What is the enthalpy change?
Find the enthalpy change of a reaction two ways. From a calorimeter, type the solution’s mass, specific heat, starting and final temperatures and the moles that reacted. From formation data, list each product and reactant with its coefficient and ΔH°f.
- Enthalpy change
- ΔH = −57.7392 kJ/mol
The enthalpy change is ΔH = −57.7392 kJ/mol: exothermic, gives off heat.
- Heat flow
- exothermic, gives off heat
- ΔH (kJ/mol of reaction)
- -57.7392
- Heat taken in by the solution (J)
- 2,886.96
- Heat of the reaction (kJ)
- -2.88696
Enthalpy change: ΔH = −57.7392 kJ/mol. The enthalpy change is ΔH = −57.7392 kJ/mol: exothermic, gives off heat.
How to calculate
Finds the enthalpy change ΔH of a reaction from coffee-cup calorimetry (mass, specific heat, temperature change and moles) or from standard enthalpies of formation of the products and reactants.
Example with the default inputs (From Calorimetry, Mass of solution 100 g, Specific heat, J/(g·°C) 4.184, Starting temperature 22 °C, Final temperature 28.9 °C, Moles reacted 0.05): The enthalpy change is ΔH = −57.7392 kJ/mol: exothermic, gives off heat.
Method: Calorimetry: q_solution = m × c × (T_final − T_start); q_rxn = −q_solution; ΔH = q_rxn ÷ moles. Formation: ΔH° = Σ n × ΔH°f(products) − Σ n × ΔH°f(reactants).
- Calorimetry: a coffee-cup calorimeter at constant pressure that loses no heat, so q_rxn = −q_solution and equals ΔH.
- Formation: ΔH°f values are standard (25 °C, 1 bar); elements in their standard state have ΔH°f = 0.
Worked examples
Each example is checked against the calculator on every build.
- From Calorimetry, Mass of solution 100 g, Specific heat, J/(g·°C) 4.184, Starting temperature 22 °C, Final temperature 28.9 °C, Moles reacted 0.05 gives Heat taken in by the solution (J) 2,886.96, Heat of the reaction (kJ) -2.88696, ΔH (kJ/mol of reaction) -57.7392.Source: OpenStax, Chemistry 2e, §5.2 Calorimetry (q = c × m × ΔT; q_rxn = −q_solution), https://openstax.org/books/chemistry-2e/pages/5-2-calorimetry, Example 5.5 Heat Produced by an Exothermic Reaction: about 2.9 × 10³ J
- From Calorimetry, Mass of solution 53.21 g, Specific heat, J/(g·°C) 4.184, Starting temperature 24.9 °C, Final temperature 20.3 °C, Moles reacted 0.0401 gives Heat of the reaction (kJ) 1.024101, ΔH (kJ/mol of reaction) 25.538677.Source: OpenStax, Chemistry 2e, §5.2 Calorimetry (q = c × m × ΔT; q_rxn = −q_solution), https://openstax.org/books/chemistry-2e/pages/5-2-calorimetry, Example 5.6 Heat Flow in an Instant Ice Pack: q_rxn = +1.0 kJ
- From Formation, Reactants and products Product 2 -207.4; Product 1 90.25; Reactant 3 33.2; Reactant 1 -285.83 gives Σ n·ΔH°f of products (kJ) -324.55, Σ n·ΔH°f of reactants (kJ) -186.23, ΔH (kJ/mol of reaction) -138.32.Source: OpenStax, Chemistry 2e, §5.3 Enthalpy (ΔH° = Σ n × ΔH°f(products) − Σ n × ΔH°f(reactants)), https://openstax.org/books/chemistry-2e/pages/5-3-enthalpy, Example 5.15 Using Hess’s Law, ΔH°f values from the example
How it works
Calorimetry. The solution’s heat is q_solution = m × c × ΔT, where m is the solution mass in grams, c the specific heat in J/(g·°C) and ΔT = T_final − T_start (a step of 1 °C equals 1 K; a step of 1 °F is 5/9 K). The reaction’s heat is q_rxn = −q_solution, shown in kJ (÷ 1,000), and
ΔH = q_rxn ÷ moles reacted (kJ/mol).
Formation. Add n × ΔH°f over the products and over the reactants, then
ΔH° = Σ n × ΔH°f(products) − Σ n × ΔH°f(reactants) (kJ per mole of reaction).
Each row needs a coefficient above 0 and a ΔH°f from −100,000 to 100,000 kJ/mol. Masses: 1 kg = 1,000 g; 1 lb = 453.59237 g; 1 oz = 28.349523125 g. Temperatures convert to kelvins: K = °C + 273.15; K = (°F + 459.67) × 5/9.
Exact arithmetic. Each value is read as the exact decimal you typed, in its unit, so each result is an exact fraction until it is rounded for display.
Output format. The answer line is ΔH = … kJ/mol, the number to 6 significant figures with no thousands separators and the true minus sign. The heat flow row says exothermic, gives off heat (ΔH below 0), endothermic, takes in heat (above 0) or no heat change (0). The rows show ΔH, and either the heats (calorimetry) or the two sums (formation), to 6 significant figures.
When there is no answer. A nonzero ΔH too close to 0 to hold as a number.
Assumptions
- Calorimetry: no heat is lost to the cup or the air, at constant pressure.
- Formation: standard enthalpies of formation at 25 °C and 1 bar.
Worked examples by hand
Neutralization (OpenStax Example 5.5). q = 100.0 × 4.184 × (28.9 − 22.0) = 2,886.96 J; q_rxn = −2.88696 kJ (the book: 2.9 × 10³ J). 50.0 mL of 1.00 M acid is 0.0500 mol, so ΔH = −2.88696 ÷ 0.0500 = −57.7392 kJ/mol.
Instant ice pack (OpenStax Example 5.6). m = 50.0 + 3.21 = 53.21 g; q = 53.21 × 4.184 × (20.3 − 24.9) = −1,024.10 J; q_rxn = +1.02410 kJ (the book: +1.0 kJ). With 3.21 ÷ 80.04 ≈ 0.0401 mol, ΔH = 1.024100944 ÷ 0.0401 = 25.5387 kJ/mol.
Nitric acid from NO₂ (OpenStax Example 5.15). Products: 2 × (−207.4) + 1 × 90.25 = −324.55 kJ. Reactants: 3 × 33.2 + 1 × (−285.83) = −186.23 kJ. ΔH° = −324.55 − (−186.23) = −138.32 kJ. The book prints a products sum of −323.03 kJ and ΔH° = −136.80 kJ; its own values add to −324.55 kJ.
Other questions people ask
What is enthalpy change?
ΔH is the heat a reaction gives off or takes in at constant pressure. A negative ΔH means the reaction is exothermic (it gives off heat); a positive ΔH means it is endothermic (it takes heat in).
How do I find ΔH from a calorimeter?
The solution’s heat is q = m × c × ΔT. The reaction’s heat is the opposite, q_rxn = −q, and ΔH = q_rxn ÷ moles reacted. 100 g of solution at 4.184 J/(g·°C) warming 6.9 °C takes in 2,887 J, so 0.0500 mol of reaction has ΔH = −57.7 kJ/mol.
How do I use enthalpies of formation?
ΔH° = Σ n × ΔH°f(products) − Σ n × ΔH°f(reactants), with each n the coefficient in the balanced equation. Look up ΔH°f values in a table such as OpenStax Appendix G.
Why is the ΔH°f of oxygen gas zero?
An element in its standard state (O₂ gas, solid carbon as graphite, H₂ gas) is the starting point for formation enthalpies, so its ΔH°f is 0 by definition.
Why does the sign flip between the solution and the reaction?
Heat the reaction gives off goes into the solution. If the solution warms (q positive), the reaction lost that heat, so q_rxn is negative.
What units does the answer use?
kJ per mole of reaction as written. With formation data, a ΔH of −138.32 kJ/mol means 138.32 kJ given off when the equation’s amounts react, for example 3 mol of NO₂ with 1 mol of water.