Unit 6: Thermochemistry

19 terms. Say what each one means, then open the card to check.

• System and surroundings

  The system is the part being studied, usually the reacting chemicals; the surroundings are everything else, such as the water in a calorimeter. Energy lost by one is gained by the other.

  Example. When a match burns, the burning chemicals are the system and the air around the flame is the surroundings.

• Heat (q)

  Energy that moves between objects because their temperatures differ. It always flows from the warmer object to the cooler one until they reach the same temperature.

  Example. A hot spoon in cool water loses heat to the water until both are at the same temperature.

• Thermal equilibrium

  The state two objects in contact reach when they are at the same temperature, so there is no net flow of heat between them.

  Example. A cold can of juice left on a kitchen counter warms until it matches the room, and then its temperature stops changing.

• Specific heat capacity (c)

  The energy needed to raise the temperature of 1 g of a substance by 1 °C. Water's value is about 4.18 J/(g·°C), high compared with most metals.

  Example. Copper has c = 0.385 J/(g·°C), so 10 g of copper warms by 1 °C with only 3.85 J.

• Heat equation q = mcΔT

  The heat gained or lost by a substance with no phase change equals its mass times its specific heat times its change in temperature (final minus initial).

  Example. Warming 40.0 g of water from 15.0 °C to 35.0 °C takes 40.0 × 4.18 × 20.0 = 3340 J.

• Calorimetry

  Measuring the heat of a process from the temperature change of a known mass of water or of a calorimeter. The heat the process releases or absorbs equals the heat the calorimeter contents gain or lose, with the opposite sign.

  Example. If the water in a cup gains 1.50 kJ while a strip of magnesium reacts with acid, the reaction released 1.50 kJ, so its q is −1.50 kJ.

• Heat capacity of a calorimeter

  The energy needed to raise the temperature of a whole calorimeter by 1 °C, in J/°C or kJ/°C. Multiply it by ΔT to get the heat the calorimeter absorbed.

  Example. A calorimeter with a heat capacity of 8.0 kJ/°C that warms by 1.5 °C has absorbed 12 kJ.

• Heat of a phase change

  The heat absorbed or released when a substance changes phase at a constant temperature. Melting and boiling absorb heat; freezing and condensing release the same amount.

  Example. Water's enthalpy of vaporization is about 40.7 kJ/mol, so boiling 2.00 mol of water at 100 °C takes about 81.4 kJ.

• Exothermic process

  A process that releases heat to the surroundings. ΔH is negative, and the products are lower in energy than the reactants.

  Example. Stirring concentrated sulfuric acid into water releases heat, so the mixture gets hot.

• Endothermic process

  A process that absorbs heat from the surroundings. ΔH is positive, and the products are higher in energy than the reactants.

  Example. Sweat evaporating from your skin absorbs heat from the skin, which is why it cools you.

• Enthalpy change (ΔH)

  The heat absorbed or released by a reaction at constant pressure. For a thermochemical equation, ΔH goes with the moles shown by the coefficients.

  Example. For N₂(g) + O₂(g) → 2NO(g), ΔH = +180.6 kJ means the reaction absorbs 180.6 kJ for each mole of N₂ that reacts.

• Thermochemical equation

  A balanced equation written with its ΔH. Doubling the coefficients doubles ΔH, and reversing the equation changes the sign of ΔH.

  Example. If H₂ + ½O₂ → H₂O(l) has ΔH = −285.8 kJ, then 2H₂O(l) → 2H₂ + O₂ has ΔH = +571.6 kJ.

• Molar enthalpy of reaction

  The enthalpy change per mole of a named reactant or product, in kJ/mol. Find it by dividing the heat of the process by the moles of that substance and giving the sign.

  Example. If 0.0200 mol of a base is neutralized and the process releases 1.15 kJ, the molar enthalpy is −57.5 kJ/mol of base.

• Hess's law

  If a reaction can be written as the sum of other reactions, its ΔH is the sum of their ΔH values, because enthalpy depends only on the start and end states.

  Example. Adding A → B (ΔH = +20 kJ) and B → C (ΔH = −50 kJ) gives A → C with ΔH = −30 kJ.

• State function

  A property that depends only on the present state of a system, not on how it got there. Enthalpy is one, which is why Hess's law works.

  Example. The change in your elevation from a valley to a summit is the same by the steep trail or the gentle one, like ΔH by any path.

• Standard enthalpy of formation

  ΔH for making 1 mol of a compound from its elements in their standard states. It is zero for an element in its standard state, such as O₂(g) or C(graphite).

  Example. For NO(g), ΔH°f = +90.3 kJ/mol, the ΔH of ½N₂(g) + ½O₂(g) → NO(g).

• ΔH° from formation values

  ΔH°rxn = Σ n ΔH°f(products) − Σ n ΔH°f(reactants), where n is each substance's coefficient.

  Example. For 2NO(g) + O₂(g) → 2NO₂(g), with ΔH°f of +90.3 kJ/mol for NO and +33.2 kJ/mol for NO₂: 2(33.2) − 2(90.3) − 0 = −114.2 kJ/mol rxn.

• Bond energy

  The energy needed to break 1 mol of a bond in the gas phase. Breaking bonds absorbs energy, and forming the same bond releases the same amount.

  Example. The O=O bond energy is about 495 kJ/mol, so splitting 1 mol of O₂ into atoms takes about 495 kJ.

• ΔH from bond energies

  ΔH ≈ Σ(energies of bonds broken) − Σ(energies of bonds formed). It is an estimate, because tables list average bond energies.

  Example. For H₂ + F₂ → 2HF with H–H 436, F–F 155 and H–F 565 kJ/mol: 436 + 155 − 2(565) = −539 kJ/mol rxn.