What you'll be able to do: Combine given thermochemical equations to obtain the enthalpy change of a target reaction.
Best after: Enthalpy of Reaction and Thermochemical Equations
Some enthalpies cannot be measured directly, because the reaction is too slow, too messy, or produces side products. Hess's law lets you build the answer from reactions that can be measured.
These are recommended, not required. You can start this lesson at any time.
If a reaction can be written as the sum of two or more steps, its enthalpy change is the sum of the enthalpy changes of those steps, regardless of whether the steps represent what physically happens.
ΔH(overall) = ΔH₁ + ΔH₂ + ΔH₃ + ...
Start from the target equation, not from the given ones. Take one species at a time, find the given equation containing it, and reverse or scale that equation so the species appears on the correct side with the correct coefficient. Only then add everything up.
Treat each given equation as an algebraic vector of species. A positive multiplier scales it as written; a negative multiplier reverses it and scales its enthalpy by the same signed factor. Begin with a target species occurring in only one supplied equation, then resolve remaining species. Do not adjust the enthalpy independently to obtain a plausible final sign. The net chemical equation is the constraint that determines the enthalpy combination.
After manipulation, any species appearing on the left of one equation and the right of another in equal amounts cancels. A leftover species that should not be in the target is a signal that a coefficient is wrong.
For target C(graphite) + 1/2 O₂(g) → CO(g), combine:
| Manipulated equation | ΔH / kJ |
|---|---|
| C(graphite) + O₂(g) → CO₂(g) | −393.5 |
| CO₂(g) → CO(g) + 1/2 O₂(g) | +283.0 |
CO₂ cancels because it has the same state and coefficient on opposite sides. One O₂ on the left minus one-half on the right leaves one-half on the left. The target is obtained and ΔH = −110.5 kJ. A species need not vanish completely if the target contains a net amount of it; compare the remainder with the target, not with a rule that every repeated species must disappear.
Enthalpies of combustion are easy to measure, so a formation reaction that cannot be run directly is often built from combustion equations. Reverse the combustion of the target compound and add the combustions of its elements.
Combustion cycles work because the target reactants and products can be taken to common combustion products. Suppose C + O₂ → CO₂ has −393.5 kJ/mol, H₂ + 1/2 O₂ → H₂O(l) has −285.8 kJ/mol, and CH₄ + 2 O₂ → CO₂ + 2 H₂O(l) has −890.5 kJ/mol. Add the first, twice the second, and the reverse of methane combustion. CO₂, water and oxygen cancel to give C + 2 H₂ → CH₄, with −393.5 − 571.6 + 890.5 = −74.6 kJ/mol. This is methane's formation enthalpy for the specified phases.
Nothing in Hess's law claims the steps really happen. The only requirement is that the manipulated equations add to the target, because enthalpy depends on the endpoints alone.
The invented route need not be a feasible laboratory mechanism. It must, however, use thermodynamic data referring to compatible temperatures, pressures and phases. A reaction through gaseous water cannot silently cancel against one producing liquid water. The unmatched phase signals a missing vaporization or condensation step, not permission to ignore its energy.
The same state-function logic works even when the given equations are formation reactions rather than a ready-made path. Sum product formation enthalpies and subtract reactant formation enthalpies, with coefficients and phases intact.
This is Hess's law expressed as a reusable shortcut, and its result should match a route built by reversing and adding individual equations.
The enthalpy change of an overall reaction is the sum of the enthalpy changes of any set of steps that sum to it.
Apply reversal and scaling to each equation and its ΔH before summing.
Species that appear on opposite sides in equal amounts drop out of the summed equation.
The summed equation must match the target exactly, coefficients and states included.
The route used is a bookkeeping device, not a proposed mechanism.
ΔH(target) = sum of manipulated step enthalpies
Find ΔH for C(s) + 1/2 O₂(g) → CO(g) given: (1) C(s) + O₂(g) → CO₂(g), ΔH = -393.5 kJ and (2) CO(g) + 1/2 O₂(g) → CO₂(g), ΔH = -283.0 kJ.
Try it first: Ask where CO has to end up in the target, then look at where it sits in the given equations.
0 of 4 steps revealed.
Find ΔH for N₂(g) + 2 O₂(g) → 2 NO₂(g) given: (1) N₂(g) + O₂(g) → 2 NO(g), ΔH = +180.5 kJ and (2) 2 NO(g) + O₂(g) → 2 NO₂(g), ΔH = -114.1 kJ.
0 of 4 steps revealed.
A student combines two equations and finds that O₂ does not cancel, leaving 1/2 O₂ on the wrong side. What went wrong and what should be checked?
0 of 3 steps revealed.
Estimate ΔH for CaCO₃(s) → CaO(s) + CO₂(g) from ΔHf values -1207, -635, and -393.5 kJ/mol, respectively.
Try it first: Subtract the reactant formation enthalpy from the sum of products.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Reversing an equation but leaving the ΔH sign unchanged. | The energy flow reverses with the reaction. | Negate ΔH the moment the arrow is flipped. |
| Scaling the equation but not the ΔH. | Enthalpy is an extensive quantity tied to the coefficients. | Apply the same multiplier to both. |
| Assuming the invented steps must be the real mechanism. | Hess law needs only that the equations sum correctly. | Judge the route by the summed equation, not by plausibility. |
| Quoting an answer without checking that intermediates cancelled. | An uncancelled species means the target was never reproduced. | Write out the sum and compare it with the target line by line. |
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After the lesson · Question bank
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Because enthalpy is a state function, the enthalpy change of an overall reaction equals the sum of the enthalpy changes of any set of steps that add up to it. The working method is mechanical: look at the target equation, decide where each species must end up, reverse or scale each given equation to put it there, then add the manipulated ΔH values. Species that appear on both sides cancel, and if anything fails to cancel, one of the manipulations is wrong.
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