Difficulty: Introductory to Intermediate | Prerequisites: mole concept, molar mass calculations, basic algebra.
This topic sits at the junction of thermochemistry and stoichiometry in a first-year General Chemistry course. You need it to understand how chemical reactions exchange energy with their surroundings, how to quantify that energy using enthalpy (ΔH), and how calorimetry experiments measure it in the lab. If you are comfortable converting between grams and moles and can read a balanced equation, you have what you need to start here.
Every chemical reaction either absorbs or releases energy. The enthalpy change (ΔH) tells you how much and in which direction: negative means energy flows out (exothermic), positive means energy flows in (endothermic). You can scale ΔH to any amount of reactant or product using mole ratios and molar masses, and you can measure it experimentally with a calorimeter using q = mcΔT.
Enthalpy (H)
A thermodynamic quantity representing the total heat content of a system at constant pressure. In simple terms, think of it as the energy stored inside a substance that can be released or absorbed during a reaction.
Enthalpy change (ΔH)
The difference in enthalpy between products and reactants for a given reaction, measured in kilojoules (kJ). A negative value means energy is released; a positive value means energy is absorbed. Think of it as the energy bill for a reaction: negative means the reaction pays out, positive means it costs.
Exothermic reaction
A reaction that releases heat to the surroundings. ΔH is negative. In simple terms, the surroundings get warmer. Burning fuel and mixing acid with water are everyday examples.
Endothermic reaction
A reaction that absorbs heat from the surroundings. ΔH is positive. In simple terms, the surroundings get cooler. Dissolving ammonium nitrate in water (cold packs) is a common example.
Specific heat capacity (c)
The amount of energy (in joules) needed to raise the temperature of one gram of a substance by one degree Celsius. For liquid water, c = 4.184 J/(g·°C). Think of it as how stubborn a material is about changing temperature. Water is very stubborn, which is why it takes a long time to boil.
Calorimetry
The experimental technique for measuring the heat exchanged during a chemical or physical process, typically using an insulated container (calorimeter) and the equation q = mcΔT. In simple terms, you let a reaction happen inside a cup of water, measure how much the water's temperature changes, and work backwards to find how much heat the reaction released or absorbed.
q (heat)
The quantity of thermal energy transferred between a system and its surroundings, measured in joules (J) or kilojoules (kJ). Think of q as the actual energy that moved, as opposed to ΔH, which is the energy change per mole written in the equation.
Molar mass
The mass of one mole of a substance, in grams per mole (g/mol). Used to convert between grams and moles in enthalpy calculations. If you know the grams and the molar mass, dividing gives you moles, which you need to use ΔH.
A negative ΔH means the reaction releases heat to the surroundings. The products sit at a lower energy level than the reactants. This is an exothermic reaction.
A positive ΔH means the reaction absorbs heat from the surroundings. The products sit at a higher energy level. This is an endothermic reaction.
The magnitude of ΔH tells you how much heat per the stoichiometric amounts written in the balanced equation. If the equation says 1 mol of product forms and ΔH = −65.5 kJ, then 65.5 kJ of heat is released per mole of that product.
ΔH as written applies to the exact mole quantities in the balanced equation.
To find the heat for a different number of moles, multiply: q = n × ΔH, where n is the number of moles of the substance whose coefficient in the equation is 1 (or scale accordingly).
Example: if ΔH = −65.5 kJ for producing 1 mol AgCl, then producing 0.45 mol AgCl releases 0.45 × 65.5 = 29.5 kJ.
Convert grams to moles first using molar mass: moles = grams ÷ molar mass.
Then multiply by ΔH as above.
Example: 9.00 g AgCl ÷ 143.32 g/mol = 0.0628 mol. Heat released = 0.0628 × 65.5 = 4.11 kJ.
When the balanced equation has coefficients greater than 1, ΔH applies to those coefficients as a set.
For 2 KClO₃ → 2 KCl + 3 O₂, ΔH = −78.0 kJ means 78.0 kJ is released when 2 mol of KClO₃ decompose.
To find how many grams of KClO₃ are needed to release 96.7 kJ: first find moles of KClO₃ (96.7 kJ × 2 mol / 78.0 kJ = 2.48 mol), then convert to grams (2.48 × 122.55 = 304 g).
Some problems give ΔH per mole and ask for heat per gram. Divide: heat per gram = ΔH ÷ molar mass.
Example: sucrose releases 5.16 × 10³ kJ/mol and has a molar mass of 342.30 g/mol. Heat per gram = 5160 ÷ 342.30 = 15.1 kJ/g.
A coffee-cup calorimeter measures heat at constant pressure. The heat gained by the water equals the heat lost by the reaction (opposite sign).
Calculate the heat absorbed by the water: q_water = m × c × ΔT, where m is the mass of water in grams, c is the specific heat of water (4.184 J/g·°C), and ΔT is the temperature change.
The heat of the reaction is q_reaction = −q_water (opposite sign, because energy gained by water was lost by the reaction).
To find ΔH per mole: convert grams of solute to moles, then divide q_reaction by that number of moles.
Example: 3.5 g of H₂SO₄ in 142 g of water, ΔT = 11.5°C. q_water = 142 × 4.184 × 11.5 = 6831 J = 6.83 kJ. Since the temperature rose, the reaction is exothermic, so q_reaction = −6.83 kJ. Moles of H₂SO₄ = 3.5 / 98.08 = 0.0357 mol. ΔH = −6.83 / 0.0357 = −191 kJ/mol.
Heat absorbed or released by a substance
q = m × c × ΔT
where q is heat in joules, m is mass in grams, c is specific heat in J/(g·°C), and ΔT is the change in temperature in °C.
Heat of reaction from calorimetry
q_reaction = −q_water
The reaction's heat is equal in magnitude but opposite in sign to the heat gained by the water.
Scaling enthalpy to a given number of moles
q = n × ΔH (adjusted for stoichiometric coefficients)
If the balanced equation shows 2 mol of a substance and ΔH applies to that pair, divide ΔH by 2 to get the value per single mole before multiplying by your actual moles.
Converting grams to moles
n = mass (g) ÷ molar mass (g/mol)
Enthalpy per gram
heat per gram = ΔH (kJ/mol) ÷ molar mass (g/mol)
Enthalpy per mole from calorimetry data
ΔH (kJ/mol) = q_reaction (kJ) ÷ moles of solute
Calorimetry is the basis of nutritional labelling: the "calories" on food packaging come from burning food in a bomb calorimeter and measuring the heat released. The enthalpy of combustion of sucrose, for instance, is how food scientists calculate the energy in sugar.
Enthalpy calculations for exothermic reactions matter in industrial process design. Engineers size cooling systems for reactors by knowing how much heat the reaction will dump into the surroundings. The decomposition of potassium chlorate is used in oxygen generators on aircraft and submarines.
Students often think a negative ΔH means the reaction "loses" energy in some vague sense. It does not disappear. The energy transfers to the surroundings as heat. The system's enthalpy decreases, the surroundings' temperature increases.
Students sometimes apply ΔH directly to any number of moles without checking the stoichiometric coefficients. If the equation is written for 2 mol of reactant and ΔH = −78.0 kJ, that value covers 2 mol, not 1. You must scale accordingly.
A common error in calorimetry is forgetting to flip the sign. If the water gets hotter, q_water is positive, but q_reaction is negative (exothermic). The reaction gave heat to the water.
Some students add the mass of the solute to the mass of water when computing q. Whether to do this depends on the problem's instructions. In a simple coffee-cup calorimeter problem at this level, many textbooks use only the mass of water unless told otherwise. Read the problem carefully.
⚠️ Sign conventions for ΔH are tested constantly. Know that negative = exothermic = heat released. Positive = endothermic = heat absorbed.
⚠️ Expect problems that give ΔH for a balanced equation and ask you to find the heat for a non-standard number of moles or grams. The two-step process (convert to moles, then multiply by ΔH per mole) appears on nearly every general chemistry exam.
⚠️ Calorimetry problems require you to compute q = mcΔT, flip the sign, and often convert to kJ/mol. Practise all three steps in sequence.
⚠️ Unit conversions between J and kJ trip students up under time pressure. 1 kJ = 1000 J. Write your units at every step.
True or false: a reaction with ΔH = +120 kJ is exothermic.
Answer: False. A positive ΔH means the reaction is endothermic (absorbs heat).
Fill in the blank: to convert grams of a substance to moles, divide grams by ________.
Answer: molar mass (g/mol).
True or false: in a coffee-cup calorimeter, if the water temperature rises, the reaction is exothermic.
Answer: True. The reaction released heat into the water, raising its temperature.
Fill in the blank: q_reaction = ________ × q_water.
Answer: −1 (q_reaction equals the negative of q_water).
True or false: if a balanced equation shows ΔH = −78.0 kJ for 2 mol of reactant, then 1 mol of that reactant releases 78.0 kJ.
Answer: False. 1 mol releases half, which is 39.0 kJ.
Q: A reaction has ΔH = −65.5 kJ per mole of product. Is heat absorbed or released? Is the reaction exothermic or endothermic?
A: Heat is released (the surroundings gain energy). The reaction is exothermic because ΔH is negative.
Q: Using the same reaction (ΔH = −65.5 kJ/mol of AgCl), calculate the heat released when 0.45 mol of AgCl is produced.
A: q = 0.45 mol × 65.5 kJ/mol = 29.5 kJ released.
Q: Calculate the heat released when 9.00 g of AgCl is produced. (Molar mass of AgCl = 143.32 g/mol, ΔH = −65.5 kJ/mol.)
A: Moles = 9.00 g ÷ 143.32 g/mol = 0.0628 mol. Heat = 0.0628 × 65.5 = 4.11 kJ released.
Q: For the reaction 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g), ΔH = −78.0 kJ. How many grams of KClO₃ are needed to produce 96.7 kJ of heat? (Molar mass of KClO₃ = 122.55 g/mol.)
A: Moles of KClO₃ = 96.7 kJ × (2 mol / 78.0 kJ) = 2.48 mol. Grams = 2.48 × 122.55 = 304 g of KClO₃.
Q: Sucrose (C₁₂H₂₂O₁₁, molar mass 342.30 g/mol) releases 5.16 × 10³ kJ/mol when oxidised. How much heat is released per gram?
A: 5160 kJ/mol ÷ 342.30 g/mol = 15.1 kJ/g.
Q: 3.5 g of H₂SO₄ is dissolved in 142 g of water in a calorimeter and the temperature rises by 11.5°C. (a) Calculate q_reaction. (b) Find ΔH in kJ/mol of H₂SO₄. (Molar mass of H₂SO₄ = 98.08 g/mol.)
A: (a) q_water = 142 g × 4.184 J/(g·°C) × 11.5°C = 6831 J = 6.83 kJ. Since the temperature rose, q_reaction = −6.83 kJ. (b) Moles of H₂SO₄ = 3.5 / 98.08 = 0.0357 mol. ΔH = −6.83 / 0.0357 = −191 kJ/mol.
This material connects directly to Hess's law and standard enthalpies of formation, which let you calculate ΔH for reactions you have never measured in a calorimeter. Once you are comfortable scaling ΔH by moles and grams, those topics are a natural extension.
The stoichiometry skills you use here (mole ratios, molar mass conversions) are the same ones needed for limiting-reagent problems, solution stoichiometry, and gas stoichiometry later in the course.
Calorimetry connects forward to thermodynamics topics such as entropy, Gibbs free energy, and the second law, where you move from asking "how much heat?" to asking "does this reaction happen on its own?"
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