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Enthalpy Calculator

Created by Bogna Szyk and Dominik Czernia, PhD
Reviewed by Steven Wooding
Last updated: Jan 18, 2024


This enthalpy calculator will help you calculate the change in enthalpy of a reaction. Read on to learn how to calculate enthalpy and its definition. We will also explain the difference between endothermic and exothermic reactions, as well as provide you with an example of calculations.

What is enthalpy?

Enthalpy measures the total energy of a thermodynamic system — either in the form of internal energy or volume multiplied by pressure. It is a state function depending only on the equilibrium state of a system.

The more interesting quantity is the change of enthalpy — the total energy that was exchanged within a system. It is a simplified description of the energy transfer (energy is in the form of heat or work done during expansion).

Endothermic or exothermic reaction?

There are two main types of thermodynamic reactions: endothermic and exothermic. An endothermic reaction causes absorption of heat from the surroundings. An exothermic one releases heat to the surroundings.

Both these reaction types cause energy level differences and, therefore, differences in enthalpy. All you need to remember for the purpose of this calculator is:

  • If the reaction is endothermic, the change in enthalpy is positive, as heat is gained (absorbed from the surroundings).
  • If the reaction is exothermic, the change in enthalpy is negative, as heat is lost (released to the surroundings).

Enthalpy formula

Enthalpy, by definition, is the sum of the internal energy of the system and the product of the system's volume and pressure:

H=U+pV\footnotesize H = U+pV

where UU stands for internal energy, pp for pressure and VV for volume. Notice that the second part closely remembers the equations we met at the combined gas law calculator: the relationship between pressure and volume allows us to find a similar connection between quantity of matter and temperature.

If you want to calculate the change in enthalpy, though, you need to consider two states — initial and final. We will assume that the pressure is constant while the reaction takes place. Then, the change in enthalpy is actually:

ΔH=(U2U1)+p(V2V1)\footnotesize ΔH = (U_2-U_1)+p\cdot(V_2-V_1)

or, simplified:

ΔH=ΔU+pΔV\footnotesize ΔH = ΔU+ p\cdot ΔV

where:

  • U2U_2 and V2V_2 — Internal energy and volume of the products of the reaction, respectively;
  • U1U_1 and V1V_1 — Internal energy and volume of the reactants, respectively;
  • pp — Constant pressure;
  • ΔUΔU — Change in internal energy;
  • ΔVΔV — Change in volume; and
  • ΔHΔH — Change in enthalpy.

Standard enthalpy of formation table and definition

For more specific problems, we can define the standard enthalpy of formation of a compound, denoted as ΔHf°ΔH_\mathrm{f}\degree. It represents the change in enthalpy during the formation of one mole of a substance under standard conditions of temperature and pressure (p=105 Pa=1 barp = 10^5\ \mathrm{Pa} = 1\ \mathrm{bar} and T=25°C=298.15 KT = 25 \degree \mathrm{C} = 298.15\ \mathrm{K}), from its pure elements in their respective reference states.

The reference state of an element is its most stable configuration under the above-mentioned conditions. Some examples of reference states are nitrogen in the form of gas molecules N2\mathrm N_{2} and carbon in the form of graphite.

The standard enthalpy of formation formula for a reaction is as follows:

ΔH°reaction=ΔHf°(products)  ΔHf°(reactants)\footnotesize \begin{split} ΔH\degree_\mathrm{reaction}& = \sum ΔH_\mathrm{f}\degree(\mathrm{products})\\ &\qquad\ \ - ΔH_\mathrm{f}\degree(\mathrm{reactants}) \end{split}

where:

  • ΔH°reactionΔH\degree_\mathrm{reaction} — Standard enthalpy change of formation expressed in kJ;
  • ΔHf°(products)ΔH_\mathrm{f}\degree(\mathrm{products}) — Sum of the standard enthalpies of formation of the products expressed in kJ/mol; and
  • ΔHf°(reactants)ΔH_\mathrm{f}\degree(\mathrm{reactants}) — Sum of the standard enthalpies of formation of the reactants, expressed in kJ/mol.

If you're paying attention, you might have observed that ΔHf°(products)ΔH_\mathrm{f}\degree(\mathrm{products}) and ΔHf°(reactants)ΔH_\mathrm{f}\degree(\mathrm{reactants}) have different units than ΔH°reactionΔH\degree_\mathrm{reaction}. This is because you need to multiply them by the number of moles, i.e., the coefficient before the compound in the reaction. We'll show you later an example that should explain it all.

But before that, you may ask, "How to calculate standard enthalpy of formation for each compound?" The most straightforward answer is to use the standard enthalpy of formation table! Here's an example one:

Substance

ΔHf°ΔH_\mathrm{f}\degree (kJ/mol\mathrm{kJ/mol})

O2(g)\mathrm O_{2\mathrm{(g)}}

00

SO2(g)\mathrm{SO}_{2\mathrm{(g)}}

296.83-296.83

SO3(g)\mathrm{SO}_{3\mathrm{(g)}}

395.72-395.72

H2O(l)\mathrm{H}_2\mathrm{O}_\mathrm{(l)}

285.8-285.8

Cu2O(s)\mathrm{Cu}_2\mathrm{O}_{\mathrm{(s)}}

168.6-168.6

Mg(aq)2+\mathrm{Mg}^{2+}_\mathrm{(aq)}

466.85-466.85

The symbols in the brackets indicate the state: s\mathrm{s} — solid, l\mathrm{l} — liquid, g\mathrm{g} — gas, and aq\mathrm{aq} — dissolved in water. If you need the standard enthalpy of formation for other substances, select the corresponding compound in the enthalpy calculator's drop-down list. We included all the most common compounds!

Let's practice our newly obtained knowledge using the above standard enthalpy of formation table. For example, we have the following reaction:

2SO3(g)2SO2(g)+O2(g)2\mathrm{SO}_{3\mathrm{(g)}}\rightarrow 2\mathrm{SO}_{2\mathrm{(g)}} + \mathrm{O}_{2\mathrm{(g)}}

What is the enthalpy change in this case? We sum ΔHf°ΔH_\mathrm{f}\degree for SO2(g)\mathrm{SO}_{2\mathrm{(g)}} and O2(g)O_{2\mathrm{(g)}} and subtract the ΔHf°ΔH_\mathrm{f}\degree for SO3(g)\mathrm{SO}_{3\mathrm{(g)}}. Remember to multiply the values by corresponding coefficients!

ΔH°reaction=2 mol(296.83 kJ/mol)+1 mol0 kJ/mol2 mol(395.72 kJ/mol)\footnotesize \begin{split} ΔH\degree_\mathrm{reaction} \!&=\! 2\ \mathrm{mol}\!\cdot\!(-296.83\ \mathrm{kJ/mol})\\ &\!+1\ \mathrm{mol} \cdot 0\ \mathrm{kJ/mol}\\ &\!-2\ \mathrm{mol}\cdot(-395.72\ \mathrm{kJ/mol}) \end{split}

Notice that the coefficient units mol\mathrm{mol} eliminates the mol\mathrm{mol} in the denominator, so the final answer is in kJ\mathrm{kJ}:

ΔH°reaction=197.78 kJΔH\degree_\mathrm{reaction} = 197.78\ \mathrm{kJ}

That's it! Still, isn't our enthalpy calculator a quicker way than all of this tedious computation?

🙋 Our pressure conversion tool will help you change units of pressure without any difficulties!

How to calculate the enthalpy of a reaction?

The enthalpy calculator has two modes. You can calculate the enthalpy change from the reaction scheme or by using the enthalpy formula. If you select the former:

  1. Look at the reaction scheme that appeared at the top of the calculator. Do you need an additional reactant/product (C or F)? If so, click the advanced mode button.

  2. Fill in the fields in the Reactants section. You need to provide the coefficient before the compound and select your substance from the drop-down list (they're ordered alphabetically). If you can't find the right one, select the Custom option and enter the standard enthalpy of formation in kJ/mol (if you don't have this on hand, check some online reference tables, like this one at Chemistry LibreTexts).

  3. Do the same thing for the Products.

  4. Verify the reaction scheme below and read the result. That's the standard enthalpy change of formation for your reaction!

  5. Optionally, check the standard enthalpy of formation table (for your chosen compounds) we listed at the very bottom.

If you want to calculate the enthalpy change from the enthalpy formula:

  1. Begin with determining your substance's change in volume. Let's assume your liquid expanded by 55 liters.

  2. Find the change in the internal energy of the substance. Let's say your substance's energy increased by 2000 J2000\ \mathrm{J}.

  3. Measure the pressure of the surroundings. We will assume 1 atmosphere.

  4. Input all of these values to the equation ΔH=ΔU+pΔVΔH = ΔU + p\cdot ΔV to obtain the change in enthalpy:

ΔH=2000 J+1 atm5 l=2000 J+101,325 Pa0.005 m3=2506.63 J\footnotesize \begin{split} ΔH &= 2000\ \mathrm{J}+1\ \mathrm{atm}\cdot 5\ \mathrm{l}\\ &=2000\ \mathrm{J}+101,\!325\ \mathrm{Pa}\cdot0.005\ \mathrm{m^3}\\ &=2506.63\ \mathrm{J} \end{split}
  1. You can also open the advanced mode of our enthalpy calculator to find the enthalpy based on the initial and final internal energy and volume.

🙋 With Omni Calculator, you can explore other interesting concepts of thermodynamics linked to enthalpy: try our entropy calculator and our Gibbs free energy calculator!

FAQ

What is enthalpy in chemistry?

In chemistry, enthalpy (at constant pressure) determines the heat transfer of a system. Roughly speaking, the change in enthalpy in a chemical reaction equals the amount of energy lost or gained during the reaction. A system often tends towards a state when its enthalpy decreases throughout the reaction.

Is negative enthalpy exothermic?

Yes. A chemical reaction that has a negative enthalpy is said to be exothermic. This means that the system loses energy, so the products have less energy than the reactants. Therefore, the term 'exothermic' means that the system loses or gives up energy.

How can I find enthalpy change?

You can calculate the enthalpy change in a basic way using the enthalpy of products and reactants:

ΔH° = ∑ΔHproducts − ΔHreactants

For example, let's look at the reaction Na+ + Cl- → NaCl. To find enthalpy change:

  1. Use the enthalpy of product NaCl (-411.15 kJ).

  2. Find the enthalpy of Na+ (-240.12 kJ) and Cl- (-167.16 kJ).

  3. Calculate enthalpy change:

    ΔH° = 1 × -411.15 kJ − (1 × -240.12 kJ − 1 × 167.16 kJ) = -3.87 kJ

    Remember corresponding coefficients!

Which substances have a standard enthalpy of formation zero?

All pure elements in their standard state (e.g., oxygen gas, carbon in graphite form, etc.) have a standard enthalpy of formation zero. Enthalpy of formation means enthalpy change during the formation of one mole of a substance. But an element formed from itself means no enthalpy change, so its enthalpy of formation will be zero.

What is the enthalpy of formation of water?

-571.7 kJ. Let's assume the formation of water, H2O, from hydrogen gas, H2, and oxygen gas, O2. To find enthalpy:

  1. Write the formation reaction: 2H2 + O2 → 2H2O.

  2. Note that the enthalpy of H2 and O2 in their elemental state is zero.

  3. Calculate the enthalpy as:

    ∑ΔHproducts − ΔHreactants = 2 × -285.83 kJ − ( 2 × 0 kJ + 0 kJ) = -571.7 kJ.

Bogna Szyk and Dominik Czernia, PhD
Calculate enthalpy change from
reaction scheme
anA + bnB + cnC → dnD + enE + fnF

By default, you can only calculate for two reactants/products. Click the advanced mode button to include more compounds in the reaction.
Reactants
aₙ coefficient
Reactant A
SO₃(g)
bₙ coefficient
Reactant B
None
Products
dₙ coefficient
Product D
SO₂(g)
eₙ coefficient
Product E
O₂(g)
Results
Change in enthalpy
kJ
Your reaction:
2 SO₃(g) → 2 SO₂(g) + O₂(g)
Standard enthalpies of formation:
SO₃(g): Hf = -395.72 kJ
SO₂(g): Hf = -296.83 kJ
O₂(g): Hf = 0 kJ
Please note that we don't check if your reaction scheme makes chemical sense.
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