Unit 6 · Topic 6.7 Beta

Bond Enthalpies

Bond enthalpies give an estimate of ΔH: add the bond enthalpies of the bonds broken, then subtract those of the bonds formed.

Practice 5: Mathematical RoutinesPractice 6: Argumentation

Question set for this topic

Part 1 · Hook

Why this matters

Why does burning methane release so much energy, while splitting water takes so much? You can predict it with a pencil: count the bonds that break, count the bonds that form, and look up how strong each one is. The answer comes out within a few percent of what a calorimeter measures.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The bond energy of a covalent bond is

  1. the energy needed to break it
  2. the energy released when it breaks
  3. the mass of the bond
  4. the length of the bond
Show the answer

It is the depth of the potential energy well: the energy needed to pull the bonded atoms apart.

  • Correct: the energy needed to break it:
  • the energy released when it breaks:
  • the mass of the bond:
  • the length of the bond:

2. How many bonds are in one CO₂ molecule (O=C=O)?

  1. Two C=O double bonds
  2. Two C–O single bonds
  3. One C=O double bond
  4. Three bonds
Show the answer

The Lewis structure has carbon double-bonded to each oxygen.

  • Correct: Two C=O double bonds:
  • Two C–O single bonds:
  • One C=O double bond:
  • Three bonds:

3. ΔH for a thermochemical equation is per

  1. mole of reaction as written
  2. gram of product
  3. molecule
  4. mole of each substance
Show the answer

It goes with the coefficients exactly as written.

  • Correct: mole of reaction as written:
  • gram of product:
  • molecule:
  • mole of each substance:

Part 4 · See it

See it first

Enthalpy ladder for H₂ + Cl₂ → 2 HCl. Breaking the H–H and Cl–Cl bonds absorbs +436 + 242 = +678 kJ and lifts the reactants to separate atoms (2 H + 2 Cl). Forming two H–Cl bonds releases −2(431) = −862 kJ and drops to the products, which lie below the reactants: ΔH ≈ 678 − 862 = −184 kJ.
Breaking the reactant bonds absorbs energy (up); forming the product bonds releases energy (down). The net change is ΔH. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Bonded atoms attract each otherbreaking a bond absorbs energy, its bond enthalpy (always positive)
  2. Separate atoms are pulled together into a bondforming the bond releases the same amount of energy
  3. A reaction breaks the reactant bonds and forms the product bondsΔH ≈ Σ(bonds broken) − Σ(bonds formed)
  4. Strong bonds form from weak onesmore energy is released than absorbed: exothermic
  5. Table values are averages over many moleculesthe result is an estimate, close to but not equal to the measured ΔH

Part 6 · Key ideas

Key ideas

  • A bond enthalpy is the energy needed to break 1 mol of a bond in gas-phase molecules; it is always positive.
  • ΔH ≈ Σ(bonds broken) − Σ(bonds formed). Broken first; this is not "products minus reactants".
  • Draw the structures and count every bond, including coefficients: 2 O₂ means 2 O=O.
  • Multiple bonds are stronger than single bonds between the same atoms.
  • Average values make the answer an estimate; it is meant for gas-phase reactions.

Part 7 · Misconception

A common mistake

The wrong idea: Energy is stored in chemical bonds and released when the bonds break.

What actually happens: Breaking a bond always absorbs energy. Energy is released when bonds form; a reaction is exothermic when the bonds formed are stronger overall than those broken.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Average bond enthalpies

Average bond enthalpies for bonds in gas-phase molecules. Use them for every item in this set.

Average bond enthalpies (kJ/mol)
BondBond enthalpy (kJ/mol)BondBond enthalpy (kJ/mol)
H–H436N≡N941
C–H413N–H391
C–C348H–Cl431
C=C614Cl–Cl242
O–H463C–Cl328
O=O495H–F567
C=O (in CO₂)799F–F155

1. Estimate ΔH, in kJ/mol, for CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g). (CO₂ is O=C=O; water is H–O–H.)

Type a number and its unit.

Show the answer

Broken: 4 C–H + 2 O=O = 4(413) + 2(495) = 2642 kJ. Formed: 2 C=O + 4 O–H = 2(799) + 4(463) = 3450 kJ. ΔH ≈ broken − formed = 2642 − 3450 = −808 kJ/mol.

  • Answer: -808 kJ/mol

2. Estimate ΔH, in kJ/mol, for the hydrogenation of ethene: H₂C=CH₂(g) + H–H(g) → H₃C–CH₃(g).

Type a number and its unit.

Show the answer

Only the changed bonds matter (the four C–H bonds of ethene stay). Broken: C=C + H–H = 614 + 436 = 1050 kJ. Formed: C–C + 2 C–H = 348 + 2(413) = 1174 kJ. ΔH ≈ 1050 − 1174 = −124 kJ/mol.

  • Answer: -124 kJ/mol

3. Estimate ΔH, in kJ/mol, for N₂(g) + 3 H₂(g) → 2 NH₃(g).

Type a number and its unit.

Show the answer

Broken: N≡N + 3 H–H = 941 + 3(436) = 2249 kJ. Formed: 2 NH₃ have 6 N–H = 6(391) = 2346 kJ. ΔH ≈ 2249 − 2346 = −97 kJ/mol.

  • Answer: -97 kJ/mol

4. The measured ΔH for N₂(g) + 3 H₂(g) → 2 NH₃(g) is −92.2 kJ/mol. Why does the estimate from the table differ slightly?

  1. The estimate counts the bonds broken but leaves out the bonds formed.
  2. Bond enthalpies give the energy released when bonds break, so the sign is reversed.
  3. The measured value includes energy to melt the ammonia, which the estimate leaves out.
  4. The table lists averages over many molecules, not the N–H bond strength in NH₃ itself.
Show the answer

Bond enthalpies are averages for a bond type across many compounds, so a bond-enthalpy ΔH is an estimate; here −97 kJ/mol versus −92.2 kJ/mol measured.

  • The estimate counts the bonds broken but leaves out the bonds formed.: Both are counted: ΔH ≈ broken − formed.
  • Bond enthalpies give the energy released when bonds break, so the sign is reversed.: Breaking a bond absorbs energy, and the sign of the estimate matches the measured value.
  • The measured value includes energy to melt the ammonia, which the estimate leaves out.: All three substances are gases at these conditions; nothing melts.
  • Correct: The table lists averages over many molecules, not the N–H bond strength in NH₃ itself.: Right: an N–H bond's strength depends on the rest of the molecule; 391 kJ/mol is an average.

5. Which expression gives ΔH from average bond enthalpies?

  1. Sum of bond enthalpies of bonds formed − sum of bond enthalpies of bonds broken
  2. Sum of bond enthalpies of bonds broken − sum of bond enthalpies of bonds formed
  3. Sum of bond enthalpies of the products − sum of bond enthalpies of the reactants
  4. Sum of bond enthalpies of the bonds in the reactants and products
Show the answer

ΔH ≈ Σ(bonds broken) − Σ(bonds formed). Breaking bonds costs energy (+); forming them pays energy back (−).

  • Sum of bond enthalpies of bonds formed − sum of bond enthalpies of bonds broken: This gives the right size with the wrong sign. Bonds broken come first.
  • Correct: Sum of bond enthalpies of bonds broken − sum of bond enthalpies of bonds formed: Right: energy in to break bonds, minus energy out from forming them.
  • Sum of bond enthalpies of the products − sum of bond enthalpies of the reactants: This is the "products minus reactants" habit; with bond enthalpies it gives the wrong sign.
  • Sum of bond enthalpies of the bonds in the reactants and products: Breaking and forming have opposite signs, so they cannot simply be added.

6. A student says, "Fuels like gasoline store energy in their bonds, and the energy is released when those bonds break." Which statement best corrects this?

  1. Breaking the fuel's bonds releases energy, but just at a high temperature.
  2. Breaking the fuel's bonds absorbs energy; burning releases energy because the bonds formed in CO₂ and H₂O are stronger.
  3. The energy comes from the oxygen's bonds breaking, not the fuel's.
  4. The energy comes from the fuel's bonds getting stronger as it burns.
Show the answer

Energy is released when bonds form. Burning a fuel is exothermic because the C=O and O–H bonds formed in CO₂ and H₂O release more energy than breaking the fuel's bonds and the O=O bonds absorbs; for methane, 2 C=O and 4 O–H formed outweigh 4 C–H and 2 O=O broken.

  • Breaking the fuel's bonds releases energy, but just at a high temperature.: Breaking bonds absorbs energy at any temperature.
  • Correct: Breaking the fuel's bonds absorbs energy; burning releases energy because the bonds formed in CO₂ and H₂O are stronger.: Right: the energy comes from the difference, with strong product bonds forming.
  • The energy comes from the oxygen's bonds breaking, not the fuel's.: Breaking O=O bonds also absorbs energy.
  • The energy comes from the fuel's bonds getting stronger as it burns.: The fuel molecules are taken apart; new, stronger bonds form in the products.

7. Using H–H 436, O=O 495 and O–H 463 kJ/mol, what is the estimated ΔH for 2 H₂(g) + O₂(g) → 2 H₂O(g)?

  1. +485 kJ/mol
  2. +5 kJ/mol
  3. −485 kJ/mol
  4. +441 kJ/mol
Show the answer

Count every bond: 2 H–H and 1 O=O broken; 2 × 2 = 4 O–H formed. ΔH ≈ 1367 − 1852 = −485 kJ/mol.

  • +485 kJ/mol: Size right, sign wrong: you subtracted broken from formed.
  • +5 kJ/mol: You counted one H–H bond and two O–H bonds; two H₂ react and two H₂O (four O–H) form.
  • Correct: −485 kJ/mol: Right: broken 2(436) + 495 = 1367; formed 4(463) = 1852; ΔH ≈ 1367 − 1852 = −485 kJ/mol.
  • +441 kJ/mol: Each water molecule has two O–H bonds, so four O–H bonds form.

Part 9 · Summary

Summary

Bond enthalpies give an estimate of ΔH: add the bond enthalpies of the bonds broken, then subtract those of the bonds formed. Breaking bonds absorbs energy and forming them releases it, so a reaction that forms stronger bonds than it breaks is exothermic.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections