is modest, tens of kilojoules rather than hundreds, which is typical for a simple hydrocarbon.

One honest note. The accepted literature value for ΔH°_f[CH₄(g)] is about -74.6 kJ/mol, not -77.2. If you run the calculation backwards with -74.6 you get ΔH°_rxn = -205.9 kJ rather than the -208.5 you were given. The gap is real but small, and it comes from which table your course uses; different sources round CO and H₂O(g) slightly differently. Answer with the numbers your own appendix lists. If your table gives CO as -110.5 and H₂O(g) as -241.8, then -77.2 kJ/mol is what this problem wants, and the method above is unchanged whatever values you substitute.