Enthalpy Change and Thermochemical Equations焓变与热化学方程式

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

Learning goals

  • Explain the meaning and sign of the enthalpy change ΔH, and what “per mole of reaction” means
  • Write thermochemical equations correctly, and adjust ΔH when the coefficients, direction, or physical states change
  • Explain enthalpy of combustion, and calculate the heat released or absorbed by a given amount of reactant

Before you start

  • Stoichiometry (mole ratios) (2.6.5)
  • Endothermic and exothermic reactions (2.10.2)
  • Calorimetry (2.10.4)

How much natural gas does it take to boil water?

From q = mcΔT in 2.10.3, warming 1 kg of water from 20 °C to 100 °C takes about 334 kJ. That heat comes from burning natural gas. To find how much gas you need, you have to know how much heat burning 1 mol of methane releases, written together with the balanced equation. That is a thermochemical equation.

Enthalpy change

For reactions at constant pressure, such as anything in an open beaker, chemists use a quantity called enthalpy焓, H. The heat absorbed or released at constant pressure equals the enthalpy change焓变:

ΔH=Hproducts−Hreactants\Delta H = H_\text{products} - H_\text{reactants}

As in 2.10.1, the sign is taken from the system’s point of view:

  • Exothermic reaction: the system’s enthalpy falls, so ΔH < 0.
  • Endothermic reaction: the system’s enthalpy rises, so ΔH > 0.

Thermochemical equations

A thermochemical equation热化学方程式 is a balanced equation together with its ΔH. For example, 1 mol of H₂ reacting with ½ mol of O₂ to form 1 mol of liquid water releases 285.8 kJ:

HX2(g)+12 OX2(g)→HX2O(l)ΔH∘=−285.8 kJ/molrxn\begin{aligned} &\ce{H2(g) + 1/2O2(g) -> H2O(l)} \\ &\Delta H^\circ = -285.8\ \text{kJ/mol}_\text{rxn} \end{aligned}

The rules:

  1. Give the physical state of every substance: (s), (l), (g), or (aq).
  2. ΔH applies to the equation exactly as written, “per mole of reaction.” AP writes the unit as kJ/molrxn.
  3. Coefficients mean moles, so they can be fractions. ΔH is proportional to the coefficients: double them and ΔH doubles. Written as 2H₂ + O₂ → 2H₂O, the same reaction has ΔH° = −571.6 kJ/molrxn.
  4. Reverse the equation and ΔH changes sign but keeps its size.
  5. The ° in ΔH° means standard conditions: each substance pure at 1 bar (about 1 atm), usually at 25 °C.

Physical states matter. Forming water vapor instead releases less heat:

HX2(g)+12 OX2(g)→HX2O(g)ΔH∘=−241.8 kJ/molrxn\begin{aligned} &\ce{H2(g) + 1/2O2(g) -> H2O(g)} \\ &\Delta H^\circ = -241.8\ \text{kJ/mol}_\text{rxn} \end{aligned}

The difference, 44.0 kJ, is exactly the heat needed to turn 1 mol of liquid water into vapor at 25 °C.

Enthalpy of combustion

The enthalpy of combustion燃烧热 ΔH°comb is ΔH for burning 1 mol of a substance completely in oxygen: C ends up as CO₂(g), H as H₂O(l), S as SO₂(g), and N as N₂(g). Some values:

SubstanceΔH°comb (kJ/mol)Heat released per gram (kJ)
H₂(g)−285.8141.8
C(s, graphite)−393.532.8
CO(g)−283.010.1
CH₄(g)−890.355.5
C₃H₈(g)−2219.950.3
C₂H₅OH(l)−1366.729.7

Per gram, hydrogen releases the most energy, one reason it is called an ideal fuel (1.8.2).

Once you know ΔH, heat works like any other stoichiometry problem (2.6.5): the heat released or absorbed is proportional to the moles that react.

This lesson uses the atomic masses on the AP periodic table (H 1.008, C 12.01, N 14.01, O 16.00, Ca 40.08).

Pick a thermochemical equation and try doubling, halving, or reversing it, or switching liquid water to vapor, and watch ΔH change. Below, find the heat for a given mass:

Try it: rewriting thermochemical equations

    How Chinese textbooks differ

    • Chinese textbooks always write the unit as kJ/mol, with the same “per mole of reaction” meaning; they don’t use molrxn. Many US textbooks simply write ΔH = −571.6 kJ for the equation as written.
    • Chinese textbooks state the heat of combustion (燃烧热) as a positive amount of heat, “the heat of combustion of CH₄ is 890.3 kJ/mol,” and add the minus sign only in ΔH. US textbooks give ΔH°comb as a negative number.
    • Chinese textbooks specify 25 °C and 101 kPa instead of using the ° symbol, and they never write heat as a term inside the equation. Some US textbooks do, as in CH₄ + 2O₂ → CO₂ + 2H₂O + 890.3 kJ.

    Example 1: writing a thermochemical equation

    Burning 4.032 g of hydrogen gas to form liquid water at 25 °C releases 571.6 kJ. Write the thermochemical equation for the combustion of 1 mol of hydrogen.

    Step-by-step solution

    1. Moles.

      n(HX2)=4.032 g2.016 g/mol=2.000 mol\begin{aligned} n(\ce{H2}) &= \frac{4.032\ \text{g}}{2.016\ \text{g/mol}} \\ &= 2.000\ \text{mol} \end{aligned}

    2. Per mole of H₂.

      571.6 kJ2.000 mol=285.8 kJ/mol\frac{571.6\ \text{kJ}}{2.000\ \text{mol}} = 285.8\ \text{kJ/mol}

    3. Sign. Heat is released, so ΔH is negative: −285.8 kJ.

    4. Coefficients. For 1 mol of H₂, forming one H₂O takes only ½ O₂; show every state:

      HX2(g)+12 OX2(g)→HX2O(l)ΔH∘=−285.8 kJ/molrxn\begin{aligned} &\ce{H2(g) + 1/2O2(g) -> H2O(l)} \\ &\Delta H^\circ = -285.8\ \text{kJ/mol}_\text{rxn} \end{aligned}

    5. Another correct form. Written for 2 mol, 2H₂(g) + O₂(g) → 2H₂O(l), ΔH° = −571.6 kJ/molrxn is also correct. ΔH scales with the coefficients.

    6. A common mistake. Writing H₂O(g) instead of H₂O(l), leaving out states, or giving ΔH as +285.8. The states change ΔH, and heat released means a negative sign.

    Example 2: how much methane?

    You want to heat 1.000 kg of water from 20.0 °C to 100.0 °C. If all the heat from burning methane went into the water, what mass of methane would you need?

    CHX4(g)+2 OX2(g)→COX2(g)+2 HX2O(l)ΔH∘=−890.3 kJ/molrxn\begin{aligned} &\ce{CH4(g) + 2O2(g)} \\ &\ce{-> CO2(g) + 2H2O(l)} \\ &\Delta H^\circ = -890.3\ \text{kJ/mol}_\text{rxn} \end{aligned}
    Step-by-step solution

    1. The route. Heat absorbed by the water → moles of methane → grams of methane.

    2. Heat absorbed by the water. m = 1.000 kg = 1000. g:

      q=1000.×4.18×80.0=3.344×105 J=334.4 kJ\begin{aligned} q &= 1000. \times 4.18 \times 80.0 \\ &= 3.344 \times 10^{5}\ \text{J} \\ &= 334.4\ \text{kJ} \end{aligned}

      One extra digit is kept for now.

    3. Moles of methane. ΔH° = −890.3 kJ/molrxn means each mole of CH₄ burned releases 890.3 kJ:

      n(CHX4)=334.4890.3=0.3756 mol\begin{aligned} n(\ce{CH4}) &= \frac{334.4}{890.3} \\ &= 0.3756\ \text{mol} \end{aligned}

    4. Mass. M(CH₄) = 12.01 + 4(1.008) = 16.04 g/mol:

      m=0.3756×16.04=6.02 g\begin{aligned} m &= 0.3756 \times 16.04 \\ &= \boxed{6.02\ \text{g}} \end{aligned}

    5. Check. 6.02 g ÷ 16.04 g/mol × 890.3 kJ/mol = 334 kJ, the heat the water needs.

    6. Think about it. A real stove loses much of its heat to the air, so it burns a lot more methane than this.

    Practice: given H₂(g) + ½O₂(g) → H₂O(l), ΔH° = −285.8 kJ/mol, what is ΔH° for each reaction?
    1. 2H₂(g) + O₂(g) → 2H₂O(l)
    2. H₂O(l) → H₂(g) + ½O₂(g)
    3. 2H₂O(l) → 2H₂(g) + O₂(g)
    4. ½H₂(g) + ¼O₂(g) → ½H₂O(l)
    5. ½H₂O(l) → ½H₂(g) + ¼O₂(g)
    6. The enthalpy of combustion of H₂
    7. ΔH for forming 1 mol of liquid water
    8. ΔH for decomposing 2 mol of liquid water

    Common mistakes

    Mistake 1: “ΔH is in kJ/mol, so it always refers to 1 mol of some substance.” “Per mole” means per mole of reaction, the equation run once as written. Double the coefficients and ΔH doubles.

    Mistake 2: “States don’t matter in a thermochemical equation.” ΔH depends on them: forming H₂O(l) gives −285.8 kJ/mol, but forming H₂O(g) gives only −241.8 kJ/mol.

    Mistake 3: “Reversing an equation leaves ΔH the same.” If a reaction releases 92.2 kJ going forward, the reverse absorbs 92.2 kJ. ΔH keeps its size but changes sign.

    Quick check

    Given N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = −92.2 kJ/mol, what is ΔH° for 2NH₃(g) → N₂(g) + 3H₂(g)?

    Burning propane, C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), has ΔH° = −2219.9 kJ/mol. How much heat is released by burning 2.00 mol of propane?

    Summary

    • At constant pressure the heat of reaction equals the enthalpy change, ΔH = Hproducts − Hreactants. Exothermic ΔH < 0, endothermic ΔH > 0, in kJ per mole of reaction.
    • Thermochemical equations show states and ΔH. Coefficients can be fractions. Multiply the equation and you multiply ΔH; reverse it and ΔH changes sign.
    • ΔH°comb is for burning 1 mol of a substance completely to CO₂(g), H₂O(l), and so on. The heat released or absorbed is proportional to the moles that react.
    On to AP (for later)

    AP topic 6.6 expects you to read ΔH as kJ/molrxn for the equation as written and to use it in stoichiometry: heat from a mass of reactant, or the mass needed for a given amount of heat. Questions often double or reverse an equation to check that you change ΔH to match. Topics 6.8 and 6.9, in the next lesson, find ΔH from enthalpies of formation and from Hess’s law.

    Data sources: enthalpies of combustion calculated from this site’s table of enthalpies of formation (lesson 2.10.6, NBS data).

    Practice this chapter: 2.10 · Thermochemistry · 10 questions

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