Enthalpy as a State Function
A thermodynamic state function is a quantity whose value depends only on the present state of the system — its temperature, pressure, composition and physical state — and not on the path by which the system arrived there. Internal energy $U$, pressure $p$ and volume $V$ are all state functions. Enthalpy is defined as $H = U + pV$, and because it is built entirely from state functions, $H$ is itself a state function.
The consequence is decisive. For a chemical reaction running from a fixed set of reactants to a fixed set of products, the change $\Delta_r H = H_{\text{products}} - H_{\text{reactants}}$ is fixed by the two end states alone. It does not matter whether the change happens in one violent step or through a long chain of intermediates; the total enthalpy change is the same. This is the entire physical content of Hess's law — everything that follows is bookkeeping.
State function versus path function
Enthalpy change $\Delta H$ and internal-energy change $\Delta U$ depend only on the initial and final states, so they are state functions. Heat $q$ and work $w$ depend on how the change is carried out, so they are path functions. Hess's law applies to $\Delta H$ precisely because $H$ is a state function — it would fail for a path-dependent quantity.
Only path-independent quantities (here $\Delta H$) may be summed over alternative routes.
The Statement of Hess's Law
NCERT states the law in the following form:
If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided, at the same temperature.
Symbolically, if an overall reaction $\ce{A -> B}$ proceeds along an alternative route through intermediate steps with enthalpy changes $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3, \dots$, then the overall enthalpy change is the sum
$$\Delta_r H = \Delta_r H_1 + \Delta_r H_2 + \Delta_r H_3 + \cdots = \sum_i \Delta_r H_i$$
The phrase "at the same temperature" is essential — all the steps must be compared under the same conditions, conventionally the standard state. The law lets us compute an inaccessible enthalpy change from a set of accessible ones, which is why it is sometimes called the law of constant heat summation.
Manipulating Thermochemical Equations
To combine reactions you treat their thermochemical equations as algebraic objects. NCERT lists three operations that follow from the extensive, state-function nature of enthalpy. Each operation has a matched effect on the value of $\Delta_r H$.
| Operation on the equation | Effect on $\Delta_r H$ | Why |
|---|---|---|
| Reverse the equation (swap reactants and products) | Reverse the sign; magnitude unchanged | Heat released forward is absorbed backward |
| Multiply all coefficients by a factor $n$ | Multiply $\Delta_r H$ by $n$ | Enthalpy is an extensive quantity |
| Add two or more equations | Add the corresponding $\Delta_r H$ values | State-function additivity (Hess's law) |
The sign reversal is illustrated by ammonia synthesis. The forward reaction is exothermic, and reversing it simply flips the sign:
$$\ce{N2(g) + 3H2(g) -> 2NH3(g)} \qquad \Delta_r H^\ominus = -91.8~\text{kJ mol}^{-1}$$ $$\ce{2NH3(g) -> N2(g) + 3H2(g)} \qquad \Delta_r H^\ominus = +91.8~\text{kJ mol}^{-1}$$
The scaling rule appears whenever a balanced equation is rewritten with fractional coefficients. Reduction of iron(III) oxide by hydrogen, $\ce{Fe2O3(s) + 3H2(g) -> 2Fe(s) + 3H2O(l)}$, has $\Delta_r H_1 = -33.3~\text{kJ mol}^{-1}$. If every coefficient is halved, the enthalpy change becomes $\Delta_r H_2 = \tfrac{1}{2}\Delta_r H_1 = -16.6~\text{kJ mol}^{-1}$, confirming that enthalpy scales with the amount of reaction.
The Classic CO Formation Cycle
The textbook's defining illustration is the enthalpy of formation of carbon monoxide:
$$\ce{C(graphite,\, s) + 1/2 O2(g) -> CO(g)} \qquad \Delta_r H^\ominus = ?$$
This cannot be measured directly. Whenever carbon burns, some $\ce{CO2}$ is always formed alongside $\ce{CO}$, so the heat released never corresponds to a pure CO-forming reaction. Hess's law sidesteps the problem by using two reactions that can be measured cleanly.
| Step | Equation | $\Delta_r H^\ominus$ / kJ mol⁻¹ |
|---|---|---|
| (i) Given | $\ce{C(graphite,\, s) + O2(g) -> CO2(g)}$ | $-393.5$ |
| (ii) Given | $\ce{CO(g) + 1/2 O2(g) -> CO2(g)}$ | $-283.0$ |
| (iii) Reverse (ii) | $\ce{CO2(g) -> CO(g) + 1/2 O2(g)}$ | $+283.0$ |
| (i) + (iii) = target | $\ce{C(graphite,\, s) + 1/2 O2(g) -> CO(g)}$ | $-110.5$ |
To place one mole of $\ce{CO}$ on the right we reverse equation (ii); the heat that was released is now absorbed, so its sign flips to $+283.0~\text{kJ mol}^{-1}$. Adding (i) and (iii), the $\ce{CO2}$ on both sides cancels and the algebra of the enthalpies gives
$$\Delta_r H^\ominus = (-393.5) + (+283.0) = -110.5~\text{kJ mol}^{-1}$$
Figure 1. Two routes from C(s) + O₂(g) down to CO₂(g): the direct combustion (−393.5) and the indirect path through CO (−110.5 then −283.0). The two routes drop by the same total, so −110.5 + (−283.0) = −393.5, which rearranges to give the enthalpy of CO formation.
Energy Cycles and the Triangle
The CO calculation is most cleanly visualised as a closed energy cycle. Arrange reactants, the final products, and an intermediate as the corners of a triangle. Because enthalpy is a state function, the net enthalpy change around the complete cycle is zero — going out along one route and back along another returns the system to its starting enthalpy.
Figure 2. A Hess triangle. The direct route A → B and the indirect route A → C → B reach the same products, so ΔH = ΔH₁ + ΔH₂. Reversing an arrow flips the sign of its enthalpy, which is how unknown legs are extracted.
Reading the triangle gives $\Delta H = \Delta H_1 + \Delta H_2$ directly. If instead the target is one of the legs, you reverse the arrow you do not want and rearrange — exactly the manoeuvre used for CO above.
Hess's law summations are only as reliable as the named enthalpies you feed in. Revise formation, combustion and bond enthalpies in Enthalpies of Reactions.
Worked Hess's Law Problems
The standard NEET pattern is to be given several thermochemical equations and asked for a fourth, target reaction. The procedure never changes: write the target, line the given equations up so each species lands on the correct side in the correct amount (reversing and scaling as needed), then add the equations and add the adjusted enthalpies.
Calculate the standard enthalpy of formation of $\ce{CH3OH(l)}$ from the data:
(i) $\ce{CH3OH(l) + 3/2 O2(g) -> CO2(g) + 2H2O(l)}$, $\;\Delta_r H = -726~\text{kJ mol}^{-1}$
(ii) $\ce{C(graphite) + O2(g) -> CO2(g)}$, $\;\Delta_c H = -393~\text{kJ mol}^{-1}$
(iii) $\ce{H2(g) + 1/2 O2(g) -> H2O(l)}$, $\;\Delta_f H = -286~\text{kJ mol}^{-1}$
Target (formation of 1 mol methanol):
$$\ce{C(graphite) + 2H2(g) + 1/2 O2(g) -> CH3OH(l)} \qquad \Delta_f H = ?$$
Methanol is a product in the target but a reactant in (i), so reverse (i) and flip its sign. Carbon and hydrogen must appear as reactants on the left, so use (ii) as written and use (iii) scaled by 2 (the target needs $\ce{2H2}$ and two waters):
| Adjusted equation | $\Delta H$ / kJ mol⁻¹ |
|---|---|
| (i) reversed: $\ce{CO2(g) + 2H2O(l) -> CH3OH(l) + 3/2 O2(g)}$ | $+726$ |
| (ii) as is: $\ce{C(graphite) + O2(g) -> CO2(g)}$ | $-393$ |
| (iii) ×2: $\ce{2H2(g) + O2(g) -> 2H2O(l)}$ | $2(-286) = -572$ |
Adding the three equations, $\ce{CO2}$ cancels, the two waters cancel, and the oxygen tallies to $\tfrac{1}{2}\ce{O2}$ net on the left, leaving exactly the target. Summing the enthalpies:
$$\Delta_f H = (+726) + (-393) + (-572) = \boxed{-239~\text{kJ mol}^{-1}}$$
The standard enthalpy of formation of liquid methanol is therefore $-239~\text{kJ mol}^{-1}$.
Enthalpies of formation of $\ce{CO(g)}$, $\ce{CO2(g)}$, $\ce{N2O(g)}$ and $\ce{N2O4(g)}$ are $-110$, $-393$, $81$ and $9.7~\text{kJ mol}^{-1}$ respectively. Find $\Delta_r H$ for:
$$\ce{N2O4(g) + 3CO(g) -> N2O(g) + 3CO2(g)}$$
Here the formation enthalpies are themselves the building-block reactions. Each formation equation can be reversed (for reactants) or scaled (for the stoichiometric coefficients) and summed. The bookkeeping collapses into the standard formula
$$\Delta_r H = \sum_i a_i\,\Delta_f H(\text{products}) - \sum_i b_i\,\Delta_f H(\text{reactants})$$
which is just Hess's law applied to formation steps. Substituting:
$$\Delta_r H = \big[\Delta_f H(\ce{N2O}) + 3\,\Delta_f H(\ce{CO2})\big] - \big[\Delta_f H(\ce{N2O4}) + 3\,\Delta_f H(\ce{CO})\big]$$ $$= \big[81 + 3(-393)\big] - \big[9.7 + 3(-110)\big]$$ $$= (81 - 1179) - (9.7 - 330) = (-1098) - (-320.3) = \boxed{-777.7~\text{kJ mol}^{-1}}$$
The reaction is strongly exothermic, releasing $777.7~\text{kJ}$ per mole of reaction.
The Formation-Enthalpy Shortcut
Worked Example 2 hints at the most efficient form of Hess's law for tabulated data. Because the formation reaction of every species is itself a step from the elements, any reaction enthalpy can be assembled from formation enthalpies using equation 5.15 of NCERT:
$$\Delta_r H^\ominus = \sum_i a_i\,\Delta_f H^\ominus(\text{products}) - \sum_i b_i\,\Delta_f H^\ominus(\text{reactants})$$
where $a_i$ and $b_i$ are the stoichiometric coefficients. By convention the standard formation enthalpy of any element in its reference state is zero, which is why $\Delta_f H^\ominus(\ce{Fe,\, s})$ and $\Delta_f H^\ominus(\ce{H2,\, g})$ vanish in the iron-oxide example. Applied to the decomposition of limestone, $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$, the formula gives
$$\Delta_r H^\ominus = [(-635.1) + (-393.5)] - [(-1206.9)] = +178.3~\text{kJ mol}^{-1}$$
an endothermic process, confirming that calcium carbonate must be heated to break it down. NCERT's benzene calculation (Problem 5.9) is the same idea run in reverse: the formation enthalpy of benzene is recovered from its combustion enthalpy ($-3267~\text{kJ mol}^{-1}$) together with the formation enthalpies of $\ce{CO2}$ and $\ce{H2O}$, simply by treating the combustion equation as one leg of a Hess cycle.
Combustion enthalpy is not formation enthalpy
A formation enthalpy is defined for one mole of compound formed from its elements in their most stable states. The reaction $\ce{CaO(s) + CO2(g) -> CaCO3(s)}$ has a perfectly good $\Delta_r H$, but it is not $\Delta_f H$ of $\ce{CaCO3}$ because the carbonate is made from compounds, not elements. Likewise $\ce{H2(g) + Br2(l) -> 2HBr(g)}$ gives $2\Delta_f H$, so you must halve it to quote the formation enthalpy of $\ce{HBr}$.
Confirm "one mole, from elements, most stable states" before calling any value $\Delta_f H$.
Common Mistakes and Recap
Most lost marks in Hess's law problems come from sign and scaling errors, not from misunderstanding the principle. Reverse the sign every time you flip an equation, and scale the enthalpy by the same factor you scale the coefficients. Keep physical states consistent — $\ce{H2O(l)}$ and $\ce{H2O(g)}$ have different enthalpies, and mixing them silently corrupts the answer.
Hess's Law in one screen
- Enthalpy is a state function ($H = U + pV$), so $\Delta_r H$ depends only on initial and final states — the basis of Hess's law.
- Statement: a reaction's enthalpy is the sum of the enthalpies of any steps it is divided into, at the same temperature: $\Delta_r H = \sum_i \Delta_r H_i$.
- Reverse an equation → flip the sign of $\Delta_r H$; multiply coefficients by $n$ → multiply $\Delta_r H$ by $n$; add equations → add enthalpies.
- Use it when a reaction (like $\ce{C + 1/2 O2 -> CO}$) cannot be measured directly: $-110.5 = -393.5 + 283.0~\text{kJ mol}^{-1}$.
- For tabulated data, $\Delta_r H^\ominus = \sum a_i \Delta_f H^\ominus(\text{products}) - \sum b_i \Delta_f H^\ominus(\text{reactants})$; elements in their reference state have $\Delta_f H^\ominus = 0$.