Combining Thermochemical Equations

Introduction

With three or more supplied equations, Hess's law becomes a careful bookkeeping problem. Each supplied reaction can be reversed or multiplied, but the final sum must reproduce the target equation exactly. An intermediate appears once as a product and again as a reactant, so it cancels from the net reaction. Tracking it explicitly is safer than trying to combine enthalpy numbers by intuition.

Core explanation

Write the target equation above the data and mark species that must appear on its left and right. Then treat each supplied equation as a signed row of stoichiometric coefficients: negative for reactants, positive for products. Reversing a row negates every coefficient and ΔH. Multiplying by a factor scales the entire row and ΔH. When rows are added, any species with net coefficient zero cancels. Only after the summed row equals the target row should the transformed enthalpies be added.

Consider three illustrative standard thermochemical equations: (1) C(graphite) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ; (2) CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ; (3) H₂(g) + ½O₂(g) → H₂O(l), ΔH₃ = −285.8 kJ. Find ΔH for C(graphite) + H₂O(l) → CO(g) + H₂(g). Equation 1 already has carbon on the needed reactant side. Reverse equation 2 so CO is formed: CO₂ → CO + ½O₂, ΔH = +283.0 kJ. Reverse equation 3 so liquid water is consumed: H₂O(l) → H₂ + ½O₂, ΔH = +285.8 kJ.

Adding the three manipulated equations gives CO₂ on both sides, so it cancels. The first equation consumes one O₂; the two reversed equations together produce one O₂, so oxygen cancels as well. The remaining equation is exactly C(graphite) + H₂O(l) → CO(g) + H₂(g). Its enthalpy is −393.5 + 283.0 + 285.8 = +175.3 kJ per mole of the target equation. The positive sign means the written process is endothermic under the specified standard states.

This target uses liquid water. If steam H₂O(g) were the target reactant, the numerical result would differ because vaporising water requires enthalpy. A line that says only “H₂O” cannot be cancelled against both liquid and gas without a phase-change step. Similar care applies to carbon allotropes and dissolved versus gaseous species.

For a difficult set, solve coefficients algebraically. Assign unknown multipliers a, b, c to the supplied equations; for each species, write an equation requiring its summed coefficient to equal the target coefficient. Solve that linear system, then compute ΔH target = aΔH₁ + bΔH₂ + cΔH₃. This avoids arbitrary trial and error, especially when several intermediates must vanish at once. A row of equations might be redundant; then different combinations can produce the same target, but a consistent data set should give the same enthalpy within measurement uncertainty.

Step-by-step reasoning

1. Copy the target reaction with all phases and balance it. 2. Orient and scale each supplied equation to place distinctive target species on the correct side. 3. Add left and right sides, cancelling identical intermediates in identical phases. 4. Check every surviving coefficient and phase against the target; repair any mismatch. 5. Sum the correspondingly reversed and scaled ΔH values only after equation closure is confirmed.

Visual explanation

Draw three horizontal equation rows. In row 1, CO₂ appears on the right; in reversed row 2, CO₂ appears on the left, so connect them with a cancellation line. Place O₂ consumed in row 1 opposite two half-O₂ products in the reversed rows, then cross those out. Highlight the uncancelled C and H₂O on the left and CO and H₂ on the right.

Real-world analogy

A traveller combines three legs of a journey. If one leg ends at a station and another begins there, that station is only an intermediate stop and not part of the overall start-to-finish route. Reversing a leg changes the signed elevation change; repeating a leg scales it. Hess-law equation rows use the same route-accounting idea.

Real-world example

Directly measuring the enthalpy of an industrially relevant reaction can be difficult when its reactants are hard to handle or its products mix. Related combustion measurements may be more accessible. Combining their equations gives the desired reaction enthalpy, provided phases and reference conditions are consistent.

Why?

Why do intermediates cancel in both chemistry and enthalpy accounting? A substance made in one imagined step and consumed in another is absent from the net initial and final states. Its internal energy contribution is included in one step and removed in the other, leaving only the overall state difference.

Common misconception

“If a species appears somewhere on both sides, erase it completely.” Cancel only the smaller common coefficient; a remaining excess may survive. Also cancel only identical chemical states, not H₂O(l) against H₂O(g). The final equation itself is the decisive validation.

Worked example

Given A → B, ΔH = +10 kJ; B → C, ΔH = −25 kJ; and D → C, ΔH = −5 kJ, find ΔH for A + D → 2C. First two equations sum to A → C with ΔH = −15 kJ, cancelling B. Add D → C with ΔH = −5 kJ. The net equation is A + D → 2C and ΔH = −20 kJ. The two C terms must remain because the target requires two product C units; cancelling C against a C in a supplied intermediate line before considering the full sum could hide this coefficient.

Quick check

1. Equations A → B (+12 kJ) and B → C (−20 kJ) are added. What net equation and enthalpy result? Answer: B cancels, leaving A → C with ΔH = +12 − 20 = −8 kJ.

Exam focus

Show transformed equations, not just arithmetic on ΔH values. Check physical states, partial cancellation and coefficient factors. A correct enthalpy sum attached to the wrong net equation is not an answer to the target question.

Advanced insight

The coefficient-matching linear system reveals whether the supplied equations span the target reaction. If no combination solves all species-balance equations, the desired ΔH cannot be determined from that data set alone. Additional independent thermochemical information is then required.

Summary

Multi-equation Hess problems are solved by orienting and scaling each reaction, cancelling identical intermediates, and verifying the exact target equation. Enthalpy changes obey the same signed multipliers. Phase labels and leftover coefficients are part of the calculation, not decoration.

Practice questions

1. Add A → B (+5 kJ), B → C (+7 kJ) and C → D (−4 kJ). What is A → D? Answer: B and C cancel; ΔH = 5 + 7 − 4 = +8 kJ. 2. If only half of A → B is needed, what happens to its ΔH? Answer: Every coefficient and its ΔH are multiplied by ½. 3. Why would substituting H₂O(g) for H₂O(l) in the three-equation carbon cycle change the answer? Answer: Gas and liquid water have different enthalpies; a vaporisation or condensation step is needed to connect them.