Exercises — Hess's law — enthalpy is a state function; enthalpy cycles

WHAT we do: the target reaction is the given one written backwards. WHY: $\Delta H=H_{\text{final}}-H_{\text{initial}}$. Swapping start and end negates the difference. $\Delta H_{\text{reverse}}=-(-286)=\boxed{+286\text{kJ}}$ Splitting water back into its elements costs energy — positive, endothermic. Makes sense: you had to add energy to un-burn it.

WHAT we do: the target is the given reaction ×$\frac{1}{2}$ (one $N{H}_3$ instead of two). WHY: enthalpy is extensive — half the substance, half the enthalpy change. $\Delta H=\frac{1}{2}\times (-92)=\boxed{-46\text{kJ}}$

Recall ${\Delta }_f{H}^{\circ }$ = $\Delta H$ of forming the substance from its elements. Both $O_2(g)$ and $B{r}_2(l)$ are elements in their standard state (the most stable form at 1 bar, 298 K). WHY zero: forming an element from itself is no change — the start and end are identical, so $H_{\text{final}}-H_{\text{initial}}=0$. ${\Delta }_f{H}^{\circ }(O_2)=0,{\Delta }_f{H}^{\circ }(B{r}_2,l)=0$ Careful: $B{r}_2(g)$ is not the standard state (liquid is), so its formation enthalpy is not zero. See Standard states and conventions.

WHAT — orient the steps: target needs $CO$ as a product; in (ii) it is a reactant, so reverse (ii) (its $\Delta H$ becomes $+283.0$). Keep (i) as-is. WHY reverse only (ii): the target already has $C+\frac{1}{2}{O}_2$ on the left, matching (i)'s left side, so (i) needs no change; only $CO$'s position is wrong, and that lives in (ii). WHY add: once oriented, stacking (i) + reversed (ii) is a two-leg path $C\to C{O}_2\to CO$; Hess says the path total equals the direct $\Delta H$. $\Delta H=\Delta H_1-\Delta H_2=-393.5-(-283.0)=\boxed{-110.5\text{kJ}}$ WHY $C{O}_2$ vanishes: it is a product in (i) and a reactant in reversed (ii) — same species, same state, so it is a "middleman" that cancels (telescoping).

WHY use the formation formula: we have ${\Delta }_f{H}^{\circ }$ (formation) data, not the reaction directly. The cycle "reactants → elements → products" lets us reach the target: tear reactants down into elements (that costs $-\sum {\Delta }_f{H}_{\text{react}}$, a reversal) then build products from elements ($+\sum {\Delta }_f{H}_{\text{prod}}$). ${\Delta }_r{H}^{\circ }=[{\Delta }_f{H}^{\circ }(C{O}_2)+2{\Delta }_f{H}^{\circ }(H_{2}O)]-[{\Delta }_f{H}^{\circ }(C{H}_4)+2{\Delta }_f{H}^{\circ }(O_2)]$ WHY the coefficient 2 on water: ${\Delta }_f{H}^{\circ }$ is per mole; there are 2 mol of $H_{2}O$, and enthalpy is extensive, so multiply. WHY $O_2$ drops out: it is an element in its standard state, ${\Delta }_f{H}^{\circ }=0$ (L1.3). $=[(-393.5)+2(-285.8)]-[(-74.8)+0]=(-965.1)-(-74.8)=\boxed{-890.3\text{kJ/mol}}$ See Enthalpy of combustion.

WHY this path: we want to travel $A\to D$. We can reach it by hopping through the states we have data for: $A\to B\to C\to D$. Hess guarantees the hop-total equals the direct $A\to D$. WHY flip the last leg: our route needs $C\to D$, but the datum is $D\to C(+30)$; reversing an arrow flips the sign, so $C\to D=-30$. WHY the others are added as-is: $A\to B$ and $B\to C$ already point the way our path runs, so no sign change. $\Delta H=(+50)+(-120)+(-30)=\boxed{-100\text{kJ}}$

Target: $2C(s)+3H_2(g)\to C_2{H}_6(g)$ (that is exactly the formation reaction of ethane, hence ${\Delta }_f{H}^{\circ }$). WHY the combustion formula: all three data are ${\Delta }_{c}H$ (combustion) values. Everything burns down to the same products, $C{O}_2$ and $H_{2}O$, so those act as a common reference state. Using ${\Delta }_{r}H=\sum {\Delta }_c{H}_{\text{react}}-\sum {\Delta }_c{H}_{\text{prod}}$: burn the reactants ($2C+3H_2$) down to $C{O}_2+H_{2}O$, then un-burn ethane back up from those same products. ${\Delta }_f{H}^{\circ }=\underset{\text{burn }2C}{\underbrace{2(-393.5)}}+\underset{\text{burn }3H_2}{\underbrace{3(-285.8)}}-\underset{\text{un-burn }{C}_2{H}_6}{\underbrace{(-1560.0)}}$ WHY multiply (i) by 2 and (ii) by 3: the target has $2C$ and $3H_2$; extensive property × stoichiometric coefficient. WHY subtract (iii): in the assembled cycle ethane is the product, so its combustion runs in reverse (a sign flip), giving $+1560.0$. $=-787.0-857.4+1560.0=\boxed{-84.4\text{kJ/mol}}$ See Enthalpy of formation.

WHY a difference of formations: both waters are built from the same elements ($H_2+\frac{1}{2}{O}_2$). So "elements → gas" and "elements → liquid" share a start. Subtracting the two formation legs cancels the common element start and leaves exactly liquid → gas. WHY gas minus liquid (this order): the target goes liquid → gas, so gas is the product (adds $+$) and liquid is the reactant (subtracts). $\Delta H_{\text{vap}}={\Delta }_f{H}^{\circ }[H_{2}O(g)]-{\Delta }_f{H}^{\circ }[H_{2}O(l)]=-241.8-(-285.8)=\boxed{+44.0\text{kJ/mol}}$ Positive: turning liquid into vapour costs energy. This is exactly why phase labels matter.

WHY this is a Hess cycle: route the reaction through free atoms — first snap every reactant bond (reactants → loose atoms), then let atoms fall into product bonds (atoms → products). Same start, same finish as the direct reaction, so Hess applies. Rule: $\Delta H=\sum (\text{bonds broken})-\sum (\text{bonds formed})$. WHY broken adds and formed subtracts: breaking a bond absorbs energy (positive leg, going up to atoms); forming a bond releases energy (negative leg, coming back down). Broken: 1 $H-H$ + 1 $Cl-Cl$ $=436+242=678$. Formed: 2 $H-Cl$ $=2(431)=862$ (the coefficient 2 because two $HCl$ molecules form). $\Delta H=678-862=\boxed{-184\text{kJ}}$ See Bond enthalpies.

WHY two routes: both the short direct route and the long "via gaseous ions" route start at $Na(s)+\frac{1}{2}C{l}_2(g)$ and end at $NaCl(s)$. Same endpoints ⟹ same total $\Delta H$ (Hess). Read this off the figure below. Route 1 (direct): ${\Delta }_{f}H=-411$. Route 2 (long way): sublimation → ionisation → dissociation → electron gain → lattice (the unknown red arrow). WHY we can solve for the unknown: set the two route totals equal and isolate $\Delta H_{\text{latt}}$. $-411=108+496+121+(-349)+\Delta H_{\text{latt}}$ Sum of known long-route steps: $108+496+121-349=376$. $\Delta H_{\text{latt}}=-411-376=\boxed{-787\text{kJ/mol}}$ Strongly negative — forming the ionic crystal from gaseous ions releases huge energy. See Born–Haber cycle.

WHY scale both to "2": the target makes 2 $N{O}_2$ and consumes 2 $NO$, but the data are written per 1 mole. Scaling first makes the middle species ($N_2$) cancel cleanly.

• Use (ii) ×2 forward: $N_2+2O_2\to 2N{O}_2$, $\Delta H=2(+33.2)=+66.4$. WHY forward: $N{O}_2$ is a product in the target, matching (ii)'s direction.
• Use (i) ×2 reversed: $2NO\to N_2+O_2$, $\Delta H=-2(90.3)=-180.6$. WHY reversed: $NO$ is a reactant in the target but a product in (i); flip it (and the sign) to move $NO$ to the left. WHY add: with $N_2$ (and one $O_2$) appearing on opposite sides, they cancel and the two oriented equations sum exactly to the target. $\Delta H=66.4+(-180.6)=\boxed{-114.2\text{kJ}}$

WHY the formation formula again: we have ${\Delta }_f{H}^{\circ }$ data, so use products-minus-reactants (each side built from elements as the common reference). ${\Delta }_r{H}^{\circ }=[2(-393.5)+3(-285.8)]-[(-277.7)+3(0)]$ WHY coefficients 2 and 3: per-mole formation enthalpies × the 2 mol $C{O}_2$ and 3 mol $H_{2}O$. WHY $O_2$ is zero: element in standard state. $=[-787.0-857.4]-(-277.7)=-1644.4+277.7=\boxed{-1366.7\text{kJ/mol}}$ Watch that $H_{2}O$ is liquid, not gas — a different phase would change the answer.

WHY combustions can be subtracted: both allotropes burn to the same $C{O}_2$, a shared reference. Route: graphite → $C{O}_2$ (forward), then $C{O}_2$ → diamond (reverse of diamond's combustion). WHY graphite-minus-diamond (this order): graphite is the reactant of the transition (its combustion runs forward, $-393.5$); diamond is the product (its combustion runs in reverse, contributing $+395.4$). $\Delta H_{\text{trans}}={\Delta }_{c}H(\text{graphite})-{\Delta }_{c}H(\text{diamond})=-393.5-(-395.4)=\boxed{+1.9\text{kJ/mol}}$ Positive: diamond sits slightly higher in enthalpy — graphite is the more stable allotrope (its ${\Delta }_f{H}^{\circ }=0$ by convention).

Target: $N_2{H}_4(l)+O_2\to N_2+2H_{2}O(l)$.

• Take (ii) forward: $N_2{H}_4+3O_2\to 2N{O}_2+2H_{2}O$, $\Delta H=-534.2$. WHY forward: it already produces the $N_2{H}_4$-consumption and the 2 $H_{2}O$ we want.
• Take (i) reversed: $2N{O}_2\to N_2+2O_2$, $\Delta H=-66.4$. WHY reversed: the target has $N_2$ as a product, but in (i) $N_2$ is a reactant; flipping moves $N_2$ to the right and removes the unwanted $2N{O}_2$ (which (ii) produced). WHY add: $2N{O}_2$ cancels (product in (ii), reactant in reversed (i)); the $O_2$ tally becomes $3-2=1$ on the reactant side ✓, leaving exactly the target. $\Delta H=-534.2+(-66.4)=\boxed{-600.6\text{kJ}}$ WHY (iii) is a decoy: the target's water is already delivered and balanced by (ii); adding (iii) would double-count hydrogen. A correct cycle uses only the steps whose middlemen cancel.

• L1–L2: you can flip signs and scale, and run a given cycle.
• L3: you can pick the recipe from formation/combustion/bond data yourself.
• L4: you build multi-step and Born–Haber cycles, combining reversal and scaling.
• L5: you handle phase traps, allotropes, and reject decoy data. If any level tripped you, revisit that [!mistake] callout — the trap is the lesson.