Enthalpy of Formation生成焓

Big Idea · Energy (ENE)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 6~45 min

Learning goals

  • State the definition of the standard enthalpy of formation, decide whether ΔH for a thermochemical equation is the standard enthalpy of formation of a substance, and explain why an element in its standard state has ΔH°f = 0
  • Calculate the enthalpy of reaction with ΔH°rxn = ΣnΔH°f(products) − ΣnΔH°f(reactants)
  • Find an unknown enthalpy of formation from ΔH for a reaction and the enthalpies of formation of the other substances

Before you start

  • Introduction to Hess's law (2.10.6)
  • Introduction to enthalpy of reaction (U6.6)
  • Bond enthalpies (U6.7)

Standard enthalpy of formation

The standard enthalpy of formation标准摩尔生成焓 ΔHf° of a substance is the enthalpy change for forming 1 mol of it from its elements in their standard states, under standard conditions (usually at 25 °C). For example:

12 NX2(g)+32 HX2(g)→NHX3(g)ΔHf∘(NHX3)=−46.1 kJ/mol\begin{aligned} &\ce{1/2N2(g) + 3/2H2(g) -> NH3(g)} \\ &\Delta H_\text{f}^\circ(\ce{NH3}) = -46.1\ \text{kJ/mol} \end{aligned}

For ΔH of an equation to equal the enthalpy of formation of a substance, all three of these must hold:

  1. the only product is that substance, and exactly 1 mol of it forms (so fractional coefficients are common);
  2. every reactant is an element;
  3. every element is in its standard state, its most stable form at 25 °C, such as O₂(g), H₂(g), N₂(g), Cl₂(g), Br₂(l), C(s, graphite), Fe(s), or Hg(l).

An element in its standard state has an enthalpy of formation of 0 (forming it from itself involves no change). Other forms of the same element are not zero: O₃(g) is +142.7, C(s, diamond) is +1.9, and Br₂(g) is +30.9 kJ/mol. The state of the product matters too: H₂O(l) is −285.8 and H₂O(g) is −241.8 kJ/mol.

Enthalpy of reaction from enthalpies of formation

Any reaction can be imagined as a detour: first break all the reactants down into elements in their standard states, then build the products from those elements (2.10.6). So:

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}

Here n is the coefficient in the equation. Elements have ΔHf° = 0 and can be left out. The enthalpies of formation on this site come from the NBS tables of chemical thermodynamic properties (Wagman et al., 1982). Pick a reaction to see each substance’s contribution and the detour through the elements:

Try it: ΔH from enthalpies of formation

SubstancenΔH°f (kJ/mol)n × ΔH°f

    Judging and working backward

    The first mode asks whether ΔH for an equation is an enthalpy of formation. The second uses the equation in reverse: from ΔH for a reaction and the enthalpies of formation of the other substances, treat one substance’s enthalpy of formation as the unknown and solve for it.

    Try it: standard enthalpies of formation

    Is ΔH for this equation the standard enthalpy of formation, ΔHf°, of a substance?

    Example 1: ΔH from enthalpies of formation

    One reaction in a blast furnace is Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g). Given ΔHf° (kJ/mol): Fe₂O₃(s) −824.2, CO(g) −110.5, CO₂(g) −393.5. Find ΔH°rxn.

    Step-by-step solution

    1. Write the formula. Sum of the products’ enthalpies of formation minus the reactants’, each times its coefficient; an element in its standard state (such as Fe(s)) has ΔHf° = 0:

      ΔH∘=∑nΔHf∘(prod)−∑nΔHf∘(react)\begin{aligned} \Delta H^\circ = {} &\textstyle\sum n\Delta H_f^\circ(\text{prod}) \\ &- \textstyle\sum n\Delta H_f^\circ(\text{react}) \end{aligned}

    2. Products. In kJ:

      2×0+3×(−393.5)=−1180.5\begin{aligned} &2 \times 0 + 3 \times (-393.5) \\ &= -1180.5 \end{aligned}

    3. Reactants.

      (−824.2)+3×(−110.5)=−1155.7\begin{aligned} &(-824.2) + 3 \times (-110.5) \\ &= -1155.7 \end{aligned}

    4. Subtract.

      ΔHrxn∘=−1180.5−(−1155.7)=−24.8 kJ/mol\begin{aligned} \Delta H^\circ_\text{rxn} &= -1180.5 - (-1155.7) \\ &= \boxed{-24.8\ \text{kJ/mol}} \end{aligned}

    5. Check. Products minus reactants, every term times its coefficient, and Fe taken as 0. The result is negative, so the reaction is exothermic, but only slightly.

    Example 2: An unknown enthalpy of formation

    Ethanol burns completely: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH°rxn = −1366.7 kJ/mol. Given ΔHf° (kJ/mol): CO₂(g) −393.5, H₂O(l) −285.8. Find the enthalpy of formation of C₂H₅OH(l).

    Step-by-step solution

    1. Write the formula with an unknown. Let x be ΔHf° of C₂H₅OH(l); O₂ is an element, so its ΔHf° is 0 (in kJ/mol):

      ΔH=[2(−393.5)+3(−285.8)]−[x+3×0]=−1366.7\begin{aligned} \Delta H &= [2(-393.5) + 3(-285.8)] \\ &\quad - [x + 3 \times 0] \\ &= -1366.7 \end{aligned}

    2. Work out the known part (products).

      2(−393.5)+3(−285.8)=−787.0−857.4=−1644.4\begin{aligned} &2(-393.5) + 3(-285.8) \\ &= -787.0 - 857.4 = -1644.4 \end{aligned}

    3. Solve for x.

      −1366.7=−1644.4−xx=−1644.4+1366.7=−277.7 kJ/mol\begin{aligned} -1366.7 &= -1644.4 - x \\ x &= -1644.4 + 1366.7 \\ &= \boxed{-277.7\ \text{kJ/mol}} \end{aligned}

    4. Substitute back. −1644.4 − (−277.7) = −1366.7, matching the given enthalpy of combustion; the result also matches the table value for C₂H₅OH(l).

    5. Watch the sign. The whole reactant side is subtracted, so x carries a minus sign; O₂’s value is 0, but that doesn’t mean the “minus reactants” step can be skipped.

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

    Common mistakes

    Mistake 1: “Every element has an enthalpy of formation of 0.” Only elements in their standard states are 0. O₃(g), C(s, diamond), and Br₂(g) are elements too, and none of them is 0.

    Mistake 2: “ΔH = ΣΔH°f(reactants) − ΣΔH°f(products), with no coefficients needed.” It is products minus reactants, and every term is multiplied by its coefficient.

    Mistake 3: “H₂O(l) and H₂O(g) have the same enthalpy of formation, since both are water.” Different states have different enthalpies of formation. Forming water vapor releases 44.0 kJ/mol less than forming liquid water (the enthalpy of vaporization at 25 °C).

    Quick check

    For which equation does ΔH equal the standard enthalpy of formation of CO₂(g)?

    Use enthalpies of formation (kJ/mol: CaO(s) −635.1, CO₂(g) −393.5, CaCO₃(s) −1206.9) to find ΔH for CaO(s) + CO₂(g) → CaCO₃(s).

    For 2H₂O₂(l) → 2H₂O(l) + O₂(g), ΔH = −196.0 kJ/mol, and ΔH°f(H₂O, l) = −285.8 kJ/mol. What is the enthalpy of formation of H₂O₂(l)?

    The enthalpy of formation of O₂(g) is 0, while that of O₃(g) is +142.7 kJ/mol. Which is the best explanation?

    Which has the more negative enthalpy of formation, H₂O(l) or H₂O(g), and why?

    Use enthalpies of formation (kJ/mol: Fe₂O₃(s) −824.2, Al₂O₃(s) −1675.7) to find ΔH for the thermite reaction Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s).

    AP-style free response

    AP-style free response

    Methanol can be used as a fuel: 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l). Enthalpies of formation (kJ/mol): CH₃OH(l) −238.7, CO₂(g) −393.5, H₂O(l) −285.8.

    (a) Write the thermochemical equation that represents the standard enthalpy of formation of CH₃OH(l).

    (b) The table gives no enthalpy of formation for O₂(g). Why not?

    (c) Calculate ΔH°rxn for the combustion reaction above.

    (d) How much heat is released when 16.0 g of CH₃OH (M = 32.04 g/mol) burns completely?

    (e) If the combustion formed H₂O(g) instead (ΔH°f = −241.8 kJ/mol), would ΔH°rxn be more negative or less negative? Calculate the value.

    Model answer and scoring

    (a) C(s, graphite) + 2H₂(g) + ½O₂(g) → CH₃OH(l), ΔH = −238.7 kJ/mol.

    (b) O₂(g) is the standard state of oxygen, and an element in its standard state has an enthalpy of formation of 0.

    (c) ΔH°rxn = [2 × (−393.5) + 4 × (−285.8)] − [2 × (−238.7) + 3 × 0] = −1930.2 + 477.4 = −1452.8 kJ/mol.

    (d) n = 16.0 ÷ 32.04 = 0.4994 mol, so 0.4994 ÷ 2 = 0.2497 mol of reaction; q = 0.2497 × (−1452.8) = −363 kJ, so 363 kJ is released.

    (e) Less negative. ΔH°rxn = [2 × (−393.5) + 4 × (−241.8)] − 2 × (−238.7) = −1276.8 kJ/mol. Forming water vapor leaves out the heat released when 4 mol of water condenses (4 × 44.0 = 176.0 kJ).

    Scoring: (a) 1 mol of product, elements in standard states, physical states; (b) standard-state elements are 0; (c) formula, coefficients, and sign; (d) conversion and “released”; (e) the judgment and the value.

    Summary

    • Standard enthalpy of formation: ΔH for forming 1 mol of a substance from its elements in their standard states. Elements in their standard states have ΔHf° = 0; other forms of the same element do not.
    • A formation equation has 1 mol of product, only elements as reactants, and every element in its standard state.
    • ΔH°rxn = ΣnΔHf°(products) − ΣnΔHf°(reactants): multiply each term by its coefficient, and subtract reactants from products.
    • From ΔH and the enthalpies of formation of the other substances, you can solve for an unknown enthalpy of formation.

    AP tip: The AP equation sheet gives ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants). Write the formula first, then substitute the values with their coefficients; for elements, state that the enthalpy of formation is 0. When writing a formation equation, check for 1 mol of product and the states of the elements, such as C(s, graphite).

    Practice this chapter: Unit 6 · Thermochemistry · 12 questionsPractice this topic: U6.8 · Enthalpy of Formation · 5 questions

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