Enthalpy of combustion, neutralization, hydration, solution: Exercises

One symbol reminder so nothing here is unearned:

Before the exercises, we must earn the two relations we will lean on again and again: the solution cycle and the Hess-law sign rule.

Intuition Why dissolving a salt splits into "break the lattice, then hug the ions"

Enthalpy is a state function — $Δ H$ depends only on where you start and end, never the route. So we are free to invent a convenient two-step route between the same start (solid salt) and end (hydrated ions in water) and know its total must equal the real one-step dissolving:

Step up — tear the lattice apart into free gaseous ions. This costs energy (bonds break) ⇒ $+ Δ_{l a tt i ce} H$ , endothermic. See the cyan up-arrow in the master map below.
Step down — wrap each gaseous ion in water dipoles. Ion–dipole bonds form, releasing energy ⇒ $Δ_{h y d} H$ , exothermic. See the amber down-arrow.

The real dissolving (the white arrow) is the net of these two: $Δ_{so l} H = Δ_{l a tt i ce} H + Δ_{h y d} H_{total}$ The signs already point in opposite directions, so you simply add with signs — never subtract by hand.

Figure — Enthalpy of combustion, neutralization, hydration, solution

The figure above is our master map. In each solution below we will tell you which arrow you are standing on so the picture carries the algebra.

Level 1 — Recognition

L1.1

State the sign (positive or negative) of each, and one word why: (a) $Δ_{c} H$ of ethanol, (b) $Δ_{h y d} H$ of $N a^{+}$ , (c) $Δ_{l a tt i ce} H$ of $N a C l$ (defined as solid → gaseous ions).

Recall Solution L1.1

(a) Negative — combustion always releases heat (strong $C O_{2}$ / $H_{2} O$ bonds form). $Δ_{c} H < 0$ by definition. (b) Negative — hydration is water dipoles hugging an ion; ion–dipole bonds form, releasing energy. On the master map this is the amber down-arrow. (c) Positive — pulling a solid lattice apart into gaseous ions costs energy (bonds break); this is the cyan up-arrow. No number needed; just the sign.

L1.2

For a strong acid + strong base in dilute solution, what is $Δ_{n e u t} H$ per mole of water, and why is it the same for HCl+NaOH and HNO₃+KOH?

Recall Solution L1.2

$Δ_{n e u t} H = - 57.1 kJ mol^{- 1}$ . Both strong acids and strong bases are fully dissociated in water, so the only real reaction in every case is $H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l) .$ The $N a^{+}, K^{+}, C l^{-}, N O_{3}^{-}$ are spectators — same reaction ⇒ same $Δ H$ .

Level 2 — Application

L2.1

$Δ_{l a tt i ce} H (K C l) = + 715$ , $Δ_{h y d} H (K^{+}) = - 322$ , $Δ_{h y d} H (C l^{-}) = - 378 kJ mol^{- 1}$ . Find $Δ_{so l} H (K C l)$ . Is dissolving exo- or endothermic?

Recall Solution L2.1

You are standing on the white "direct solution" arrow of the master map — compute it as the cyan up-arrow plus the amber down-arrow: $Δ_{so l} H = Δ_{l a tt i ce} H + Δ_{h y d} H_{total} .$ Total hydration = sum over both ions (both take the amber step down): $Δ_{h y d} H_{total} = - 322 + (- 378) = - 700.$ $Δ_{so l} H = + 715 + (- 700) = + 15 kJ mol^{- 1} .$ Positive ⇒ endothermic (solution cools slightly). The cyan lattice cost just exceeds the amber hydration payback.

L2.2

$100$ mL of $0.5$ M $H C l$ is fully neutralized by $N a O H$ . How much heat is released? ( $Δ_{n e u t} H = - 57.1$ .)

Recall Solution L2.2

Moles of $H^{+}$ = moles of water formed: $n = 0.100 L \times 0.5 mol L^{- 1} = 0.05 mol .$ $q = 0.05 \times (- 57.1) = - 2.855 kJ .$ Sign convention: $q < 0$ means heat leaves the reacting system and warms the surroundings. So " $q = - 2.855 kJ$ (system loses)" and " $2.855 kJ$ of heat released to the surroundings" describe the same event — the minus lives with the system, the magnitude is what a thermometer in the beaker registers as warming.

L2.3

Given $Δ_{c} H$ : $C = - 393.5$ , $H_{2} = - 285.8$ , $C_{2} H_{6} = - 1560$ . Find $Δ_{f} H (C_{2} H_{6})$ for $2 C (s) + 3 H_{2} (g) \to C_{2} H_{6} (g)$ .

Recall Solution L2.3

Why this formula (walk the two legs): to form $C_{2} H_{6}$ from its elements, first burn the elements ( $2 C + 3 H_{2}$ , releasing their combustion enthalpies), then un-burn $C_{2} H_{6}$ (run its combustion backwards, which flips the sign of its $Δ_{c} H$ ). Note $O_{2}$ is not a fuel — it contributes $0$ , so it never appears in the sum. $Δ_{f} H = \sum Δ_{c} H (reactants) - \sum Δ_{c} H (products) .$ Reactants burnt: $2 C + 3 H_{2}$ ; product un-burnt: $C_{2} H_{6}$ . $Δ_{f} H = [2 (- 393.5) + 3 (- 285.8)] - (- 1560)$ Why the $\times 2$ and $\times 3$ : each $Δ_{c} H$ is per mole, and forming one $C_{2} H_{6}$ consumes 2 mol C and 3 mol H₂. $= (- 787 - 857.4) + 1560 = - 1644.4 + 1560 = - 84.4 kJ mol^{- 1} .$

Level 3 — Analysis

L3.1

A weak acid $C H_{3} C O O H$ neutralized by $N a O H$ releases only $- 55.2 kJ mol^{- 1}$ . What is its enthalpy of ionization?

Recall Solution L3.1

A weak acid must first ionize (endothermic, $+ Δ H_{i o n}$ ) before its $H^{+}$ can meet $O H^{-}$ : $Δ_{n e u t} H_{weak} = - 57.1 + Δ H_{i o n} .$ Solve: $- 55.2 = - 57.1 + Δ H_{i o n} \Rightarrow Δ H_{i o n} = - 55.2 + 57.1 = + 1.9 kJ mol^{- 1} .$ Positive, as it must be — ionization costs energy, so less net heat is released.

L3.2

Two salts, both giving $Δ_{so l} H$ from lattice/hydration. Salt X: $Δ_{l a tt i ce} H = + 780, Δ_{h y d} H = - 820$ . Salt Y: $Δ_{l a tt i ce} H = + 700, Δ_{h y d} H = - 640$ . Which dissolves exothermically, and by how much do the two $Δ_{so l} H$ differ?

Recall Solution L3.2

Both salts walk the same master-map route (cyan up + amber down): $Δ_{so l} H (X) = 780 - 820 = - 40 kJ mol^{- 1} (exothermic) .$ $Δ_{so l} H (Y) = 700 - 640 = + 60 kJ mol^{- 1} (endothermic, like a cold pack) .$ X dissolves exothermically. Difference: $Δ_{so l} H (Y) - Δ_{so l} H (X) = 60 - (- 40) = 100 kJ mol^{- 1} .$

Level 4 — Synthesis

L4.1 (build the cycle yourself)

You are given $Δ_{so l} H (N a B r) = - 1 kJ mol^{- 1}$ , $Δ_{l a tt i ce} H (N a B r) = + 742$ , and $Δ_{h y d} H (N a^{+}) = - 406$ . Find $Δ_{h y d} H (B r^{-})$ .

Recall Solution L4.1

Read the master map backwards: you know the white direct arrow and the cyan up-arrow, and part of the amber down-arrow — solve for the missing piece: $Δ_{so l} H = Δ_{l a tt i ce} H + [Δ_{h y d} H (N a^{+}) + Δ_{h y d} H (B r^{-})] .$ $- 1 = 742 + (- 406) + Δ_{h y d} H (B r^{-}) .$ $- 1 = 336 + Δ_{h y d} H (B r^{-}) \Rightarrow Δ_{h y d} H (B r^{-}) = - 1 - 336 = - 337 kJ mol^{- 1} .$ Negative — correct for a hydration enthalpy. $B r^{-}$ is larger than $C l^{-}$ , so its hydration is less exothermic than $C l^{-}$ 's $- 378$ ; the sign of our answer is physically sensible.

L4.2

Using combustion data $Δ_{c} H$ : $C = - 393.5$ , $H_{2} = - 285.8$ , $C_{2} H_{5} O H (l) = - 1367$ , and $Δ_{f} H$ of $C O_{2} = - 393.5$ , $H_{2} O (l) = - 285.8$ , verify by two routes that $Δ_{f} H (C_{2} H_{5} O H)$ for $2 C (s) + 3 H_{2} (g) + \frac{1}{2} O_{2} (g) \to C_{2} H_{5} O H (l)$ is the same. Compute it.

Recall Solution L4.2

Route 1 — via combustion data. Why: form the elements by burning them, then un-burn ethanol (reverse its combustion ⇒ subtract it). $O_{2}$ contributes $0$ , so it drops out: $Δ_{f} H = \sum Δ_{c} H (reactants) - \sum Δ_{c} H (products) = [2 (- 393.5) + 3 (- 285.8)] - (- 1367) .$ $= - 1644.4 + 1367 = - 277.4 kJ mol^{- 1} .$ Route 2 — cross-check via the combustion equation of ethanol. Why: the products $C O_{2}$ and $H_{2} O$ have known $Δ_{f} H$ , and $Δ_{c} H = \sum Δ_{f} H (prod) - \sum Δ_{f} H (react)$ , with $Δ_{f} H (O_{2}) = 0$ : $C_{2} H_{5} O H + 3 O_{2} \to 2 C O_{2} + 3 H_{2} O, - 1367 = [2 (- 393.5) + 3 (- 285.8)] - Δ_{f} H (C_{2} H_{5} O H) .$ $- 1367 = - 1644.4 - Δ_{f} H (C_{2} H_{5} O H) \Rightarrow Δ_{f} H (C_{2} H_{5} O H) = - 1644.4 - (- 1367) = - 277.4 kJ mol^{- 1} . ✓$ Both routes agree — a demonstration of Hess's Law and Thermochemical Cycles.

Level 5 — Mastery

L5.1 (full multi-step, exam-grade)

A student prepares a "cold pack" by dissolving $8.0$ g of $N H_{4} N O_{3}$ (molar mass $80 g mol^{- 1}$ ) in $100$ g of water. Given $Δ_{so l} H (N H_{4} N O_{3}) = + 25.7 kJ mol^{- 1}$ and the solution's specific heat $c = 4.18 J g^{- 1} K^{- 1}$ (treat the $108$ g solution as absorbing all the heat), find the temperature change of the solution.

Recall Solution L5.1

Step 1 — moles of salt. $n = 8.0/80 = 0.10 mol$ . Step 2 — heat absorbed by dissolving. $Δ_{so l} H > 0$ means the salt walks a net uphill route on the master map (lattice cost beats hydration payback), so heat is taken from the water: $q_{absorbed} = n Δ_{so l} H = 0.10 \times 25.7 = 2.57 kJ = 2570 J .$ Step 3 — that heat comes out of the solution, cooling it. Using $q = m c Δ T$ with $m = 108 g$ : $Δ T = \frac{- q _{absorbed}}{m c} = \frac{- 2570}{108 \times 4.18} = \frac{- 2570}{451.44} = - 5.69 K .$ The solution cools by about $5.7^{\circ} C$ — exactly how a cold pack works. This ties to Spontaneity and Gibbs Free Energy: it still dissolves spontaneously because the entropy gain outweighs the endothermic $Δ H$ .

L5.2 (sign-and-cycle capstone)

For $M g C l_{2}$ : $Δ_{l a tt i ce} H = + 2526$ , $Δ_{h y d} H (M g^{2 +}) = - 1920$ , $Δ_{h y d} H (C l^{-}) = - 378 kJ mol^{- 1}$ . (a) Compute $Δ_{so l} H$ . (b) Explain, using charge density, why $Δ_{h y d} H (M g^{2 +})$ is far more negative than $Δ_{h y d} H (N a^{+}) = - 406$ .

Recall Solution L5.2

(a) On the master map the amber down-arrow now bundles two chloride ions plus one magnesium ion: $Δ_{h y d} H_{total} = - 1920 + 2 (- 378) = - 1920 - 756 = - 2676.$ $Δ_{so l} H = + 2526 + (- 2676) = - 150 kJ mol^{- 1} (exothermic) .$ (b) Hydration strength $\propto q / r$ (charge density). $M g^{2 +}$ has charge $+ 2$ (vs $+ 1$ ) and a smaller radius than $N a^{+}$ , so its charge is far more concentrated. Water dipoles are pulled in much harder ⇒ far more energy released ⇒ much more negative $Δ_{h y d} H$ . (See Lattice Enthalpy and Born–Haber Cycle.)

Recall One-line self-test

A salt has $Δ_{l a tt i ce} H = + 600$ and total $Δ_{h y d} H = - 590$ . Exo or endo? ::: Endothermic: $Δ_{so l} H = + 600 - 590 = + 10 kJ mol^{- 1}$ .

Connections

Hess's Law and Thermochemical Cycles — every route/cross-check above is Hess's law.
Standard Enthalpy of Formation — the $Δ_{f} H$ from combustion problems.
Lattice Enthalpy and Born–Haber Cycle — the lattice/hydration split.
Bond Enthalpies — why combustion is exothermic at the bond level.
First Law of Thermodynamics — $Δ H = q_{p}$ .
Spontaneity and Gibbs Free Energy — why endothermic salts still dissolve.