Hess's Law

Hess's Law says that enthalpy change depends only on start and end states. It does not depend on the path taken between them. This single idea turns thermochemistry into an accounting exercise. If a reaction is hard or dangerous to run directly, it can often be built from a sequence of reactions whose enthalpies are already known. Adding those known $\Delta H$ values, with correct signs and stoichiometric scaling, gives the $\Delta H$ of the target reaction exactly. The law rests on the fact that enthalpy is a state function. Once that is accepted, Hess's Law becomes almost algebraic: reactions can be added, reversed, multiplied, and cancelled like equations, and their $\Delta H$ values follow the same operations. This article develops the rules for combining reactions and works through the arithmetic carefully, since sign errors are the single most common mistake.

Key takeaways

Hess's Law: total $\Delta H$ for a reaction equals the sum of $\Delta H$ values for any set of steps that add up to it, because enthalpy is a state function.
Reversing a reaction flips the sign of $\Delta H$ ; scaling a reaction by factor $c$ scales $\Delta H$ by $c$ .
To combine reactions, arrange them so intermediate species cancel exactly, leaving only the target's reactants and products.
Path independence means a multi-step route and a hypothetical direct route give identical $\Delta H$ , even if the direct reaction is never actually performed.
Hess's Law is the logical basis for tabulating Enthalpies of Formation, since any reaction can be built from formation reactions of elements.
Key idea: Enthalpy changes add like vectors along a reaction path — only the net displacement (reactants → products) matters, not the route.

The core insight

Because enthalpy is a state function, $\Delta H$ for A → C is the same whether you go directly or via A → B → C. So $\Delta H_{A\to C} = \Delta H_{A\to B} + \Delta H_{B \to C}$ , and this additivity extends to any combination of known reactions that sums to the target.

The physical picture

Think of enthalpy as an altitude on a landscape, where each chemical state — a particular set of substances at specified conditions — sits at a fixed elevation. A reaction is a hike from one location (reactants) to another (products). The change in elevation, $\Delta H$ , depends only on the starting and ending points, never on which trail you took. This means two different hiking routes between the same two points must show the same net altitude change. If a direct trail is unmapped or impossible to walk, but two connecting trails through a waypoint are known, the elevation change for the direct route can be computed by adding the elevation changes of the two known trails. Chemically, the "waypoint" is an intermediate compound. A common example: carbon can burn straight to $\mathrm{CO_2}$ , or it can first form $\mathrm{CO}$ and then burn further to $\mathrm{CO_2}$ . Both routes start at C(s) + O $_2$ (g) and end at CO $_2$ (g), so both must release the same total heat.

Figure

Two routes between the same start and end states give equal net ΔH

The dashed line is the direct combustion of carbon to CO₂; the solid two-segment line goes through CO. Both drops in enthalpy total the same number, $-393.5\ \mathrm{kJ}$ , because the start and end platforms are fixed regardless of the trail.

The core relations

\Delta H_{\text{target}} = \sum_i n_i \, \Delta H_i

where:

$\Delta H_{\text{target}}$ is the enthalpy change of the reaction being sought,
$\Delta H_i$ is the enthalpy change of a known reference reaction $i$ ,
$n_i$ is the integer or fractional multiplier applied to reaction $i$ (positive if used as written, negative if reversed) so that the weighted sum of reactions equals the target reaction.

\Delta H_{\text{reverse}} = -\Delta H_{\text{forward}}

where reversing the direction of a reaction changes the sign of its enthalpy change but not its magnitude.

\Delta H_{c \times \text{rxn}} = c \, \Delta H_{\text{rxn}}

where multiplying every stoichiometric coefficient in a reaction by a constant $c$ multiplies $\Delta H$ by the same constant.

Derivation AWhy enthalpy changes simply add

Goal. Show that if reaction A → B → C is broken into steps, the total $\Delta H$ for A → C equals the sum of the step enthalpies, and derive the sign/scaling rules used to combine reactions.

Step 1. Enthalpy $H$ is a state function, meaning its value depends only on the current state of the system (composition, temperature, pressure), not on the history of how that state was reached. This is inherited directly from the definition of enthalpy as $H = U + PV$ , where $U$ , $P$ , $V$ are all themselves state functions.

Step 2. Consider states A, B, C with enthalpies $H_A$ , $H_B$ , $H_C$ . By definition,

\Delta H_{A\to B} = H_B - H_A, \qquad \Delta H_{B\to C} = H_C - H_B.

Step 3. Add these two expressions:

\Delta H_{A\to B} + \Delta H_{B\to C} = (H_B - H_A) + (H_C - H_B) = H_C - H_A.

Step 4. The right-hand side is exactly the definition of $\Delta H_{A \to C}$ , so

\Delta H_{A\to B} + \Delta H_{B\to C} = \Delta H_{A\to C}.

This is Hess's Law for a two-step path; the same telescoping argument extends to any number of intermediate states by induction.

Step 5 (reversal rule). Since $\Delta H_{A\to B} = H_B - H_A$ and $\Delta H_{B \to A} = H_A - H_B$ , it follows immediately that

\Delta H_{B\to A} = -\Delta H_{A\to B}.

Running a reaction backward simply negates $\Delta H$ , because it swaps which state is "start" and which is "end."

Step 6 (scaling rule). Enthalpy is an extensive property — it scales with the amount of substance present. If reaction R is carried out on $c$ times as much material (same reactants, same products, coefficients all multiplied by $c$ ), the enthalpy change scales proportionally:

\Delta H_{cR} = c\,\Delta H_R.

This follows because $\Delta H$ for a reaction is defined per the mole quantities written in the balanced equation; doubling all coefficients doubles the extent of reaction and hence doubles the heat released or absorbed.

Step 7 (combination rule). Given known reactions with enthalpies $\Delta H_1, \Delta H_2, \dots$ , any linear combination

n_1(\text{rxn}_1) + n_2(\text{rxn}_2) + \cdots = \text{target reaction}

(where reactions may be reversed, corresponding to negative $n_i$ ) has enthalpy

\Delta H_{\text{target}} = n_1 \Delta H_1 + n_2 \Delta H_2 + \cdots

This follows by repeated application of Steps 4–6: each reversal or scaling of a known reaction transforms its $\Delta H$ by the same operation, and the telescoping sum from Steps 2–4 guarantees that when the chemical equations add up correctly (all intermediates cancelling), the enthalpies add up correctly too.

Worked examples

Example 1 — Carbon combustion through carbon monoxide. Target reaction:

\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)} \qquad \Delta H_{\text{target}} = \, ?

Given data:

\text{(1)}\quad \mathrm{C(s) + \tfrac12 O_2(g) \rightarrow CO(g)} \qquad \Delta H_1 = -110.5\ \mathrm{kJ}

\text{(2)}\quad \mathrm{CO(g) + \tfrac12 O_2(g) \rightarrow CO_2(g)} \qquad \Delta H_2 = -283.0\ \mathrm{kJ}

Step 1. Add reaction (1) and reaction (2) directly, since both are already written in the forward, unscaled direction:

\mathrm{C(s) + \tfrac12 O_2(g) + CO(g) + \tfrac12 O_2(g) \rightarrow CO(g) + CO_2(g)}

Step 2. Cancel species appearing identically on both sides. CO(g) appears as a product in (1) and a reactant in (2), so it cancels. The two $\tfrac12\mathrm{O_2}$ terms combine to one full $\mathrm{O_2}$ :

\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}

This is exactly the target reaction, so no further scaling is needed; $n_1 = 1$ , $n_2 = 1$ .

Step 3. Sum the enthalpies with these multipliers:

\Delta H_{\text{target}} = (1)(-110.5\ \mathrm{kJ}) + (1)(-283.0\ \mathrm{kJ}) = -393.5\ \mathrm{kJ}

This matches the accepted enthalpy of combustion of graphite to CO₂, confirming the additivity used in the figure above.

Example 2 — Forming propane from graphite and hydrogen (asymmetric combination). Target reaction:

\mathrm{3\,C(s) + 4\,H_2(g) \rightarrow C_3H_8(g)} \qquad \Delta H_{\text{target}} = \,?

Given combustion data (all per mole as written, all exothermic):

\text{(1)}\quad \mathrm{C(s) + O_2(g) \rightarrow CO_2(g)} \qquad \Delta H_1 = -393.5\ \mathrm{kJ}

\text{(2)}\quad \mathrm{H_2(g) + \tfrac12 O_2(g) \rightarrow H_2O(l)} \qquad \Delta H_2 = -285.8\ \mathrm{kJ}

\text{(3)}\quad \mathrm{C_3H_8(g) + 5\,O_2(g) \rightarrow 3\,CO_2(g) + 4\,H_2O(l)} \qquad \Delta H_3 = -2220.0\ \mathrm{kJ}

Step 1. The target has 3 carbons and 4 hydrogens on the left, and one mole of $\mathrm{C_3H_8}$ on the right. Reaction (3) is the only one containing $\mathrm{C_3H_8}$ , and it is a product in the target but a reactant in (3), so (3) must be reversed. Reversed (3):

\mathrm{3\,CO_2(g) + 4\,H_2O(l) \rightarrow C_3H_8(g) + 5\,O_2(g)} \qquad \Delta H_{3,\text{rev}} = +2220.0\ \mathrm{kJ}

Step 2. Scale reaction (1) by 3, to supply 3 moles of CO₂ (needed to cancel with the reversed reaction 3):

\mathrm{3\,C(s) + 3\,O_2(g) \rightarrow 3\,CO_2(g)} \qquad 3\,\Delta H_1 = 3(-393.5) = -1180.5\ \mathrm{kJ}

Step 3. Scale reaction (2) by 4, to supply 4 moles of H₂O:

\mathrm{4\,H_2(g) + 2\,O_2(g) \rightarrow 4\,H_2O(l)} \qquad 4\,\Delta H_2 = 4(-285.8) = -1143.2\ \mathrm{kJ}

Step 4. Add all three adjusted reactions:

\mathrm{3\,C(s) + 3\,O_2(g) + 4\,H_2(g) + 2\,O_2(g) + 3\,CO_2(g) + 4\,H_2O(l)}

\rightarrow \mathrm{3\,CO_2(g) + 4\,H_2O(l) + C_3H_8(g) + 5\,O_2(g)}

Cancel $3\,\mathrm{CO_2}$ and $4\,\mathrm{H_2O}$ from both sides (they appear identically on left and right). Combine oxygen: left has $3+2 = 5$ mol O₂; right has $5$ mol O₂ — these cancel completely too:

\mathrm{3\,C(s) + 4\,H_2(g) \rightarrow C_3H_8(g)}

This is exactly the target reaction.

Step 5. Sum the enthalpies with the multipliers used ( $n_1 = 3$ , $n_2 = 4$ , $n_3 = -1$ ):

\Delta H_{\text{target}} = 3(-393.5) + 4(-285.8) + (-1)(-2220.0)

\Delta H_{\text{target}} = -1180.5 - 1143.2 + 2220.0 = -103.7\ \mathrm{kJ}

So forming one mole of propane gas from graphite and hydrogen gas releases $103.7\ \mathrm{kJ}$ . As a limiting-case check: if reaction (3) were used unreversed by mistake, the sign of the $2220.0$ term would flip, giving a wildly different (and wrong) large negative answer — a useful reminder that the reversal step, dictated purely by which side $\mathrm{C_3H_8}$ appears on, is where most errors creep in.

Limiting cases & common traps

Situation	What to check	Consequence if ignored
Species appears on both sides after adding	Must cancel exactly, same phase and state	Leftover species means the equations don't actually sum to the target
Reaction used in reverse	Flip the sign of that $\Delta H$	Sign error propagates directly into final answer
Reaction scaled by $c \ne 1$	Multiply $\Delta H$ by same $c$ , not just the coefficients	Answer off by a wrong factor
Phase labels (s, l, g, aq) differ between given and target	Treat as different states — cannot cancel or match freely	Silent inconsistency; answer may look plausible but is wrong
Fractional coefficients (e.g. $\tfrac12\mathrm{O_2}$ )	Perfectly valid — enthalpy is extensive and scales continuously	Avoiding fractions by doubling everything is fine, provided $\Delta H$ is also doubled

The chemist behind it

Germain Henri Hess formulated this law from careful calorimetry in the 1830s, well before the concept of a state function had been placed on a rigorous thermodynamic footing. He measured heats of reaction for related sequences of acid-base and dissolution processes and observed empirically that the total heat evolved was the same regardless of the route or number of intermediate steps — an experimental discovery that anticipated the first law of thermodynamics. His result gave chemists a practical tool: heats of reactions too slow, too dangerous, or too impractical to measure directly could be obtained by combining measurable ones. This is precisely the strategy later formalized through tables of standard Enthalpies of Formation, where every reaction's $\Delta H$ is built from a common set of reference values.

Origins · 1 minGermain Henri Hess and how this was discovered

Germain Henri Hess

1802–1850 · Russian Empire, mid-19th century

Trained originally as a physician, Hess turned to chemistry and became a professor in St. Petersburg. His 1840 summation of thermochemical measurements — now called Hess's Law — was among the first clear statements of path-independence for a chemical quantity, predating a full formal theory of energy conservation.

Ask the tutor: How do I decide which given reaction to reverse or scale first when the target reaction has several unfamiliar intermediates?

Does Hess's Law require the intermediate steps to be physically realizable?

No. The steps only need to be chemically balanced and thermodynamically valid on paper; the reaction sequence never has to actually occur in the lab. Hess's Law is a statement about state functions, so any consistent set of hypothetical steps connecting the same start and end states gives the correct total $\Delta H$ .

Can Hess's Law be applied to other state functions besides enthalpy?

Yes. Any state function — internal energy, entropy, Gibbs free energy — obeys the same additivity, since the underlying argument (Step 1 through Step 4 in the derivation) only used the fact that the quantity depends solely on the state, not the path.

Why must phases be tracked so carefully when combining reactions?

Enthalpy values are specific to the exact physical state of each substance (solid, liquid, gas, aqueous, and often a specified temperature and pressure). Two entries like H₂O(l) and H₂O(g) represent different states with different enthalpies, differing by the enthalpy of vaporization, so they cannot be treated as identical when cancelling terms.

How does Hess's Law connect to enthalpies of formation tables?

Every reaction can be rewritten as a combination of formation reactions (elements in their standard states forming one mole of a compound). Hess's Law guarantees that summing the formation enthalpies of products and subtracting those of reactants gives the correct reaction enthalpy — this is the basis of the standard formula used with tabulated Enthalpies of Formation values.

What is the most common arithmetic mistake when applying Hess's Law?

Forgetting to scale $\Delta H$ when a reaction's coefficients are multiplied, or forgetting to flip the sign of $\Delta H$ when a reaction is reversed. Both errors are easy to catch by explicitly writing out the multiplier $n_i$ next to each reaction before summing.