Introduction to Enthalpy of Reaction反应焓入门

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

Learning goals

  • Explain that the enthalpy of reaction ΔH refers to the reaction as written, so ΔH is multiplied when the coefficients are multiplied, changes sign when the equation is reversed, and changes when the physical states change
  • Calculate the heat released or absorbed from the amounts of reactants, identifying the limiting reactant when needed
  • Compare the heat released per mole and per gram by different fuels

Before you start

  • Enthalpy change and thermochemical equations (2.10.5)
  • Limiting and excess reactants (2.6.6)
  • Energy diagrams (U6.2)

Enthalpy of reaction: one “run” of the equation as written

At constant pressure, the heat a reaction absorbs or releases equals the system’s enthalpy of reaction反应焓变, ΔH. A negative ΔH means exothermic; a positive ΔH means endothermic (U6.1). The ΔH written after a thermochemical equation refers to the reaction as written:

2 HX2(g)+OX2(g)→2 HX2O(l)ΔH=−571.6 kJ/mol\begin{aligned} &\ce{2H2(g) + O2(g) -> 2H2O(l)} \\ &\Delta H = -571.6\ \text{kJ/mol} \end{aligned}

This means that when 2 mol of H₂ react completely with 1 mol of O₂ to form 2 mol of liquid water, 571.6 kJ of heat is released. The “mol” in the unit means “per mole of reaction,” not per mole of any one substance (AP writes kJ/molrxn). From it you can get the heat for each substance: 285.8 kJ is released per mole of H₂ consumed or per mole of H₂O(l) formed, and 571.6 kJ per mole of O₂ consumed.

Change the equation, change ΔH

The enthalpy change is proportional to the amount of reaction, so:

  • if every coefficient is multiplied by n, ΔH is multiplied by n (coefficients can be fractions);
  • if the equation is reversed, ΔH keeps its size but changes sign;
  • if the physical states change, ΔH changes. Forming H₂O(g) releases less heat than forming H₂O(l), because condensing water vapor to liquid water releases more heat.

Use the buttons to rewrite an equation and watch ΔH; then enter a mass and see whether the calculated heat depends on how the equation is written:

Try it: rewriting thermochemical equations

    Heat from amounts of reactants

    First find how many moles of reaction (as written) occur, then multiply by ΔH:

    q=n(substance)its coefficient×ΔHq = \frac{n(\text{substance})}{\text{its coefficient}} \times \Delta H

    If amounts of two reactants are given, first find the limiting reactant限制反应物 (2.6.6): calculate n ÷ coefficient for each, and the smaller value sets how far the reaction goes. The leftover excess reactant doesn’t react, so it neither releases nor absorbs heat. Only the limiting reactant can be used for the calculation.

    Drag the masses of the two reactants and watch the limiting reactant and the heat change:

    Try it: heat of reaction from the masses of reactants
    Substancem (g)M (g/mol)n (mol)n ÷ coefficient
    Heat per mole of each substance and per gram of reactant

      Per mole and per gram

      When comparing fuels, check whether the comparison is “per mole” or “per gram.” The table shows the heat released by complete combustion (forming CO₂(g) and H₂O(l)), calculated from the table of enthalpies of formation (NBS):

      FuelReleased per mole (kJ)M (g/mol)Released per gram (kJ)
      H₂(g)285.82.016141.8
      CH₄(g)890.316.0455.5
      C₃H₈(g)2219.944.0950.3
      C₂H₅OH(l)1366.746.0729.7

      Per mole, H₂ releases the least heat; per gram, H₂ releases the most, because its molar mass is so small.

      Example 1: One reactant in excess

      For 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s), ΔH = −1648.4 kJ/mol. How much heat is released when 11.2 g of iron powder reacts completely with excess oxygen?

      Step-by-step solution

      1. What ΔH means. ΔH = −1648.4 kJ/mol means “per mole of reaction as written”: 4 mol of Fe reacting completely with 3 mol of O₂ releases 1648.4 kJ. O₂ is in excess, so work from Fe.

      2. Moles of iron.

        n(Fe)=11.255.85=0.2005 moln(\text{Fe}) = \frac{11.2}{55.85} = 0.2005\ \text{mol}

      3. Convert to moles of reaction. The coefficient of Fe is 4:

        0.20054=0.05013 mol\frac{0.2005}{4} = 0.05013\ \text{mol}

      4. Multiply by ΔH. In kJ:

        q=0.05013×(−1648.4)=−82.6\begin{aligned} q &= 0.05013 \times (-1648.4) \\ &= -82.6 \end{aligned}

        82.6 kJ\boxed{82.6\ \text{kJ}} of heat is released.

      5. Check another way. Per mole of Fe, 1648.4 ÷ 4 = 412.1 kJ is released, and 412.1 × 0.2005 = 82.6 kJ. Same result.

      Example 2: Find the limiting reactant first

      For 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = −571.6 kJ/mol. A mixture of 3.00 g of H₂ and 32.0 g of O₂ is ignited and reacts completely. How much heat is released?

      Step-by-step solution

      1. Moles of each reactant.

        n(HX2)=3.002.016=1.488 moln(OX2)=32.032.00=1.00 mol\begin{aligned} n(\ce{H2}) &= \frac{3.00}{2.016} = 1.488\ \text{mol} \\ n(\ce{O2}) &= \frac{32.0}{32.00} = 1.00\ \text{mol} \end{aligned}

      2. Find the limiting reactant. Divide moles by coefficients; the smaller runs out first:

        HX2:1.488÷2=0.744OX2:1.00÷1=1.00\begin{aligned} \ce{H2} &: 1.488 \div 2 = 0.744 \\ \ce{O2} &: 1.00 \div 1 = 1.00 \end{aligned}

        H₂ gives the smaller value, so it is the limiting reactant; 0.744 mol of reaction occurs.

      3. Heat. In kJ:

        q=0.744×(−571.6)=−425\begin{aligned} q &= 0.744 \times (-571.6) \\ &= -425 \end{aligned}

        425 kJ\boxed{425\ \text{kJ}} of heat is released.

      4. Check the excess O₂. Only 0.744 mol of O₂ is used, leaving 0.256 mol (8.19 g), which doesn’t react and releases no heat.

      5. Common mistake. Using O₂ by mistake (1.00 × 571.6) gives 572 kJ, which is too large: the heat must come from the limiting reactant.

      Practice: for N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.2 kJ/mol. What is ΔH for each of the following?
      1. 2NH₃(g) → N₂(g) + 3H₂(g)
      2. ½N₂(g) + 3/2H₂(g) → NH₃(g)
      3. 2N₂(g) + 6H₂(g) → 4NH₃(g)
      4. NH₃(g) → ½N₂(g) + 3/2H₂(g)
      5. 1.5 mol of H₂ reacts completely with excess N₂
      6. 4 mol of NH₃ forms from the elements
      7. 1 mol of N₂ reacts completely with 3 mol of H₂
      8. 1 mol of NH₃ decomposes into N₂ and H₂

      Common mistakes

      Mistake 1: “ΔH is in kJ/mol, so it is the heat per mole of some substance.” ΔH refers to one run of the reaction as written (per mole of reaction). For the same reaction, the heat per mole of each substance has to be worked out from its coefficient.

      Mistake 2: “When amounts of both reactants are given, either one can be used.” Only the limiting reactant can be used. The leftover excess reactant doesn’t react, so it doesn’t release or absorb heat.

      Mistake 3: “A fuel that releases more heat per mole also releases more per gram.” To compare per gram, divide by the molar mass. H₂ releases the least heat per mole but the most per gram.

      Quick check

      For 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −197.8 kJ/mol. How much heat is released when 1.00 mol of SO₂ reacts completely with excess O₂?

      For CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178.3 kJ/mol. How much heat is needed to decompose 20.0 g of CaCO₃ (M = 100.09 g/mol)?

      For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890.3 kJ/mol. About how much heat is released when 2.0 mol of CH₄ reacts completely with 3.0 mol of O₂?

      In the reaction A + B → C, A and B react 1 : 1. Trial 1 uses 0.10 mol of A and 0.10 mol of B; trial 2 uses 0.10 mol of A and 0.20 mol of B. Both trials release the same amount of heat. Which statement is correct?

      For H₂(g) + ½O₂(g) → H₂O(g), ΔH = −241.8 kJ/mol. What is ΔH for 2H₂O(g) → 2H₂(g) + O₂(g)?

      A student says, “Reaction X has ΔH = −200 kJ/mol and reaction Y has ΔH = −100 kJ/mol, so in any experiment X releases more heat than Y.” What is wrong with this claim?

      Using the fuel table above, which fuel releases the most heat per gram burned?

      AP-style free response

      AP-style free response

      The thermite reaction can be used to weld railroad tracks: Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s), ΔH = −851.5 kJ/mol. In one experiment, 32.0 g of Fe₂O₃ (M = 159.70 g/mol) and 16.2 g of Al (M = 26.98 g/mol) are mixed and ignited, and the reaction goes to completion.

      (a) Determine by calculation which reactant is limiting.

      (b) Calculate the heat released by the reaction.

      (c) Calculate the mass of the excess reactant that remains.

      (d) Another student writes the equation as 2Fe₂O₃(s) + 4Al(s) → 2Al₂O₃(s) + 4Fe(s). Give ΔH for this equation. Would using it change the answer to (b)? Justify your answer.

      (e) Which has the higher enthalpy, the reactants or the products of this reaction?

      Model answer and scoring

      (a) n(Fe₂O₃) = 32.0 ÷ 159.70 = 0.2004 mol, and 0.2004 ÷ 1 = 0.2004; n(Al) = 16.2 ÷ 26.98 = 0.6004 mol, and 0.6004 ÷ 2 = 0.3002. Fe₂O₃ has the smaller value, so it is the limiting reactant.

      (b) q = 0.2004 mol × (−851.5 kJ/mol) = −171 kJ, so 171 kJ of heat is released (170.6 kJ).

      (c) Al used: 0.2004 × 2 = 0.4008 mol, or 10.81 g; 16.2 − 10.81 = 5.4 g of Al remains.

      (d) ΔH = 2 × (−851.5) = −1703.0 kJ/mol. The answer doesn’t change: by this equation only 0.2004 ÷ 2 = 0.1002 mol of reaction occurs, and 0.1002 × 1703.0 = 170.6 kJ.

      (e) The reactants. The reaction is exothermic (ΔH < 0), so the products have lower enthalpy than the reactants.

      Scoring: (a) compares n ÷ coefficient; (b) uses the limiting reactant, with units and sign; (c) mass of excess; (d) doubled ΔH, unchanged heat, and the reason; (e) reactants are higher.

      Summary

      • ΔH refers to one run of the reaction as written (kJ/mol means per mole of reaction). Multiplying the coefficients by n multiplies ΔH by n; reversing the equation changes the sign; different physical states give different ΔH.
      • Heat: q = [n(substance) ÷ coefficient] × ΔH. When amounts of two reactants are given, find the limiting reactant first and use only it.
      • When comparing fuels, separate “per mole” from “per gram”: heat per gram = heat per mole ÷ molar mass.

      AP tip: Show three steps when calculating heat: find n, divide by the coefficient (finding the limiting reactant), and multiply by ΔH; then say whether heat is released or absorbed. When a question asks how much heat is released, give a positive value and say it is released.

      Practice this chapter: Unit 6 · Thermochemistry · 12 questionsPractice this topic: U6.6 · Introduction to Enthalpy of Reaction · 5 questions

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