The definition of enthalpy is properly described in here.

The principal objective of this question is mainly academic. In order to any students with internet connection can find a definition and an example of the enthalpy concept applications.

What are the main differences of:

  • Standard enthalpy
  • Enthalpy of atomization
  • Enthalpy of formation
  • Enthalpy of hydration
  • Enthalpy of reaction
  • Enthalpy of combustion
  • Enthalpy of solution
  • Enthalpy of lattice dissociation
  • Enthalpy of lattice formation
  • Enthalpy of mixture
  • Enthalpy of excess mixture
  • Enthalpy of (any phase transition)
  • Law of Hess
  • Law of Kirchhoff
    • Any other definition not listed here

2 Answers 2


Explanation of notation:

$H$ is the enthalpy of the system.

$\Delta$ means change of, so $\Delta H$ means change of the enthalpy

Image of standard conditions

This symbol means standard condition, standard condition is defined as a pressure of $100\text{kPa}$ and reactants and products are in their standard state, or concentration of solutions are $1\text{M}$. However, since LaTeX doesn't have this symbol, I'll substitute it for $^\ominus$.

Enthalpy changes:

Enthalpy of dilution – The enthalpy change when a solution containing one mole of a solute is diluted from one concentration to another.

Enthalpy of ($n\text{th}$) electron affinity – The enthalpy change when $n$ electrons are added to one mole of gaseous atoms. $$\ce{Li(g) + e-(g) -> Li-(g) +60\ \text{kJ}}$$ $$\ce{F(g) + e-(g) -> F-(g) +328\ \text{kJ}}$$

Enthalpy of ($n\text{th}$) ionization – The enthalpy change when $n$ electrons are removed from one mole of gaseous atoms. It is always positive.

$$\ce{Li(g) +520\ \text{kJ} -> Li+(g) +e-(g)}$$ $$\ce{He(g) +2372\ \text{kJ} -> He+(g) +e-(g)}$$

Enthalpy of lattice dissociation – The enthalpy change when one mole of an ionic lattice dissociates into isolated gaseous ions. An example for sodium chloride which have an enthalpy of lattice dissociation, also known as lattice energy, of $787\ \mathrm{kJ/mol}$ $$\ce{NaCl(s) +787\ \text{kJ} -> Na+(g) + Cl-(g)}\qquad\Delta H=787\ \text{kJ/mol}$$

Enthalpy of lattice formation – The enthalpy change when one mole of solid crystal is formed from its scattered gaseous ions.

$$\ce{Na+(g) + Cl-(g) -> NaCl(s)} +787\ \text{kJ}\qquad\Delta H=-787\ \text{kJ/mol}$$

Which means Enthalpy of lattice dissociation$=-$Enthalpy of lattice formation Enthalpy of hydration($\Delta_\text{hyd}H^\ominus$) – Enthalpy change when when one mole of ions undergo hydration.

Enthalpy of mixing – The enthalpy change from a substance when mixed.

Enthalpy of neutralisation – The enthalpy change when an acid is completely neutralised by a base.

For a strong acid, like $\ce{HCl}$ and strong base, like $\ce{NaOH}$, they disassociate almost completely $\ce{Cl-}$ and $\ce{Na+}$ are spectator ions so what is actually happening is $$\ce{H+(aq) + OH-(aq) -> H2O(l) + 58\ \text{kJ/mol}}$$

However, using a weak acid/base will have a lower enthalpy of neutralisation as normally most of the acid/base does not disassociate.

For example, mixing ethanoic acid and potassium hydroxide only has a enthalpy of neutralisation of $-11.7\ \text{kJ/mol}$

Enthalpy of precipitation – The enthalpy change when one mole of a sparingly soluble substance precipitates by mixing dilute solutions of suitable electrolytes.

Enthalpy of solution – Enthalpy change when 1 mole of an ionic substance dissolves in water to give a solution of infinite dilution.

Enthalpy of solution can be positive or negative as when a ionic substance dissolves, the dissolution can be broken into three steps

  1. Breaking of solute-solute attraction (endothermic)
  2. Breaking solvent-solvent attraction (endothermic), eg. hydrogen bonds, LDF
  3. Forming solvent-solute attraction (exothermic)

An example of a positive enthalpy of solution is potassium chlorate which has an enthalpy of solution of $41.38\ \text{kJ/mol}$

Enthalpy of (Solid$\rightarrow$Liquid: $\Delta_\text{fus}H^\ominus$, Liquid$\rightarrow$Solid:$\Delta_\text{freezing}H^\ominus$, Liquid$\rightarrow$Gas: $\Delta_\text{vap}H^\ominus$, Gas$\rightarrow$Liquid: $\Delta_\text{cond}H^\ominus$, Solid$\rightarrow$Gas: $\Delta_\text{sub}H^\ominus$, Gas$\rightarrow$Solid: $\Delta_\text{deposition}H^\ominus$) – The enthalpy change from providing energy, to a specific quantity of the substance to change its state. $$\Delta_\text{fus}H^\ominus=-\Delta_\text{freezing}H^\ominus,\Delta_\text{vap}H^\ominus=-\Delta_\text{cond}H^\ominus,\Delta_\text{sub}H^\ominus=-\Delta_\text{deposition}H^\ominus$$ Standard enthalpy of atomization($\Delta_\text{at}H_T^\ominus$) – Change when a compound's bonds are broken and the component atoms are reduced to individual atoms at at $T^\circ K$. $$\ce{S_8 -> 8S}\qquad\Delta_{at}H^\ominus=278.7\ \text{kJ/mol}$$

Standard enthalpy of combustion($\Delta_\text{c}H_T^\ominus$) – The enthalpy change which occurs when one mole of the compound is burned completely in oxygen at $T^\circ K$ and $10\ \mathrm{kPa}$. $$\ce{H2(g) +\frac{1}{2}O2(g) -> H2O(g)}+572\ \text{kJ}\qquad\Delta_cH^\ominus = -286\ \text{kJ/mol}$$

Standard enthalpy of formation($\Delta_\text{f}H_T^\ominus$) – Change in enthalpy during the formation of one mole of the compound from its constituent elements, with all substances in their standard states, and at a pressure of $100\ \mathrm{kPa}$ at $T^\circ K$.

It can be calculated using Hess's law if the reaction is hypothetical. An example is methane, $\ce{C}$ and $\ce{H2}$ will not normally react but the standard enthalpy of formation of methane is determined by Hess's law to be $-74.8\ \text{kJ/mol}$ $$\ce{\frac{1}{2}N2(g) +\frac{1}{2}O2(g) -> NO(g)}\qquad\Delta_\text{f}H^\ominus=90.25\ \text{kJ/mol}$$

Standard enthalpy of reaction($\Delta_\text{r}H^\ominus_T$) – Enthalpy change that when matter is transformed by a chemical reaction at $T^\circ K$ and $10\ \mathrm{kPa}$. $$\ce{H2(g) +\frac{1}{2}O2(g) -> H2O(g)} +572\ \text{kJ}\qquad\Delta_\text{r}H^\ominus = −572\ \text{kJ/mol}$$ $$\Delta_\text{r}H^\ominus=\sum H^\ominus_\text{products}-\sum H^\ominus_\text{reactants}$$ $$\Delta_\text{r}H_\text{forward}=-\Delta_\text{r}H_\text{backwards}$$ Laws:

Le Chatelier's Principle – When an external change is made to a system in dynamic equilibrium, the system responds to minimise the effect of the change.

Yellow $\ce{Fe^3+}$ reacting with colorless thiocyanate ions $\ce{SCN-}$ to form deep red $\ce{[Fe(SCN)]^2+}$ ions: $$\ce{Fe^3+(aq) + SCN-(aq) -> [Fe(SCN)]^2+(aq)}$$ When $\ce{NH4Cl}$ is added, $\ce{Cl-}$ reacts with $\ce{Fe^3+}$ ions to form $\ce{[FeCl4]-}$ ions. By the Le Chatelier's Principle, when $\ce{Cl-}$ is added into a solution of deep red $\ce{[Fe(SCN)]^2+(aq)}$ ions, the equilibrium will shift to $\ce{Fe^3+(aq) + SCN-(aq)}$, turning the solution pale red.

Kirchhoff's Law – Enthalpy of any substance increases with temperature.

Hess's Law – Total enthalpy change of a chemical reaction is independent of the number of steps the reaction takes.

Henry's Law – Amount of dissolved gas is proportional to its partial pressure in the gas phase.


Constant Entropy – Isoentropic Constant Pressure – Isobaric

Constant Volume – Isovolumetric

Side note: This is probably not a complete list as I may have missed some

  • 3
    $\begingroup$ Standard state does not include any stipulation of temperature. $\endgroup$ Apr 23, 2017 at 11:47
  • $\begingroup$ @orthocresol Shouldn't it include, as due to Kirchhoff's law, the enthalpy of ___ will change with the temperature, so the temperature and pressure should be specified $\endgroup$
    – Ariana
    Apr 23, 2017 at 12:05
  • 2
    $\begingroup$ Yes, you are correct that the enthalpy change will vary with temperature. Therefore, a temperature has to be specified. However, there is no condition that it has to be 273 K in particular (or 298 K or anything else). $\endgroup$ Apr 23, 2017 at 13:08
  • 1
    $\begingroup$ @ArianaGrande I have edited your post using the \ce notation used for chemical equations. For example, write $\ce{A + B -> C}$ to get $\ce{A + B -> C}$. Furthermore, add \text{Input-text-here} on the units you write. Pretty good answer by the way. $\endgroup$ Apr 29, 2017 at 16:36
  • $\begingroup$ @Ariana Standard conditions usually means a pressure of 1 atm or 1 bar ie 100 kPa (the value in the text is wrong). $\endgroup$
    – Buck Thorn
    Feb 15, 2021 at 20:12

Most of the chemistry happens in open containers. Open containers have constant pressure. At constant pressure work is done to carry out the reaction and also to "push" the air up (called as expansion work). This expansion work is $${\Delta PV}$$. Thus some part of the heat leaks as expansion work without being converted into Internal Energy.

So we define a new term $${\Delta H ~(Enthalpy) = \Delta U + \Delta PV}$$. It accounts for expansion work and internal energy. As the heat is used only for internal energy and expansion work, Enthalpy is equal to heat supplied. There are various types of enthalpies depending upon Internal Energy is used to do what kind of work.

Phase is something that is uniform in composition and physical state. It is similar to state of matter (solid, liquid and gas), but these is only one state of matter (solid carbon) but there can be many phases (graphite, diamond)

Enthalpy of phase transition

It is the heat supplied at constant 1 atm pressure to bring about a change in phase. It can be $$\ce{H2O (liquid) <=> H2O (gas) -> +44KJ}$$ or $$\ce{C (graphite) <=> C (diamond) -> +4KJ}$$ Conversion of graphite into diamond is thermodynamically possible but kinetics does not allow it, as there are very strong bond to break and rearrange. This is true for every thermodynamical reaction.

Enthalpy of vapourisation and Enthalpy of Condensation

Enthalpy of vapourisation is the heat required to vaporise 1 mole of liquid into gas. (a phase change) Energy needs to be supplied and hence enthalpy is positive.

As Enthalpy is a state function, the reaction can proceed from the final state (gas) to initial state (liquid) by absorbing the same amount of energy. This heat absorbed at constant pressure is called as enthalpy of condensation and is numerically equal to negative of enthalpy of vapourisation.

Enthalpy of Fusion and Enthalpy of freezing

The heat required to melt 1 gm of solid into liquid at a constant pressure of 1 atm is enthalpy of fusion. The enthalpy required for the reverse reaction is (liquid to solid) is called enthalpy of freezing. It follows the same lines as of above two enthalpies.

Enthalpy of sublimation and Enthalpy of vapour deposition

Some substances directly vaporise into gas, without forming liquid and liquid is not stable at any temperature and pressure. The enthalpy required to do so is enthalpy of sublimation. As enthalpy is a state function, the enthalpy required to deposit the gas into a solid is called enthalpy of vapour deposition.

Enthalpy of atomic and molecular changes

Enthalpy of ionisation and Enthalpy of electron capture

The energy required to remove the first electron of one mole of gas at constant pressure is called enthalpy of ionisation. The energy required to remove the second electron is called Second enthalpy of ionisation. The enthalpy of ionisation of 1 mole of gas at 273.15 Kelvin and 1 atm pressure is called "Ionisation Energy".

The enthalpy of capturing an electron by a neutral atom (neutral, not positive ion) is called enthalpy of electron capture. At 273.15 K and 1 atm, for 1 mole this energy is also called "Standard electron gain Enthalpy"

Enthalpy of disassociation and Enthalpy of bond formation

The heat required to disassociate a molecule into 2 entities is called enthalpy of bond dissociation Eg $$\ce{H2O -> H2 + O ~~~~ \Delta H = +927 KJ}$$

the opposite of it is enthalpy of bond formation Eg. $$\ce{H2 + O2 -> H2O}$$ Here the enthalpy of (H-O-H) is not equal to 2 times enthalpy of (O-H)

for this particular reaction, $$\ce{H2O(gas) -> H + HO^. ~~~~\Delta H = +428KJ }$$ and $$\ce{OH^. -> H + O^. ~~~~ \Delta H = +428 KJ}$$. Therefore total enthalpy is +927KJ. As Enthalpy of disassociation is dependent upon reaction itself $$\ce{\Delta H(O-H)~is~the~average~value~of~\Delta H(O-H)~of~H-OH~,~\Delta H(O-H)~of~CH3OH~etc.}$$


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