Introduction to Hess's Law盖斯定律入门

Big Idea · Energy (ENE)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 6~40 min

Learning goals

  • State Hess's law: ΔH depends only on the initial and final states, not on the path
  • Reverse, multiply, and add known thermochemical equations to find a ΔH that can’t be measured directly
  • Calculate ΔH for a reaction from standard enthalpies of formation

Before you start

  • Enthalpy change and thermochemical equations (2.10.5)

A ΔH you can’t measure

Burning carbon in too little oxygen makes carbon monoxide: C(s) + ½O₂(g) → CO(g). Its ΔH is very hard to measure directly. With too little oxygen some carbon doesn’t burn, and with a little more some CO burns on to CO₂, so you always get a mixture. But ΔH for carbon burning to CO₂, and for CO burning to CO₂, are both easy to measure. Can you combine them to get the first one?

Hess’s law

In 1840, Germain Hess found that the heat of a reaction is the same whether it happens in one step or several. ΔH depends only on the initial and final states, not on the path between them. This is Hess’s law盖斯定律.

Think of climbing a mountain: whichever trail you take from the base to the summit, you gain the same height. Enthalpy is like altitude, and ΔH is the change in altitude.

So you can treat thermochemical equations like algebra:

  1. Find where each substance in the target equation appears in the given equations, and on which side.
  2. Reverse given equations (change the sign of ΔH) or multiply them (multiply ΔH by the same factor) until those substances are on the correct sides with the correct coefficients.
  3. Add the equations, cancelling anything that appears on both sides, and add the ΔH values.

For the CO reaction, you are given:

C(s)+OX2(g)→COX2(g)ΔH1=−393.5 kJ/mol(1)CO(g)+12 OX2(g)→COX2(g)ΔH2=−283.0 kJ/mol(2)\begin{aligned} &\ce{C(s) + O2(g) -> CO2(g)} \\ &\qquad \Delta H_1 = -393.5\ \text{kJ/mol} \quad \text{(1)} \\ &\ce{CO(g) + 1/2O2(g) -> CO2(g)} \\ &\qquad \Delta H_2 = -283.0\ \text{kJ/mol} \quad \text{(2)} \end{aligned}

C is a reactant in the target, as in (1). CO is a product in the target, so reverse (2). Adding (1) and the reversed (2), CO₂ cancels and ½O₂ is left: ΔH = ΔH₁ − ΔH₂ = −393.5 − (−283.0) = −110.5 kJ/mol.

Try it yourself: reverse and multiply the steps until they add up to exactly the target equation:

Try it: Hess's law puzzle
Target equation
Sum of the steps

Standard enthalpies of formation

The standard enthalpy of formation标准摩尔生成焓 ΔHf° of a substance is ΔH for forming 1 mol of it from its elements in their most stable forms, under standard conditions. For example,

C(s)+OX2(g)→COX2(g)ΔHf∘(COX2)=−393.5 kJ/mol\begin{aligned} &\ce{C(s) + O2(g) -> CO2(g)} \\ &\Delta H_\text{f}^\circ(\ce{CO2}) = -393.5\ \text{kJ/mol} \end{aligned}

An element in its most stable form, such as O₂(g), H₂(g), N₂(g), C(s, graphite), or Fe(s), has ΔHf° = 0.

With formation enthalpies, Hess’s law gives the ΔH of any reaction by a detour through the elements. First break the reactants down into elements (ΔH = −ΣnΔHf° of the reactants), then build the products from the elements (ΔH = +ΣnΔHf° of the products). Adding the two steps:

ΔHrxn∘=Σ n ΔHf∘(products)−Σ n ΔHf∘(reactants)\begin{aligned} \Delta H^\circ_\text{rxn} &= \Sigma\, n\,\Delta H_\text{f}^\circ(\text{products}) \\ &\quad - \Sigma\, n\,\Delta H_\text{f}^\circ(\text{reactants}) \end{aligned}

Pick a reaction and see each substance’s contribution, with the detour through the elements drawn on the right:

Try it: ΔH from enthalpies of formation

SubstancenΔH°f (kJ/mol)n × ΔH°f

    How Chinese textbooks differ

    • Chinese textbooks call it 盖斯定律, from a different transliteration of Hess.
    • The AP equation sheet gives ΔH°rxn = ΣΔHf°(products) − ΣΔHf°(reactants), and calculations with formation enthalpies are a core AP skill. Chinese problem sets concentrate on combining thermochemical equations.
    • Chinese textbooks write the standard enthalpy of formation as ΔfHm⊖, which means the same as ΔHf°.

    Example 1: graphite to diamond

    You are given the thermochemical equations for burning graphite and diamond:

    C(s, graphite)+OX2(g)→COX2(g)ΔH1=−393.5 kJ/mol(1)C(s, diamond)+OX2(g)→COX2(g)ΔH2=−395.4 kJ/mol(2)\begin{aligned} &\ce{C}(\text{s, graphite}) \ce{+ O2(g) -> CO2(g)} \\ &\qquad \Delta H_1 = -393.5\ \text{kJ/mol} \quad \text{(1)} \\ &\ce{C}(\text{s, diamond}) \ce{+ O2(g) -> CO2(g)} \\ &\qquad \Delta H_2 = -395.4\ \text{kJ/mol} \quad \text{(2)} \end{aligned}

    Find ΔH for turning graphite into diamond. Which form of carbon is more stable?

    Step-by-step solution

    1. Match the target. The target is C(s, graphite) → C(s, diamond). Graphite is a reactant, as in (1), so keep (1) as written; diamond must be a product, so reverse (2).

    2. Reversing flips the sign. Reversed, (2) is CO₂(g) → C(s, diamond) + O₂(g), with ΔH = +395.4 kJ/mol.

    3. Add. Adding (1) and the reversed (2), which is (1) − (2), cancels CO₂ and O₂:

      ΔH=ΔH1−ΔH2=−393.5−(−395.4)=+1.9 kJ/mol\begin{aligned} \Delta H &= \Delta H_1 - \Delta H_2 \\ &= -393.5 - (-395.4) \\ &= \boxed{+1.9\ \text{kJ/mol}} \end{aligned}

    4. Stability. ΔH > 0, so diamond is higher in energy than graphite. Lower energy means more stable, so graphite is the more stable form.

    5. A common mistake. Adding ΔH₁ and ΔH₂ directly, or reversing (2) without changing the sign of its ΔH.

    Example 2: the combustion of propane from formation enthalpies

    Given ΔHf° values of −103.8 kJ/mol for C₃H₈(g), −393.5 kJ/mol for CO₂(g), and −285.8 kJ/mol for H₂O(l), find ΔH° for

    CX3HX8(g)+5 OX2(g)→3 COX2(g)+4 HX2O(l)\begin{aligned} &\ce{C3H8(g) + 5O2(g)} \\ &\ce{-> 3CO2(g) + 4H2O(l)} \end{aligned}
    Step-by-step solution

    1. The formula. ΔH° = ΣnΔHf°(products) − ΣnΔHf°(reactants), where n is each coefficient.

    2. Products. Multiply each ΔHf° by its coefficient:

      Σproducts=3(−393.5)+4(−285.8)=−1180.5−1143.2=−2323.7 kJ/mol\begin{aligned} \Sigma_\text{products} &= 3(-393.5) + 4(-285.8) \\ &= -1180.5 - 1143.2 \\ &= -2323.7\ \text{kJ/mol} \end{aligned}

    3. Reactants. O₂ is an element in its standard state, so its ΔHf° is 0: Σreactants = −103.8 + 5(0) = −103.8 kJ/mol.

    4. Subtract. Products minus reactants:

      ΔH∘=−2323.7−(−103.8)=−2219.9 kJ/mol\begin{aligned} \Delta H^\circ &= -2323.7 - (-103.8) \\ &= \boxed{-2219.9\ \text{kJ/mol}} \end{aligned}

    5. Conclusion. This matches the table in 2.10.5: burning 1 mol of propane releases 2219.9 kJ.

    6. A common mistake. Subtracting reactants minus products (giving +2219.9), or forgetting to multiply by the coefficients 3 and 4.

    Practice: is the standard enthalpy of formation of this substance zero?
    1. O₂(g)
    2. O₃(g)
    3. H₂O(l)
    4. C(s, graphite)
    5. C(s, diamond)
    6. Br₂(l)
    7. Br₂(g)
    8. Fe(s)
    9. N₂(g)

    Common mistakes

    Mistake 1: “Doing a reaction in several steps releases more heat than doing it in one.” By Hess’s law, ΔH depends only on the initial and final states, so the total is the same however many steps there are.

    Mistake 2: “Every element has ΔH°f = 0.” Only an element in its most stable form does. O₃(g), C(s, diamond), and Br₂(g) are elements, but their formation enthalpies aren’t zero.

    Mistake 3: “Reactants minus products.” It is products minus reactants, and each ΔH°f must be multiplied by that substance’s coefficient.

    Quick check

    From Example 1, graphite → diamond has ΔH = +1.9 kJ/mol. Which statement is correct?

    Given ΔH°f(SO₂(g)) = −296.8 kJ/mol and ΔH°f(SO₃(g)) = −395.7 kJ/mol, what is ΔH° for 2SO₂(g) + O₂(g) → 2SO₃(g)?

    Summary

    • Hess’s law: ΔH depends only on the initial and final states, not the path. You can reverse an equation (ΔH changes sign), multiply it (ΔH is multiplied too), and add equations (ΔH values add).
    • The standard enthalpy of formation is ΔH for making 1 mol of a substance from its elements in their most stable forms; for those elements ΔH°f = 0.
    • ΔH°rxn = ΣnΔH°f(products) − ΣnΔH°f(reactants).
    On to AP (for later)

    AP topic 6.8 calculates ΔH°rxn from enthalpies of formation, with the formula on the equation sheet, and topic 6.9 combines equations with Hess’s law. AP questions also like to compare all three routes: an estimate from bond enthalpies, an exact value from formation enthalpies, and a measurement from calorimetry. Be ready to explain why the three don’t agree exactly.

    Data sources: enthalpies of formation from the NBS tables of chemical thermodynamic properties (Wagman et al., 1982), rounded to 0.1 kJ/mol; most US and Chinese textbook values come from the same tables.

    Practice this chapter: 2.10 · Thermochemistry · 10 questions

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