Standard Enthalpy of Formation and How to Find Reaction Enthalpy
In Class 11 thermodynamics, one question appears again and again: “Calculate the enthalpy change of this reaction using the standard enthalpies of formation given in the table.” The method is short, but students lose marks by missing a coefficient or a sign. This page shows the method step by step, with every number computed.
What is standard enthalpy of formation?
The standard enthalpy of formation (ΔfH°) of a compound is the enthalpy change when 1 mole of the compound is formed from its elements, each element in its most stable form (its standard state) at 1 bar pressure and a stated temperature, usually 298 K.
Two points decide everything:
- By definition, ΔfH° of an element in its standard state is zero. So O2(g), H2(g), N2(g), C(graphite) and Na(s) all have ΔfH° = 0.
- The state matters. H2O(l) and H2O(g) have different values, because turning liquid water into vapour needs energy.
The main formula
Here np and nr are the stoichiometric coefficients from the balanced equation. The idea is simple: imagine every reactant being broken down into elements (the reverse of formation, so signs flip), and then the elements being built up into the products. This is Hess’s law in its most useful form.
Values used in this page
The values below are typical textbook values at 298 K. Different books round them slightly, so always use the table given in your own question or book.
| Substance | ΔfH° (kJ/mol) |
|---|---|
| CH4(g) | −74.8 |
| CO2(g) | −393.5 |
| H2O(l) | −285.8 |
| NH3(g) | −46.1 |
| O2(g), N2(g), H2(g) | 0 (elements) |
Worked example 1: combustion of methane
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l). Find ΔrH°.
Products: 1 × (−393.5) + 2 × (−285.8) = −393.5 − 571.6 = −965.1 kJ.
Reactants: 1 × (−74.8) + 2 × (0) = −74.8 kJ.
ΔrH° = −965.1 − (−74.8) = −965.1 + 74.8 = −890.3 kJ per mole of CH4.
The negative sign means the reaction is exothermic: heat is released.
Worked example 2: heat from a given mass
How much heat is released on burning 8.0 g of methane completely? (C = 12.011, H = 1.008)
Molar mass of CH4 = 12.011 + 4 × 1.008 = 16.043 g/mol.
Moles = 8.0 / 16.043 = 0.4987 mol.
Heat released = 0.4987 × 890.3 = 444.0 kJ (about 444 kJ).
Always convert mass to moles first, because ΔrH° is given per mole of reaction as written.
Worked example 3: finding an unknown enthalpy of formation
The enthalpy of combustion of liquid ethanol, C2H5OH(l), is −1367 kJ/mol. Find its standard enthalpy of formation.
Equation: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l).
Use the formula and keep the unknown x = ΔfH°(ethanol):
−1367 = [2(−393.5) + 3(−285.8)] − [x + 0]
−1367 = (−787.0 − 857.4) − x = −1644.4 − x
x = −1644.4 + 1367 = −277.4 kJ/mol.
Check by putting x back: −1644.4 − (−277.4) = −1367.0. Correct.
Worked example 4: the “per mole” trap
For N2(g) + 3 H2(g) → 2 NH3(g), find ΔrH° and the heat released when 5.1 g of NH3 forms (N = 14.007).
ΔrH° = 2 × (−46.1) − (0 + 0) = −92.2 kJ for the equation as written (that is, for 2 mol NH3).
Molar mass of NH3 = 14.007 + 3 × 1.008 = 17.031 g/mol. Moles = 5.1 / 17.031 = 0.2995 mol.
Heat = 0.2995 × 46.1 = 13.8 kJ released.
Notice that −92.2 kJ belongs to 2 mol, but ΔfH° of NH3 itself is −46.1 kJ per mole. Do not mix these.
Common mistakes
- Reversing the subtraction. It is products minus reactants, never reactants minus products.
- Forgetting the coefficient. Water appears twice in the methane equation, so its value must be multiplied by 2.
- Putting a non-zero value for an element such as O2(g). It is zero by definition.
- Using the wrong state. Using H2O(g) value when the equation says H2O(l) changes the answer by about 88 kJ for 2 mol of water.
- Using an unbalanced equation. Balance first, then calculate.
- Forgetting the unit. The answer is in kJ for the reaction as written, not kJ per gram.
Exam relevance
| Question type | What to do |
|---|---|
| ΔH of a reaction from a table | Products minus reactants, with coefficients |
| Unknown ΔfH° | Put x in the formula and solve |
| Heat for a given mass | Mass → moles → multiply by ΔH per mole |
| Exothermic or endothermic | Negative = exothermic, positive = endothermic |
| Why elements are zero | Formation of an element from itself involves no change |
Your board or entrance exam may give a slightly different table of values. Use the table printed in your question paper, and read the current official syllabus for what is required.
Use the calculator suite to check the mole and mass arithmetic in problems like these, then keep the enthalpy steps in your own notebook.
Open the Chemistry Calculator Suite →Reaction mechanisms stick only with practice: attempt previous-year CSIR-NET questions and a timed quiz.
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