FE Chemical Practice Question Solution included after investigation
CASE E-066 · FE

The Methane Flame That Released 890 kJ per Mol

The reactant methane formation enthalpy was only -74.8 kJ/mol. The balanced combustion released nearly 890 kJ/mol.

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Case E-066 engineering failure visual
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Case Brief

SAME PRINCIPLES. HIGHER STANDARDS.
PROBLEM STATEMENT

Methane undergoes complete standard combustion: CH4 + 2O2 → CO2 + 2H2O(l). Use ΔHf° values: CH4=-74.8 kJ/mol, O2=0, CO2=-393.5 kJ/mol, H2O(l)=-285.8 kJ/mol. A calculation reports -74.8 kJ/mol as the reaction heat. Determine the correct standard reaction enthalpy.

Thermodynamics Chemical Design Verification Root Cause Analysis Safe Correction
E-066
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CASE CLOSED · ENGINEERING VERDICT

What actually failed — and what should change.

What failed

The fuel's formation enthalpy was mistaken for the reaction enthalpy.

What was missed

Product enthalpies and stoichiometric coefficients.

Corrected model

Use Hess's law: products minus reactants.

QUICK ANSWER

For CH4+2O2→CO2+2H2O(l), ΔHrxn°=[-393.5+2(-285.8)]-[-74.8]=-890.3 kJ/mol CH4. The negative sign indicates an exothermic reaction.

ENGINEERING TAKEAWAY

Use ΔHrxn=sum(νΔHf° products)-sum(νΔHf° reactants), including stoichiometric coefficients and signs.

CASE QUESTIONS

Questions this case should settle.

What is the standard heat of methane combustion to liquid water using these data?

About -890.3 kJ/mol CH4.

What formula uses standard heats of formation?

ΔHrxn°=ΣνΔHf°products-ΣνΔHf°reactants.

NEXT FE PRACTICE QUESTION

The Compressor That Heated the Gas to 446 K

Thermodynamics · 5–10 min

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