The enthalpy change doesn’t depend on the path
Climbing from the foot of a mountain to the top, you gain the same height whichever trail you take. Enthalpy works the same way: when a reaction goes from reactants (initial state) to products (final state), the enthalpy change depends only on those two states, not on the steps in between. This is Hess’s law盖斯定律:
If a reaction can be written as the sum of several reactions, its ΔH is the sum of their ΔH values.
So a reaction enthalpy that can’t be measured directly can be “assembled” from reactions that can (2.10.6). Only three rules are needed:
- reverse an equation, and ΔH changes sign;
- multiply an equation by a number (all coefficients together), and ΔH is multiplied by the same number;
- add equations, and their ΔH values add; a substance appearing on both sides cancels.
Hess’s law cycles
Drawing several states and the routes between them on an energy diagram gives a Hess’s law cycle. You can go from A to C directly or by way of B, and both routes have the same ΔH. Make one route the unknown and see how the other two give it:
Make one route the unknown and find it from the other two:
Combining equations
A more common task gives several known thermochemical equations and asks for a target equation. Compare with the target: decide which side each substance must be on and how many moles are needed, so you know whether to reverse each known equation and what to multiply it by; then add. Try the puzzle:
Example 1: A hydrogenation from enthalpies of combustion
Find ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g), given:
- C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l), ΔH₁ = −1410.9 kJ/mol
- H₂(g) + ½O₂(g) → H₂O(l), ΔH₂ = −285.8 kJ/mol
- C₂H₆(g) + 7/2O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH₃ = −1559.7 kJ/mol
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Compare each equation with the target. Target: C₂H₄(g) + H₂(g) → C₂H₆(g).
- C₂H₄ is on the left of the target: keep equation 1
- H₂ is on the left of the target: keep equation 2
- C₂H₆ is on the right of the target: reverse equation 3 and flip the sign of its ΔH
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Write the adjusted equations.
- ① C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l), ΔH = −1410.9
- ② H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8
- ③′ 2CO₂(g) + 3H₂O(l) → C₂H₆(g) + 7/2O₂(g), ΔH = +1559.7
All values in kJ/mol.
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Add them and check the cancellations.
- O₂: 3 + ½ = 7/2 on the left, 7/2 on the right, cancels
- CO₂: 2 mol on each side, cancels
- H₂O: 2 + 1 = 3 on the left, 3 on the right, cancels
What’s left is C₂H₄(g) + H₂(g) → C₂H₆(g), the target.
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Add the ΔH values the same way.
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Compare with other methods. Enthalpies of formation also give −137.0 kJ/mol; average bond enthalpies give an estimate of about −124 kJ/mol (U6.7), off because bond enthalpies are averages.
Example 2: Multiplying by a fraction
Given S(s) + O₂(g) → SO₂(g), ΔH₁ = −296.8 kJ/mol, and 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH₂ = −197.8 kJ/mol, find ΔH for S(s) + 3/2O₂(g) → SO₃(g).
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Compare with the target. The target S(s) + 3/2O₂(g) → SO₃(g) has 1 mol of SO₃, but the second equation forms 2 mol, so multiply it, and its ΔH, by ½ (in kJ/mol):
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Keep the first equation. S is on the left of the target with the right coefficient: S(s) + O₂(g) → SO₂(g), ΔH₁ = −296.8 kJ/mol.
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Add and check the cancellations. SO₂ is on the right of the first equation and the left of the second, 1 mol each, so it cancels; O₂ totals 1 + ½ = 3/2 mol, matching the target.
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Add the ΔH values.
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Check. This is the enthalpy of formation of SO₃(g): the reactants are elements in their standard states and 1 mol of SO₃ forms, matching the table value.
Hess’s law is everywhere
- ΔH from enthalpies of formation (U6.8): the route is reactants → elements → products.
- ΔH estimated from bond enthalpies (U6.7): the route is reactants → gaseous atoms → products.
- Reaction mechanisms (U5.7): the elementary steps add up to the overall reaction, so their ΔH values add up to the overall ΔH. On a multistep energy profile, the overall ΔH is just the height difference between the starting reactants and the final products, however many peaks and intermediates lie in between.
- Catalysts (U5.11): a catalyst changes the path but not the initial and final states, so ΔH doesn’t change.
Enrichment (the Born–Haber cycle): The lattice energy of an ionic solid can’t be measured directly, but a cycle can be built: Na(s) sublimes to Na(g), Na(g) loses an electron to become Na⁺(g), ½Cl₂(g) breaks apart to Cl(g), Cl(g) gains an electron to become Cl⁻(g), and finally the gaseous ions combine to form NaCl(s). These five steps add up to forming NaCl(s) from its elements, so their ΔH values add up to the enthalpy of formation of NaCl, and the lattice energy can be calculated from it.
- Target: C(s) + ½O₂(g) → CO(g); known: CO(g) + ½O₂(g) → CO₂(g)
CO is on the right in the target but on the left in the known equation.
- Target: 2SO₂(g) + O₂(g) → 2SO₃(g); known: SO₂(g) + ½O₂(g) → SO₃(g)
Every substance in the target is twice the known equation.
- Target: N₂(g) + 2O₂(g) → 2NO₂(g); known: N₂(g) + O₂(g) → 2NO(g)
N₂ is on the left in both, 1 mol each.
- Target: C₂H₄(g) + H₂(g) → C₂H₆(g); known: C₂H₆(g) + 7/2O₂(g) → 2CO₂(g) + 3H₂O(l)
C₂H₆ must be on the right.
- Target: C(s) + 2H₂(g) → CH₄(g); known: H₂(g) + ½O₂(g) → H₂O(l)
The target needs 2 mol of H₂ on the left.
- Target: S(s) + 3/2O₂(g) → SO₃(g); known: 2SO₂(g) + O₂(g) → 2SO₃(g)
The target needs only 1 mol of SO₃.
- Target: 2CO₂(g) → 2CO(g) + O₂(g); known: CO(g) + ½O₂(g) → CO₂(g)
CO₂ must be on the left, and 2 mol of it.
- Target: C(s, graphite) → C(s, diamond); known: C(s, diamond) + O₂(g) → CO₂(g)
Diamond must be on the right.
Common mistakes
Mistake 1: “Reversing an equation leaves ΔH unchanged.” Reversing keeps the size of ΔH but changes its sign. If the forward reaction is exothermic, the reverse reaction is endothermic.
Mistake 2: “To multiply an equation by 2, double only the substance you need.” The whole equation is multiplied: every coefficient and ΔH are multiplied by the same number.
Mistake 3: “More steps, or adding a catalyst, changes the heat released.” ΔH depends only on the initial and final states. A different path, including the new mechanism a catalyst provides, gives the same overall ΔH.
Quick check
Given N₂(g) + O₂(g) → 2NO(g), ΔH = +180.6 kJ/mol, and 2NO(g) + O₂(g) → 2NO₂(g), ΔH = −114.2 kJ/mol, what is ΔH for N₂(g) + 2O₂(g) → 2NO₂(g)?
Add the two equations directly; 2NO cancels: +180.6 + (−114.2) = +66.4 kJ/mol. C is the value for 1 mol of NO₂.
Given H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ/mol, and H₂O(l) → H₂O(g), ΔH = +44.0 kJ/mol, what is ΔH for 2H₂(g) + O₂(g) → 2H₂O(g)?
Adding the two equations gives H₂(g) + ½O₂(g) → H₂O(g), ΔH = −241.8; multiplying by 2 gives −483.6 kJ/mol.
The enthalpy change for A → B is ΔH₁, for B → C is ΔH₂, and for A → C is ΔH₃. Which relationship must be true?
A → B plus B → C is A → C. Going around the loop A → B → C → A, ΔH₁ + ΔH₂ − ΔH₃ = 0.
A reaction occurs in two steps: step 1 has ΔH = +35 kJ/mol and step 2 has ΔH = −110 kJ/mol. What is ΔH for the overall reaction?
The two steps add up to the overall reaction, so ΔH = +35 + (−110) = −75 kJ/mol.
When a catalyst is added, a reaction proceeds by a different mechanism. What happens to ΔH for the reaction?
A catalyst lowers the activation energy and changes the path, but ΔH depends only on the initial and final states.
The enthalpy change for turning graphite directly into diamond can’t be measured, yet Hess’s law gives it. Why?
Both C(graphite) → CO₂ and C(diamond) → CO₂ can be measured; subtracting the second from the first (reversing it and adding) gives graphite → diamond.
AP-style free response
Methanol can be made industrially from CO and H₂: CO(g) + 2H₂(g) → CH₃OH(l). Given:
- CO(g) + ½O₂(g) → CO₂(g), ΔH₁ = −283.0 kJ/mol
- H₂(g) + ½O₂(g) → H₂O(l), ΔH₂ = −285.8 kJ/mol
- CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(l), ΔH₃ = −726.4 kJ/mol
(a) Describe how to combine the three equations to get the target equation.
(b) Calculate ΔH for the synthesis of methanol.
(c) Given ΔH°f(CO, g) = −110.5 kJ/mol and ΔH°f(CH₃OH, l) = −238.7 kJ/mol, use enthalpies of formation to check your answer to (b).
(d) How much heat is released when 1.50 mol of CH₃OH(l) forms?
(e) If gaseous methanol, CH₃OH(g), formed instead, would ΔH be more negative or less negative? Justify your answer.
Model answer and scoring
(a) Use equation 1 as written, multiply equation 2 by 2, reverse equation 3, and add the three. CO₂, H₂O, and O₂ (½ + 1 = 3/2 mol on the left and 3/2 mol on the right) all cancel.
(b) ΔH = −283.0 + 2 × (−285.8) + 726.4 = −128.2 kJ/mol.
(c) ΔH = −238.7 − [(−110.5) + 2 × 0] = −128.2 kJ/mol, the same as (b).
(d) q = 1.50 × (−128.2) = −192 kJ, so 192 kJ of heat is released (192.3 kJ).
(e) Less negative. Vaporizing CH₃OH(l) to CH₃OH(g) absorbs heat, so less heat is released when gaseous methanol forms.
Scoring: (a) as written, × 2, reversed; (b) value and sign; (c) the enthalpy of formation formula; (d) conversion and “released”; (e) the judgment, with vaporization absorbing heat as the reason.
Summary
- The enthalpy change depends only on the initial and final states, not on the path (Hess’s law).
- Combining equations: reversing changes the sign of ΔH, multiplying by a number multiplies ΔH by that number, and adding equations adds their ΔH values.
- In a Hess’s law cycle, every route from the same start to the same end has the same ΔH; knowing two routes gives the third.
- The enthalpy of formation method, the bond enthalpy method, and adding the steps of a mechanism are all applications of Hess’s law; a catalyst doesn’t change ΔH.
AP tip: When combining equations, write what you did beside each known equation (“reversed,” “× 2”) along with its new ΔH, then show which substances cancel when you add them, to confirm you get the target equation. When a question asks only for a relationship, remember that ΔH(overall) is the sum of the ΔH values of the steps.