Enthalpy of combustion, neutralization, hydration, solution: Worked examples

The scenario matrix

Everything these four enthalpies can ask you falls into one of the cells below. Each cell has at least one worked example further down, tagged like [C1].

Cell	Scenario class	What is tricky about it	Example
C1	Combustion → formation via Hess's law	signs of "reactants minus products"	Ex 1
C2	Combustion where product state matters ( $H_{2} O (l)$ vs $g$ )	degenerate/wrong-state trap	Ex 2
C3	Solution enthalpy = lattice + hydration, positive result	endothermic dissolving (cold pack)	Ex 3
C4	Solution enthalpy, negative result	exothermic dissolving	Ex 4
C5	Solution ≈ zero (borderline / degenerate)	why entropy still drives it	Ex 5
C6	Hydration comparison across ions (charge & radius)	qualitative sign/magnitude ordering	Ex 6
C7	Neutralization, strong + strong (limiting reagent)	real-world calorimetry word problem	Ex 7
C8	Neutralization, weak acid	subtracting ionization energy	Ex 8
C9	Neutralization with excess of one reagent	zero/degenerate: extra reagent does nothing	Ex 9
C10	Exam twist: dibasic acid / 2 mol water	scaling per-mole $Δ H$	Ex 10

We use one master formula from the parent repeatedly, so pin it up:

C1 — Combustion data → enthalpy of formation

Worked example Ex 1

[C1] — $Δ_{f} H$ of ethane $C_{2} H_{6}$ Given combustion enthalpies (kJ mol⁻¹): $Δ_{c} H (C_{2} H_{6}) = - 1560$ , $Δ_{c} H (C, graphite) = - 393.5$ , $Δ_{c} H (H_{2}) = - 285.8$ . Find $Δ_{f} H$ for $2 C (s) + 3 H_{2} (g) \to C_{2} H_{6} (g)$ .

Forecast: ethane is a stable fuel — guess whether $Δ_{f} H$ comes out small-negative, big-negative, or positive.

List reactants of formation and their combustions. We form ethane from 2 carbons and 3 hydrogens. Why this step? $Δ_{f} H = \sum Δ_{c} H (reactants) - Δ_{c} H (product)$ needs the combustion of each element used. $\sum Δ_{c} H (reactants) = 2 (- 393.5) + 3 (- 285.8) = - 787 - 857.4 = - 1644.4$
Subtract the product's combustion. Why? Building ethane then burning it must equal burning the raw elements (same end point, $C O_{2} + H_{2} O$ ). So we "un-burn" ethane by subtracting. $Δ_{f} H = - 1644.4 - (- 1560) = - 1644.4 + 1560 = - 84.4 kJ mol^{- 1}$

Verify: result is a small negative number — matches the forecast for a stable-but-not-hugely-stable molecule. Units kJ mol⁻¹ ✓. Look at

: the two downward burns from the elements go deeper than the single burn from ethane, and the gap between them (dashed) is exactly the formation enthalpy.

Figure — Enthalpy of combustion, neutralization, hydration, solution

C2 — Product-state trap in combustion

Worked example Ex 2

[C2] — same reaction, wrong water state A student writes hydrogen combustion producing steam: $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$ and reads $Δ H = - 241.8$ kJ. The standard value with liquid water is $- 285.8$ kJ. Explain the $44.0$ kJ gap.

Forecast: which is more negative — the one making liquid water or steam?

Identify what standard combustion demands. Why? Standard $Δ_{c} H$ requires products in their standard states; water's standard state at 298 K is liquid, not gas.
Account for the missing step. Steam → liquid water releases the latent heat of condensation, $\approx - 44.0$ kJ mol⁻¹. Why? Condensation is exothermic; the "steam" path stopped short of it. $Δ_{c} H (l) = Δ H (to steam) + Δ H_{condense} = - 241.8 + (- 44.0) = - 285.8 kJ mol^{- 1}$

Verify: $- 285.8$ matches the tabulated standard value ✓. The lesson: liquid-water combustion is more negative, because you count the extra heat given off when steam condenses. Writing $H_{2} O (g)$ under-counts the release.

C3 – C5 — The dissolving cycle in all three signs

The dissolving of an ionic salt is a tug-of-war (see figure): lattice enthalpy pulls the total up (energy needed to rip the crystal apart), hydration enthalpy pulls it down (energy released as water hugs the ions). Whoever wins sets the sign of $Δ_{so l} H$ .

C6 — Comparing hydration across ions (no arithmetic, pure reasoning)

Worked example Ex 6

[C6] — order these hydration enthalpies Rank $Δ_{h y d} H$ (most negative first): $N a^{+}$ , $M g^{2 +}$ , $A l^{3 +}$ , $K^{+}$ . Radii (pm): $A l^{3 +} = 54$ , $M g^{2 +} = 72$ , $N a^{+} = 102$ , $K^{+} = 138$ .

Forecast: which single factor dominates — charge or size?

Rank by charge first. Why? $Δ_{h y d} H \propto q / r$ (charge density); charge is in the numerator and jumps by whole units, so $+ 3 > + 2 > + 1$ dominates. $A l^{3 +} (most negative) > M g^{2 +} > {N a^{+}, K^{+}}$
Break the +1 tie by radius. Why? Equal charge, so smaller radius = higher charge density = more exothermic. $N a^{+} (102) < K^{+} (138)$ so $N a^{+}$ is more negative. $A l^{3 +} > M g^{2 +} > N a^{+} > K^{+}$

Verify: matches known values roughly $- 4600, - 1920, - 406, - 322$ kJ mol⁻¹ — the order is exactly $A l^{3 +} > M g^{2 +} > N a^{+} > K^{+}$ ✓. Look at : the smaller, higher-charge ion pulls more water dipoles closer, so more ion–dipole bonds form → more heat out.

C7 – C10 — Neutralization, every twist

Worked example Ex 7

[C7] — strong + strong, limiting reagent word problem Mix $250$ mL of $0.40$ M HCl with $150$ mL of $0.50$ M NaOH. $Δ_{n e u t} H = - 57.1$ kJ per mol water. How much heat is released?

Forecast: which reagent runs out first, acid or base?

Moles of each. Why? Heat scales with moles of water formed, which is set by the reagent that runs out. $n (H^{+}) = 0.250 \times 0.40 = 0.100 mol; n (O H^{-}) = 0.150 \times 0.50 = 0.075 mol$
Identify limiting reagent. $O H^{-}$ (0.075) < $H^{+}$ (0.100), so base limits → 0.075 mol water forms. Why? Each $H^{+} + O H^{-} \to H_{2} O$ consumes one of each; you cannot make more water than the smaller pile allows.
Multiply. $q = 0.075 \times (- 57.1) = - 4.28 kJ$

Verify: negative ✓ (neutralization is exothermic), and $∣ q ∣$ is less than if all 0.1 mol reacted ( $- 5.71$ kJ) ✓ — sensible because base ran out. Units kJ ✓.

Recall Which cell was hardest?

The dissolving sign (C3–C5) trips most students: lattice is always $+$ , hydration always $-$ , and you add them — the subtraction happens by itself.

Recall Reveal checks

In Ex 7, why is 0.075 mol the water formed, not 0.100? ::: $O H^{-}$ (0.075 mol) is the limiting reagent; water cannot exceed the smaller reagent. In Ex 9, why doesn't the extra NaOH release more heat? ::: The excess $O H^{-}$ has no $H^{+}$ partner; spectator ions release no neutralization enthalpy. In Ex 10, why multiply moles of acid by 2? ::: $H_{2} S O_{4}$ is dibasic — two $H^{+}$ per molecule → two water molecules per mole of acid. In Ex 4, why is hydration multiplied for chloride? ::: $C a C l_{2}$ gives two $C l^{-}$ ions, each hydrated separately.

Connections

Hess's Law and Thermochemical Cycles — the state-function logic behind every cycle above
Standard Enthalpy of Formation — target of Ex 1
Lattice Enthalpy and Born–Haber Cycle — the $+$ term in C3–C5
Bond Enthalpies — why combustion is exothermic
First Law of Thermodynamics — $Δ H = q_{p}$
Spontaneity and Gibbs Free Energy — why endothermic salts (Ex 3, Ex 5) still dissolve