/

Enthalpy Calculator — ΔH Reaction, Hess’s Law, Calorimetry & Clausius-Clapeyron

Enthalpy Calculator — ΔH Reaction, Hess's Law, Calorimetry & Clausius-Clapeyron
🔥 Thermochemistry Tool

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.

⚗️ Enthalpy of Reaction — Hess's Law Calculator

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.

2H₂+O₂→2H₂O (ΔH=−571.66 kJ)
CH₄+2O₂→CO₂+2H₂O (ΔH=−890.30 kJ)
N₂+3H₂→2NH₃ (ΔH=−92.22 kJ)
CaCO₃→CaO+CO₂ (ΔH=+178.32 kJ)
Fe₂O₃+2Al→Al₂O₃+2Fe (ΔH=−851.5 kJ)
Reactants → Products

Reactants

Products

⚠️Error

Standard Enthalpy of Reaction

ΔH°rxn = — kJ/mol

Energy level diagram — reactant enthalpy level (left) vs. product enthalpy level (right). Downward red arrow = exothermic; upward green arrow = endothermic.

📋 Step-by-Step Working — Hess's Law
🌡️ Calorimetry Calculator — q=mcΔT

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.

⚠️Error

Calorimetry Result

📋 Step-by-Step Working — q=mcΔT
❄️🔥 Phase Change Calculator — Enthalpy of Fusion & Vaporization
⚠️Error

Enthalpy of Phase Change

Heating curve (per 1 mole): heating solid → melting plateau (ΔH_fus) → heating liquid → boiling plateau (ΔH_vap) → heating gas.

📋 Step-by-Step Working — Phase Change Enthalpy
📈 Clausius-Clapeyron Calculator — Vapor Pressure & ΔH_vap
ln(P₂/P₁) = −(ΔH_vap/R) × (1/T₂ − 1/T₁) Clausius-Clapeyron equation — R = 8.31446 J/(mol·K), temperatures in Kelvin
Water boiling point at 0.5 atm (Denver)
Water vapor pressure at 80°C
Find ΔH_vap of ethanol from 2 points
⚠️Error

Clausius-Clapeyron Result

📋 Step-by-Step Working — Clausius-Clapeyron
📚 Enthalpy Reference Tables — Constants & Standard Values
Table A — Standard Enthalpies of Formation ΔHf° at 298.15 K
SubstanceFormulaΔHf° (kJ/mol)State
HydrogenH₂0g
OxygenO₂0g
NitrogenN₂0g
Carbon (graphite)C0s
Carbon (diamond)C1.895s
IronFe0s
WaterH₂O−285.83l
Water vaporH₂O−241.82g
Carbon monoxideCO−110.53g
Carbon dioxideCO₂−393.51g
MethaneCH₄−74.87g
EthaneC₂H₆−84.68g
EthyleneC₂H₄52.47g
AcetyleneC₂H₂227.40g
PropaneC₃H₈−103.85g
ButaneC₄H₁₀−126.15g
BenzeneC₆H₆49.04l
GlucoseC₆H₁₂O₆−1274.0s
MethanolCH₃OH−238.66l
EthanolC₂H₅OH−277.69l
AmmoniaNH₃−46.11g
Nitric oxideNO90.25g
Nitrogen dioxideNO₂33.18g
Nitrous oxideN₂O82.05g
Dinitrogen tetroxideN₂O₄9.16g
Hydrochloric acidHCl−92.31g
Hydrobromic acidHBr−36.40g
Hydrofluoric acidHF−273.30g
Hydroiodic acidHI26.48g
Sulfur dioxideSO₂−296.83g
Sulfur trioxideSO₃−395.72g
Sulfuric acidH₂SO₄−813.99l
Sodium chlorideNaCl−411.15s
Sodium hydroxideNaOH−425.93s
Potassium chlorideKCl−436.75s
Calcium chlorideCaCl₂−795.42s
Calcium oxideCaO−635.09s
Calcium hydroxideCa(OH)₂−986.09s
Calcium carbonateCaCO₃−1206.92s
Iron(III) oxideFe₂O₃−824.2s
Iron(II,III) oxideFe₃O₄−1118.4s
Aluminum oxideAl₂O₃−1675.7s
Magnesium oxideMgO−601.6s
Magnesium hydroxideMg(OH)₂−924.54s
Zinc oxideZnO−350.5s
Copper(I) oxideCu₂O−168.6s
Copper(II) oxideCuO−157.3s
Silver chlorideAgCl−127.01s
Lead(II) oxidePbO−219.0s
Silicon dioxideSiO₂−910.7s
Table B — Specific Heat Capacities (for q=mcΔT calculations)
Substancec (J/g·°C)Notes
Water (liquid)4.184Highest common specific heat
Water (ice)2.090Used in heating-curve calculations
Steam (water vapor)2.010Used in heating-curve calculations
Aluminum0.897Common metal
Copper0.385Common metal, low specific heat
Iron0.449Common metal
Gold0.129Very low specific heat
Silver0.233
Lead0.128Lowest common metal
Zinc0.388
Glass0.840
Granite0.790
Sand0.835
Wood1.760
Ethanol2.440
Glycerin2.430
Air1.005
Seawater3.993Slightly less than pure water
Table C — Enthalpy of Phase Changes
SubstanceΔH_fus (kJ/mol)ΔH_vap (kJ/mol)T_melt (°C)T_boil (°C)
Water6.01040.650.00100.00
Ethanol5.0238.56−114.178.37
Methanol3.21535.21−97.764.7
Benzene9.8730.725.580.1
Ammonia5.65723.33−77.7−33.4
Acetone5.7729.10−94.756.2
Iron13.81340.015382861
Aluminum10.71293.4660.32519
Mercury2.29559.11−38.8356.7
Table D — Common Enthalpy Values Quick Reference
ReactionΔH (kJ/mol)Type
CH₄ + 2O₂ → CO₂ + 2H₂O−890.4Exothermic
N₂ + 3H₂ → 2NH₃−92.2Exothermic
CaCO₃ → CaO + CO₂+178.3Endothermic
H₂ + ½O₂ → H₂O(l)−285.8Exothermic
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe−851.5Exothermic

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.

ΔH = H_products − H_reactants Negative ΔH = exothermic (heat released) · Positive ΔH = endothermic (heat absorbed)

How to Calculate Enthalpy Change — ΔH°rxn Formula

The standard method to calculate enthalpy change of a reaction uses standard enthalpies of formation:

ΔH°rxn = Σ[n·ΔHf°(products)] − Σ[m·ΔHf°(reactants)] n, m = stoichiometric coefficients; ΔHf° = standard enthalpy of formation (kJ/mol)

Follow this four-step method to find enthalpy of reaction every time:

  1. Write the balanced chemical equation — coefficients must be correct before any enthalpy calculation is meaningful.
  2. Look up ΔHf° for every substance — elements in their standard state (H₂(g), O₂(g), Fe(s), C-graphite) have ΔHf° = 0 by definition.
  3. 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.
  4. 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)

  1. Products: 1×(−393.51) + 2×(−285.83) = −393.51 − 571.66 = −965.17 kJ
  2. Reactants: 1×(−74.87) + 2×(0) = −74.87 kJ
  3. ΔH°rxn = −965.17 − (−74.87) = −890.30 kJ/mol
  4. 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)

  1. Products: 2×(−46.11) = −92.22 kJ
  2. Reactants: 0 + 3×(0) = 0 kJ
  3. 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)

  1. Products: (−635.09) + (−393.51) = −1028.60 kJ
  2. Reactants: (−1206.92) kJ
  3. 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

  1. Reverse the second equation: CO₂(g)→CO(g)+½O₂(g), ΔH=+282.98 kJ
  2. Add to the first: C(s)+O₂(g)+CO₂(g)→CO₂(g)+CO(g)+½O₂(g)
  3. Cancel common terms: C(s)+½O₂(g)→CO(g)
  4. 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).

  1. Multiply the first equation by 2: 2S(s)+2O₂(g)→2SO₂(g), ΔH=2×(−296.83)=−593.66 kJ
  2. Add the second equation as given: 2SO₂(g)+O₂(g)→2SO₃(g), ΔH=−197.78 kJ
  3. 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.

  1. ΔT = 31.5 − 22.0 = 9.5°C
  2. q = mcΔT = 100.0 × 4.184 × 9.5 = 3974.8 J = 3.975 kJ
  3. q_rxn = −q_solution = −3.975 kJ (exothermic)
  4. 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

  1. q_cal = C_cal×ΔT = 5.00×2.36 = 11.8 kJ
  2. q_rxn = −q_cal = −11.8 kJ (exothermic)
  3. If 0.550 g of a compound (M=180.16 g/mol) burned: n = 0.550/180.16 = 0.003053 mol
  4. 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).

  1. ΔT = 19.2 − 25.0 = −5.8°C
  2. q_solution = 50.0 × 4.184 × (−5.8) = −1213.4 J
  3. 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:

  1. n = m/M = 50.0/18.015 = 2.776 mol
  2. 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:

ln(P₂/P₁) = −(ΔH_vap/R) × (1/T₂ − 1/T₁) Temperatures MUST be in Kelvin · R = 8.31446 J/(mol·K) · ΔH_vap in J/mol

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. 1/T₂ = 1/T₁ + (R/ΔH)ln(P₁/P₂) = 1/373.15 + (8.314/40650)×ln(2.000)
  2. = 0.002680 + 0.0002046×0.6931 = 0.002680 + 0.0001418 = 0.002822
  3. T₂ = 1/0.002822 = 354.4 K = 81.3°C
  4. 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

  1. 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.
  2. 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.
  3. 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)).
  4. 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.
  5. 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

Frequently Asked Questions

What is enthalpy?
Enthalpy (symbol H) is a thermodynamic state function representing the total heat content of a system at constant pressure. ΔH = H_products − H_reactants, measured in kJ/mol. At constant pressure, ΔH equals the heat transferred (q_p).
What is Hess's law?
Hess's law states the total enthalpy change for a reaction is the same regardless of pathway, since enthalpy is a state function. ΔH°rxn = Σ[n·ΔHf°(products)] − Σ[m·ΔHf°(reactants)]. Reversing a reaction flips the sign of ΔH; multiplying scales it.
How do you calculate enthalpy of reaction?
Use Hess's law: ΔH°rxn = Σ[n·ΔHf°(products)] − Σ[m·ΔHf°(reactants)]. Write the balanced equation, look up ΔHf° for each substance (elements = 0), multiply by coefficients, and subtract reactant sum from product sum.
What is the difference between enthalpy and heat?
Heat (q) is energy transferred due to a temperature difference; enthalpy (H) is a state function of the system. At constant pressure, ΔH = q_p — numerically equal, but heat is path-dependent while enthalpy change is path-independent.
What is specific heat capacity?
Specific heat capacity (c) is the heat needed to raise 1 gram of a substance by 1°C, in J/(g·°C). Water is 4.184 J/(g·°C). Used in q=mcΔT for calorimetry calculations.
What is the Clausius-Clapeyron equation?
ln(P₂/P₁) = −(ΔH_vap/R)(1/T₂−1/T₁) relates vapor pressure to temperature using enthalpy of vaporization. Used to find vapor pressure, boiling point at new pressure, or ΔH_vap from two measurements. Temperatures must be in Kelvin.
How do you calculate heat in calorimetry?
Use q=mcΔT. For a coffee cup calorimeter, q_rxn=−q_solution. For a bomb calorimeter, q_rxn=−C_cal×ΔT, where C_cal is the calorimeter's total heat capacity in kJ/°C.
What is enthalpy of fusion vs vaporization?
Enthalpy of fusion (ΔH_fus) melts 1 mole of solid into liquid (water: 6.010 kJ/mol). Enthalpy of vaporization (ΔH_vap) vaporizes 1 mole of liquid into gas (water: 40.65 kJ/mol). Both use q=n×ΔH_phase.

Related Calculators

🔗 Share This Enthalpy Calculator

Share Hess's law, calorimetry & Clausius-Clapeyron tools with students!

Free chemistry, physics, biology & math calculators with step-by-step solutions. Trusted by 100,000+ students. Solve any science problem instantly!

Newsletter

Subscribe to our Newsletter to be updated. We promise not to spam.

Copyright © 2026 SciSolveLab. All Rights Reserved

Scroll to Top