Bond enthalpies — estimating ΔH_rxn from bond energies: Exercises

Unless a problem gives its own table, use this shared one (all kJ·mol⁻¹, gas phase):

Level 1 — Recognition

L1.1

Is the process $Cl-Cl (g) \to 2 Cl (g)$ endothermic or exothermic, and what is $Δ H$ for it?

Recall Solution

This is literally the definition of a bond enthalpy: breaking one mole of Cl–Cl in the gas phase into free atoms. Breaking a bond always costs energy (you pull the atoms apart against their attraction), so it is endothermic and $Δ H = + 242$ kJ·mol⁻¹ (positive).

L1.2

In the master equation, which term carries the positive contribution and which the negative?

Recall Solution

Bonds broken → positive (you spend energy). $+ \sum Δ H_{broken}$
Bonds formed → negative (energy is paid back). $- \sum Δ H_{formed}$

L1.3

Which state-function law lets us pretend the reaction goes through free gaseous atoms even though it never really does?

Recall Solution

Hess's Law — because enthalpy is a state function, $Δ H$ depends only on start and end states, not the path. So the imaginary "atomize everything, then rebuild" path gives the same answer as the real one.

Level 2 — Application

L2.1

Estimate $Δ H$ for $H_{2} (g) + Cl_{2} (g) \to 2 HCl (g)$

Recall Solution

Why subtract this way? First we spend energy tearing the reactant molecules into free atoms (that's the broken sum, uphill on the see-saw). Then those atoms snap into the product bonds, giving energy back (the formed sum, downhill). The net height change is broken − formed. Broken: 1×H–H + 1×Cl–Cl $= 436 + 242 = 678$ Formed: 2×H–Cl $= 2 (431) = 862$ $Δ H = 678 - 862 = - 184 kJ\cdotmol^{- 1}$ The downhill (862) is bigger than the uphill (678), so we finish below where we started: exothermic. Strong H–Cl bonds are built. ✅

L2.2

Estimate $Δ H$ for the hydrogenation of ethyne to ethene: $HC \equiv CH (g) + H_{2} (g) \to H_{2} C = CH_{2} (g)$

Recall Solution

Why these bonds? On the reactant side we must break everything that changes: the triple bond of ethyne and the H–H we add. On the product side we count every bond ethene now has. The subtraction (broken − formed) then gives the net energy: uphill to atoms, downhill to products. Ethyne has 1 C≡C and 2 C–H. Ethene has 1 C=C and 4 C–H. Broken: 1×C≡C + 2×C–H + 1×H–H $= 839 + 826 + 436 = 2101$ Formed: 1×C=C + 4×C–H $= 614 + 1652 = 2266$ $Δ H = 2101 - 2266 = - 165 kJ\cdotmol^{- 1}$ Formed (2266) exceeds broken (2101), so exothermic.

L2.3

Estimate $Δ H$ for the synthesis of ammonia: $N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g)$

Recall Solution

Draw the Lewis structures first: N₂ has 1 N≡N; each NH₃ has 3 N–H bonds, so 2 NH₃ → 6 N–H. Broken: 1×N≡N + 3×H–H $= 945 + 3 (436) = 945 + 1308 = 2253$ Formed: 6×N–H $= 6 (391) = 2346$ $Δ H = 2253 - 2346 = - 93 kJ\cdotmol^{- 1}$

Level 3 — Analysis

L3.1

For the ethene hydrogenation from the parent note, $Δ H = - 124$ kJ·mol⁻¹. Without recomputing the whole thing, explain in bond-accounting terms why it is negative. What single net change dominates?

Recall Solution

The bookkeeping cancels most bonds. Look at the figure below: 4 C–H bonds sit on both sides, unchanged, so they cancel and can be ignored. Strip them away and only these change:

Lose: 1 C=C (614) and 1 H–H (436) → cost 1050.
Gain: 1 C–C (348) and 2 extra C–H (2×413 = 826) → payback 1174.

Figure — Bond enthalpies — estimating ΔH_rxn from bond energies

Payback (1174) beats cost (1050) by 124. So $Δ H = - 124$ kJ·mol⁻¹. The dominating effect: the double bond's "extra" second bond (614 − 348 = 266 over a single C–C) is weaker than the two fresh C–H bonds worth of new bonding — the H₂'s energy ends up redistributed into stronger single bonds. Net: exothermic.

L3.2

The reaction $H_{2} (g) + F_{2} (g) \to 2 HF (g)$ is far more exothermic than $H_{2} + Cl_{2}$ . Using the table, compute both and explain the difference.

Recall Solution

H₂ + F₂: broken $= 436 + 158 = 594$ ; formed $= 2 (567) = 1134$ . $Δ H = 594 - 1134 = - 540 kJ\cdotmol^{- 1}$ H₂ + Cl₂ (from L2.1) $= - 184$ kJ·mol⁻¹. Analysis: two effects stack. (1) F–F is very weak (158) — cheap to break. (2) H–F is very strong (567) — huge payback. Both push toward exothermic, so HF wins by a landslide ( $- 540$ vs $- 184$ ).

L3.3

Combustion of methane is $CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (g)$ First compute the bond-enthalpy estimate (which produces gaseous water). Then explain why the standard value, with liquid water, is $- 890$ kJ·mol⁻¹ — which assumption causes the gap, and in which direction?

Recall Solution

Compute the estimate (gaseous water).

Broken: 4 C–H + 2 O=O $= 4 (413) + 2 (498) = 1652 + 996 = 2648$ .
Formed: 2 C=O (in CO₂) + 4 O–H (2 H₂O × 2 each) $= 2 (799) + 4 (463) = 1598 + 1852 = 3450$ . $Δ H_{est} = 2648 - 3450 = - 802 kJ\cdotmol^{- 1}$ Why the gap to −890? Bond enthalpies are defined for gas-phase species only, so our estimate produces $H_{2} O (g)$ . Condensing steam to liquid releases extra energy (enthalpy of condensation ≈ −44 kJ·mol⁻¹ per mole of water, ×2 waters ≈ −88). Adding that: $- 802 + (- 88) \approx - 890$ — matching the liquid-water value. Direction: the true (liquid) reaction is more exothermic than the gas-only bond-enthalpy estimate. See Standard Enthalpy of Formation for the accurate route.

Level 4 — Synthesis

L4.1

You are given the standard enthalpy of formation of methane, $Δ H_{f}^{\circ}$ — the heat change when one mole of a compound forms from its elements in their standard states (here: solid graphite carbon and gaseous H₂): $C (s) + 2 H_{2} (g) \to CH_{4} (g), Δ H_{f}^{\circ} = - 75 kJ\cdotmol^{- 1}$ You are also given the enthalpy to atomize graphite, $C (s) \to C (g)$ , $Δ H = + 715$ kJ·mol⁻¹, and H–H = 436 kJ·mol⁻¹. Derive the average C–H bond enthalpy in methane.

Recall Solution

Build an atomic-path cycle (Hess's law). We reach the same free atoms $C (g) + 4 H (g)$ by two routes starting from $C (s) + 2 H_{2} (g)$ :

Route A: form CH₄ ( $Δ H = - 75$ kJ·mol⁻¹), then atomize it by breaking all 4 C–H bonds ( $+ 4 x$ , where $x = Δ H_{C - H}$ ).
Route B: atomize the graphite ( $+ 715$ kJ·mol⁻¹) and break the 2 H–H bonds ( $+ 2 \times 436 = + 872$ kJ·mol⁻¹).

Since both routes end at the same atoms, Hess's law makes them equal: $- 75 + 4 x = 715 + 872$ Add the right-hand side first, step by step: $715 + 872 = 1587$ $- 75 + 4 x = 1587 ⟹ 4 x = 1587 + 75 = 1662$ $x = \frac{1662}{4} = 415.5 kJ\cdotmol^{- 1}$ Reassuringly close to the tabulated average of 413 kJ·mol⁻¹. This is how the table gets made — from formation and atomization data.

L4.2

For methanol combustion, $CH_{3} OH (g) + \frac{3}{2} O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (g)$ estimate $Δ H$ . (Methanol: 3 C–H, 1 C–O, 1 O–H.)

Recall Solution

Broken: in CH₃OH → 3 C–H + 1 C–O + 1 O–H $= 3 (413) + 358 + 463 = 1239 + 358 + 463 = 2060$ . Plus $\frac{3}{2}$ O=O $= 1.5 (498) = 747$ . Total broken $= 2060 + 747 = 2807$ . Formed: CO₂ has 2 C=O $= 2 (799) = 1598$ ; 2 H₂O → 4 O–H $= 4 (463) = 1852$ . Total formed $= 1598 + 1852 = 3450$ . $Δ H = 2807 - 3450 = - 643 kJ\cdotmol^{- 1}$

Level 5 — Mastery

L5.1 (resonance correction)

The bond-enthalpy estimate for hydrogenating benzene, $C_{6} H_{6} (g) + 3 H_{2} (g) \to C_{6} H_{12} (g)$ treats benzene as one Kekulé structure (3 C=C + 3 C–C + 6 C–H) and cyclohexane as 6 C–C + 12 C–H. (a) Compute the estimate. (b) The experimental value is $- 206$ kJ·mol⁻¹. What accounts for the difference, and how big is it?

Recall Solution

(a) Count every bond on each side and apply broken − formed. Broken (benzene + 3 H₂): 3 C=C + 3 C–C + 6 C–H + 3 H–H $= 3 (614) + 3 (348) + 6 (413) + 3 (436) = 1842 + 1044 + 2478 + 1308 = 6672$ . Formed (cyclohexane): 6 C–C + 12 C–H $= 6 (348) + 12 (413) = 2088 + 4956 = 7044$ . $Δ H_{est} = 6672 - 7044 = - 372 kJ\cdotmol^{- 1}$ (Note on the C–H bookkeeping: benzene supplies 6 C–H, and the 3 added H₂ molecules supply the other 6 H atoms that become cyclohexane's remaining C–H bonds — total 12 C–H in the product. Both counts are handled automatically above by breaking benzene's 6 C–H and the 3 H–H, then forming all 12 C–H.) (b) Estimate $- 372$ vs experiment $- 206$ : benzene releases 166 kJ·mol⁻¹ less than the Kekulé model predicts. That "missing" exothermicity is the Resonance Energy — real benzene is more stable (lower in enthalpy) than the localised 3-double-bond picture, so it resists hydrogenation. Difference $= ∣ - 372 - (- 206) ∣ = 166 kJ\cdotmol^{- 1}$ .

L5.2 (forecast, compute, reconcile)

For $CH_{4} (g) + Cl_{2} (g) \to CH_{3} Cl (g) + HCl (g)$ (a) forecast the sign, then (b) compute $Δ H$ . (CH₃Cl: 3 C–H + 1 C–Cl.)

Recall Solution

(a) Forecast: we break a weak Cl–Cl (242) and one C–H (413); we form a C–Cl (328) and a strong H–Cl (431). Weak bonds broken, decent bonds formed → likely mildly exothermic. (b) Only 1 C–H breaks (the other 3 survive into CH₃Cl), so cancel the unchanged bonds: Broken: 1 C–H + 1 Cl–Cl $= 413 + 242 = 655$ . Formed: 1 C–Cl + 1 H–Cl $= 328 + 431 = 759$ . $Δ H = 655 - 759 = - 104 kJ\cdotmol^{- 1}$ Exothermic — forecast confirmed. ✅

Active Recall

Bond-enthalpy subtraction order?

Broken − Formed (B before F).

Benzene hydrogenation estimate too exothermic by how much and why?

By ≈166 kJ/mol, because of resonance stabilisation (resonance energy).

Direction of gas→liquid water correction in combustion?

Makes it more exothermic (condensation releases extra energy).

Why can a tabulated bond enthalpy be off even in the gas phase?

Values are averages; the same bond type varies with molecular context (e.g. C=O in CO₂ vs in an aldehyde).

Connections

Bond Enthalpies — Estimating ΔH_rxn from Bond Energies — the parent method these exercises drill.
Hess's Law — used explicitly in L4.1's cycle.
Standard Enthalpy of Formation — the accurate alternative referenced in L3.3 and L4.1.
Resonance Energy — the L5.1 correction.
Lewis Structures — draw them first to count bonds (L2.3, L4).
Exothermic vs Endothermic Reactions — every sign we forecast.