Enthalpy Calculator — ΔH, Hess's Law & Calorimetry
This enthalpy calculator computes the enthalpy change of a reaction using Hess's law (ΔH°rxn=ΣΔHf°products−ΣΔHf°reactants), solves calorimetry problems (q=mcΔT) for coffee cup and bomb calorimeters, finds enthalpy of fusion and vaporization for phase changes, and uses the Clausius-Clapeyron equation to relate vapor pressure to temperature — with full step-by-step working and a real-time energy level diagram for every enthalpy calculation.
Enter reactants and products with stoichiometric coefficients. Type a substance name to search — enthalpy of formation ΔHf° auto-fills from the standard enthalpies table, or enter a custom value.
Reactants
Products
Standard Enthalpy of Reaction
Energy level diagram — reactant enthalpy level (left) vs. product enthalpy level (right). Downward red arrow = exothermic; upward green arrow = endothermic.
Constant-pressure calorimetry: q = m × c × ΔT. Heat released by the reaction equals heat absorbed by the solution (q_rxn = −q_solution).
Constant-volume calorimetry: q_rxn = −C_cal × ΔT, where C_cal is the heat capacity of the entire calorimeter assembly.
Solve q=mcΔT for any one variable. Leave exactly ONE field blank (q, m, c, or T_f) — the calculator solves for it.
Calorimetry Result
Enthalpy of Phase Change
Heating curve (per 1 mole): heating solid → melting plateau (ΔH_fus) → heating liquid → boiling plateau (ΔH_vap) → heating gas.
Clausius-Clapeyron Result
| Substance | Formula | ΔHf° (kJ/mol) | State |
|---|---|---|---|
| Hydrogen | H₂ | 0 | g |
| Oxygen | O₂ | 0 | g |
| Nitrogen | N₂ | 0 | g |
| Carbon (graphite) | C | 0 | s |
| Carbon (diamond) | C | 1.895 | s |
| Iron | Fe | 0 | s |
| Water | H₂O | −285.83 | l |
| Water vapor | H₂O | −241.82 | g |
| Carbon monoxide | CO | −110.53 | g |
| Carbon dioxide | CO₂ | −393.51 | g |
| Methane | CH₄ | −74.87 | g |
| Ethane | C₂H₆ | −84.68 | g |
| Ethylene | C₂H₄ | 52.47 | g |
| Acetylene | C₂H₂ | 227.40 | g |
| Propane | C₃H₈ | −103.85 | g |
| Butane | C₄H₁₀ | −126.15 | g |
| Benzene | C₆H₆ | 49.04 | l |
| Glucose | C₆H₁₂O₆ | −1274.0 | s |
| Methanol | CH₃OH | −238.66 | l |
| Ethanol | C₂H₅OH | −277.69 | l |
| Ammonia | NH₃ | −46.11 | g |
| Nitric oxide | NO | 90.25 | g |
| Nitrogen dioxide | NO₂ | 33.18 | g |
| Nitrous oxide | N₂O | 82.05 | g |
| Dinitrogen tetroxide | N₂O₄ | 9.16 | g |
| Hydrochloric acid | HCl | −92.31 | g |
| Hydrobromic acid | HBr | −36.40 | g |
| Hydrofluoric acid | HF | −273.30 | g |
| Hydroiodic acid | HI | 26.48 | g |
| Sulfur dioxide | SO₂ | −296.83 | g |
| Sulfur trioxide | SO₃ | −395.72 | g |
| Sulfuric acid | H₂SO₄ | −813.99 | l |
| Sodium chloride | NaCl | −411.15 | s |
| Sodium hydroxide | NaOH | −425.93 | s |
| Potassium chloride | KCl | −436.75 | s |
| Calcium chloride | CaCl₂ | −795.42 | s |
| Calcium oxide | CaO | −635.09 | s |
| Calcium hydroxide | Ca(OH)₂ | −986.09 | s |
| Calcium carbonate | CaCO₃ | −1206.92 | s |
| Iron(III) oxide | Fe₂O₃ | −824.2 | s |
| Iron(II,III) oxide | Fe₃O₄ | −1118.4 | s |
| Aluminum oxide | Al₂O₃ | −1675.7 | s |
| Magnesium oxide | MgO | −601.6 | s |
| Magnesium hydroxide | Mg(OH)₂ | −924.54 | s |
| Zinc oxide | ZnO | −350.5 | s |
| Copper(I) oxide | Cu₂O | −168.6 | s |
| Copper(II) oxide | CuO | −157.3 | s |
| Silver chloride | AgCl | −127.01 | s |
| Lead(II) oxide | PbO | −219.0 | s |
| Silicon dioxide | SiO₂ | −910.7 | s |
| Substance | c (J/g·°C) | Notes |
|---|---|---|
| Water (liquid) | 4.184 | Highest common specific heat |
| Water (ice) | 2.090 | Used in heating-curve calculations |
| Steam (water vapor) | 2.010 | Used in heating-curve calculations |
| Aluminum | 0.897 | Common metal |
| Copper | 0.385 | Common metal, low specific heat |
| Iron | 0.449 | Common metal |
| Gold | 0.129 | Very low specific heat |
| Silver | 0.233 | — |
| Lead | 0.128 | Lowest common metal |
| Zinc | 0.388 | — |
| Glass | 0.840 | — |
| Granite | 0.790 | — |
| Sand | 0.835 | — |
| Wood | 1.760 | — |
| Ethanol | 2.440 | — |
| Glycerin | 2.430 | — |
| Air | 1.005 | — |
| Seawater | 3.993 | Slightly less than pure water |
| Substance | ΔH_fus (kJ/mol) | ΔH_vap (kJ/mol) | T_melt (°C) | T_boil (°C) |
|---|---|---|---|---|
| Water | 6.010 | 40.65 | 0.00 | 100.00 |
| Ethanol | 5.02 | 38.56 | −114.1 | 78.37 |
| Methanol | 3.215 | 35.21 | −97.7 | 64.7 |
| Benzene | 9.87 | 30.72 | 5.5 | 80.1 |
| Ammonia | 5.657 | 23.33 | −77.7 | −33.4 |
| Acetone | 5.77 | 29.10 | −94.7 | 56.2 |
| Iron | 13.81 | 340.0 | 1538 | 2861 |
| Aluminum | 10.71 | 293.4 | 660.3 | 2519 |
| Mercury | 2.295 | 59.11 | −38.8 | 356.7 |
| Reaction | ΔH (kJ/mol) | Type |
|---|---|---|
| CH₄ + 2O₂ → CO₂ + 2H₂O | −890.4 | Exothermic |
| N₂ + 3H₂ → 2NH₃ | −92.2 | Exothermic |
| CaCO₃ → CaO + CO₂ | +178.3 | Endothermic |
| H₂ + ½O₂ → H₂O(l) | −285.8 | Exothermic |
| Fe₂O₃ + 2Al → Al₂O₃ + 2Fe | −851.5 | Exothermic |
Physical constants used throughout this enthalpy calculator: R = 8.31446 J/(mol·K) = 0.082057 L·atm/(mol·K); all ΔHf° values reported at 298.15 K per NIST standard reference data.
Enthalpy Calculator — ΔH of Reaction, Calorimetry & Phase Changes
This enthalpy calculator computes the enthalpy change of a reaction using Hess's law (ΔH°rxn = ΣΔHf°products − ΣΔHf°reactants), solves calorimetry problems using q=mcΔT for coffee cup and bomb calorimeters, calculates enthalpy of fusion and vaporization during phase changes, and applies the Clausius-Clapeyron equation to relate vapor pressure to temperature. Every calculation includes step-by-step working and a real-time energy level diagram that visually shows whether a reaction is exothermic or endothermic.
What Is Enthalpy? — Definition, Symbol, and Units
Enthalpy is a thermodynamic state function representing the total heat content of a system at constant pressure. The enthalpy symbol is H; the change in enthalpy during a process is written ΔH (delta H). Enthalpy is measured in kJ/mol for a molar quantity in chemistry, or simply kJ for a specific amount of substance.
The defining relationship is ΔH = H_products − H_reactants. When ΔH is negative, the reaction is exothermic — the products have lower enthalpy than the reactants, and the excess energy is released as heat. When ΔH is positive, the reaction is endothermic — the products have higher enthalpy than the reactants, meaning the reaction absorbs heat from its surroundings to proceed.
Understanding enthalpy vs heat is a common point of confusion: heat (q) is energy in transit, while enthalpy (H) is a property of the system. The link between them is simple: at constant pressure, ΔH = q_p — the heat transferred at constant pressure equals the enthalpy change exactly. This is why calorimetry (which operates near constant pressure in a coffee cup calorimeter) is the standard experimental method for measuring enthalpy changes.
How to Calculate Enthalpy Change — ΔH°rxn Formula
The standard method to calculate enthalpy change of a reaction uses standard enthalpies of formation:
Follow this four-step method to find enthalpy of reaction every time:
- Write the balanced chemical equation — coefficients must be correct before any enthalpy calculation is meaningful.
- Look up ΔHf° for every substance — elements in their standard state (H₂(g), O₂(g), Fe(s), C-graphite) have ΔHf° = 0 by definition.
- Multiply each ΔHf° by its stoichiometric coefficient — a coefficient of 2 in front of H₂O means you use 2 × ΔHf°(H₂O), not ΔHf°(H₂O) alone.
- Subtract the sum of reactant enthalpies from the sum of product enthalpies to calculate delta H for the reaction.
Worked Example — How to Calculate Enthalpy of CH₄ Combustion
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
- Products: 1×(−393.51) + 2×(−285.83) = −393.51 − 571.66 = −965.17 kJ
- Reactants: 1×(−74.87) + 2×(0) = −74.87 kJ
- ΔH°rxn = −965.17 − (−74.87) = −890.30 kJ/mol
- Result: ΔH°rxn = −890.30 kJ/mol — exothermic combustion
Worked Example — How to Find Enthalpy of Formation of Ammonia
N₂(g) + 3H₂(g) → 2NH₃(g)
- Products: 2×(−46.11) = −92.22 kJ
- Reactants: 0 + 3×(0) = 0 kJ
- ΔH°rxn = −92.22 − 0 = −92.22 kJ/mol (exothermic)
Worked Example — How to Solve for ΔH of Decomposition
CaCO₃(s) → CaO(s) + CO₂(g)
- Products: (−635.09) + (−393.51) = −1028.60 kJ
- Reactants: (−1206.92) kJ
- ΔH°rxn = −1028.60 − (−1206.92) = +178.32 kJ/mol (endothermic)
Hess's Law — Path Independence of Enthalpy
Hess's law states that the total enthalpy change for a reaction is the same regardless of the pathway taken to get there. Because enthalpy is a state function, only the initial state (reactants) and final state (products) matter — not the route in between. This means thermochemical equations can be manipulated algebraically like ordinary algebraic equations:
- Reversing a reaction changes the sign of ΔH: if N₂+3H₂→2NH₃ has ΔH=−92.2 kJ, then 2NH₃→N₂+3H₂ has ΔH=+92.2 kJ.
- Multiplying a reaction by a coefficient multiplies ΔH by that same coefficient.
- Adding two (or more) reactions adds their ΔH values — this is Hess's law in action.
Hess's Law Example — Finding ΔH for C(s) + ½O₂(g) → CO(g)
Given: C(s)+O₂(g)→CO₂(g), ΔH₁=−393.51 kJ and CO(g)+½O₂(g)→CO₂(g), ΔH₂=−282.98 kJ
- Reverse the second equation: CO₂(g)→CO(g)+½O₂(g), ΔH=+282.98 kJ
- Add to the first: C(s)+O₂(g)+CO₂(g)→CO₂(g)+CO(g)+½O₂(g)
- Cancel common terms: C(s)+½O₂(g)→CO(g)
- Sum enthalpies via Hess's law: ΔH = −393.51 + 282.98 = −110.53 kJ/mol (matches ΔHf° of CO directly)
Hess's Law Example — Combining Two Combustion Steps
Given ΔH for S(s)+O₂(g)→SO₂(g) = −296.83 kJ and 2SO₂(g)+O₂(g)→2SO₃(g) = −197.78 kJ, find ΔH for 2S(s)+3O₂(g)→2SO₃(g).
- Multiply the first equation by 2: 2S(s)+2O₂(g)→2SO₂(g), ΔH=2×(−296.83)=−593.66 kJ
- Add the second equation as given: 2SO₂(g)+O₂(g)→2SO₃(g), ΔH=−197.78 kJ
- Sum by Hess's law: ΔH = −593.66 + (−197.78) = −791.44 kJ
Hess's Law Example — Reversing a Step
If 2H₂(g)+O₂(g)→2H₂O(l) has ΔH=−571.66 kJ, then the reverse reaction 2H₂O(l)→2H₂(g)+O₂(g) has ΔH=+571.66 kJ — electrolysis of water is endothermic, requiring exactly the energy that combustion releases, confirming Hess's law of path independence.
Calorimetry — Measuring Enthalpy with q = mcΔT
Calorimetry measures heat flow experimentally by measuring temperature change. The fundamental calorimetry equation is q=mcΔT, where q is heat transferred, m is mass of the substance absorbing/releasing heat, c is specific heat capacity, and ΔT = T_final − T_initial.
In a coffee cup calorimeter (constant pressure, open to atmosphere): q=mcΔT applies to the surrounding solution, using m = mass of solution and c = 4.184 J/(g·°C) for dilute aqueous solutions. The key relationship is q_rxn = −q_solution, because energy is conserved — heat released by an exothermic reaction is absorbed by the solution, so the reaction's heat and the solution's heat have opposite signs. To convert to molar enthalpy: ΔH = q_rxn/n, dividing by moles of reaction.
In a bomb calorimeter (constant volume, sealed): q_rxn = −C_cal×ΔT, where C_cal is the heat capacity of the entire calorimeter assembly in kJ/°C (not a per-gram specific heat — it already accounts for the total mass of the device).
Calorimetry Example — Neutralization Reaction (q=mcΔT)
100.0 g of solution warms from 22.0°C to 31.5°C in a coffee cup calorimeter.
- ΔT = 31.5 − 22.0 = 9.5°C
- q = mcΔT = 100.0 × 4.184 × 9.5 = 3974.8 J = 3.975 kJ
- q_rxn = −q_solution = −3.975 kJ (exothermic)
- If 0.050 mol reacted: ΔH = −3.975/0.050 = −79.5 kJ/mol
Calorimetry Example — Combustion in a Bomb Calorimeter
C_cal = 5.00 kJ/°C, ΔT = 2.36°C
- q_cal = C_cal×ΔT = 5.00×2.36 = 11.8 kJ
- q_rxn = −q_cal = −11.8 kJ (exothermic)
- If 0.550 g of a compound (M=180.16 g/mol) burned: n = 0.550/180.16 = 0.003053 mol
- ΔH_combustion = −11.8/0.003053 = −3865 kJ/mol
Calorimetry Example — Dissolution (Endothermic)
50.0 g of water cools from 25.0°C to 19.2°C when a salt dissolves (heat absorbed from solution).
- ΔT = 19.2 − 25.0 = −5.8°C
- q_solution = 50.0 × 4.184 × (−5.8) = −1213.4 J
- q_rxn = −q_solution = +1213.4 J (endothermic dissolution)
Specific Heat Capacity — The q = mcΔT Equation
The specific heat capacity c is the amount of heat required to raise the temperature of 1 gram of a substance by 1°C, expressed in J/(g·°C). Water's specific heat of 4.184 J/(g·°C) is unusually high compared to most substances — this is why oceans and large bodies of water moderate the climate of nearby land, absorbing and releasing large amounts of heat with relatively small temperature swings.
Metals generally have low specific heats: copper (0.385 J/g·°C) and aluminum (0.897 J/g·°C) heat up and cool down quickly compared to water. The q=mcΔT equation can be algebraically rearranged to solve for any one of its four variables when the other three are known — this is exactly what the specific heat solver in Tool 2 does.
The sign convention is essential: q > 0 means heat is absorbed by the system (temperature increases, ΔT is positive), while q < 0 means heat is released by the system (temperature decreases, ΔT is negative). This sign convention is identical to the exothermic/endothermic convention used for ΔH in chemical reactions.
Phase Changes — Enthalpy of Fusion and Vaporization
During a phase change, temperature remains constant while added or removed energy rearranges the physical structure of the substance (breaking or forming intermolecular forces) rather than increasing kinetic energy. Enthalpy of fusion (ΔH_fus) is the energy needed to melt 1 mole of a solid into a liquid; enthalpy of vaporization (ΔH_vap) is the energy needed to vaporize 1 mole of a liquid into a gas.
For water: ΔH_fus = 6.010 kJ/mol at 0°C, and ΔH_vap = 40.65 kJ/mol at 100°C — vaporization always requires far more energy than fusion because it must completely separate molecules rather than just loosen their rigid arrangement. For freezing and condensation (the reverse processes), enthalpy has the same magnitude but a negative sign, since energy is released rather than absorbed.
The classic heating curve for water shows five distinct segments as heat is continuously added: sloped line (heating ice, q=mcΔT with c=2.090), flat plateau (melting, q=nΔH_fus), sloped line (heating liquid water, q=mcΔT with c=4.184), flat plateau (boiling, q=nΔH_vap), and sloped line (heating steam, q=mcΔT with c=2.010).
Phase Change Example — Melting Ice
Heat required to melt 50.0 g of ice at 0°C:
- n = m/M = 50.0/18.015 = 2.776 mol
- q = n × ΔH_fus = 2.776 × 6.010 = 16.69 kJ (16,690 J or 3.99 kcal)
Clausius-Clapeyron Equation — Vapor Pressure and Boiling Point
The Clausius-Clapeyron equation relates the vapor pressure of a liquid to its temperature using its enthalpy of vaporization:
The Clausius-Clapeyron equation has three common applications: finding vapor pressure at a new temperature, finding the boiling point at a new pressure (explaining why water boils at only 81°C in Denver, Colorado at reduced atmospheric pressure), and determining ΔH_vap experimentally from two vapor pressure measurements at two different temperatures. The equation assumes ΔH_vap is approximately constant over the temperature range considered.
Clausius-Clapeyron Example — Boiling Point at Altitude
Find the boiling point of water at 0.50 atm (Denver, CO approximation): P₁=1atm at T₁=373.15K, ΔH_vap=40,650 J/mol
- 1/T₂ = 1/T₁ + (R/ΔH)ln(P₁/P₂) = 1/373.15 + (8.314/40650)×ln(2.000)
- = 0.002680 + 0.0002046×0.6931 = 0.002680 + 0.0001418 = 0.002822
- T₂ = 1/0.002822 = 354.4 K = 81.3°C
- At 0.50 atm, water boils at 81.3°C instead of 100°C — less efficient for cooking at altitude!
Common Mistakes in Enthalpy Calculations
- Wrong sign for ΔH when reversing a reaction — if N₂+3H₂→2NH₃ has ΔH=−92.2 kJ, then 2NH₃→N₂+3H₂ has ΔH=+92.2 kJ. Forgetting to flip the sign is the single most common Hess's law error.
- Forgetting to multiply by stoichiometric coefficients — ΔHf° values in reference tables are per mole of compound as written; 2×CO₂ requires 2×(−393.51), not just −393.51.
- Using ΔHf°=0 for the wrong allotrope — only elements in their standard state have ΔHf°=0 (graphite, not diamond; O₂(g), not O(g) or O₃(g)).
- Forgetting the sign convention in calorimetry — q_rxn = −q_solution; a rise in solution temperature (positive q_solution) means the reaction released heat (negative q_rxn), i.e., it was exothermic.
- Using Celsius instead of Kelvin in Clausius-Clapeyron — the equation absolutely requires Kelvin; using Celsius produces wildly wrong results because 1/T is not linear under a shifted scale.
Worked Examples — 8 Complete Enthalpy Problems
1. ΔH for CH₄ Combustion
CH₄+2O₂→CO₂+2H₂O(l) using standard enthalpies of formation → ΔH = −890.30 kJ/mol
2. ΔH for Thermite Reaction
Fe₂O₃+2Al→Al₂O₃+2Fe → ΔH = −851.5 kJ (highly exothermic, used in welding)
3. Calorimetry: 100 g Water, ΔT=9.5°C
q=3974.8 J; if 0.05 mol reacted → ΔH=−79.5 kJ/mol
4. Bomb Calorimeter
C_cal=5.00 kJ/°C, ΔT=2.36°C → q_rxn=−11.8 kJ
5. Specific Heat Solver
Find final T when 5000 J added to 200 g Al (c=0.897) starting at 25°C → ΔT=5000/(200×0.897)=27.87°C → T_f=52.9°C
6. Heat to Melt 50 g Ice
n=2.776 mol, q=2.776×6.010 = 16.69 kJ
7. Clausius-Clapeyron: Boiling Point at 0.5 atm
T=354.4 K = 81.3°C
8. Find ΔH_vap of Ethanol
From two vapor pressure measurements (P₁=40 mmHg at 292.15K, P₂=400mmHg at 336.65K) → ΔH_vap ≈ 42.3 kJ/mol
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