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Flashcards · AP® Chemistry · Unit 6

Unit 6: Thermochemistry flashcards

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6.1 Endothermic and Exothermic Processes

What is the difference between the system and the surroundings?
The system is what you study, usually the reacting particles. The surroundings are everything else, usually the water and the container. Energy is conserved: what the system loses, the surroundings gain, and the other way round.
What do exothermic and endothermic mean?
Exothermic: the system's energy decreases as energy flows out, so the surroundings warm up. Endothermic: the system's energy increases as energy flows in, so the surroundings cool down. Always name the system: "dissolving NaOH is exothermic".
What does a thermometer reading in a reaction usually tell you?
The thermometer usually reads the surroundings (the water). Warmer water means the system released energy (exothermic); cooler water means it absorbed energy (endothermic). A solution cooling from 20.4 °C to 17.1 °C as a salt dissolves shows endothermic dissolving.
Are phase changes endothermic or exothermic?
Melting, boiling and sublimation are endothermic: attractions between particles are overcome. Freezing, condensing and deposition are exothermic: attractions form. Frost forming on a car is exothermic for the water.
In an energy change for a reaction, is bond breaking endothermic or exothermic?
Breaking a bond always needs energy; forming a bond always releases energy. "Breaking bonds releases energy" is never correct.
When is dissolving exothermic and when is it endothermic?
It can be either. It is exothermic when the solute–solvent attractions formed are stronger overall than the solute–solute and solvent–solvent attractions broken, and endothermic when they are weaker.
In what two ways can energy cross the system boundary?
As heat, because of a temperature difference, or as work, for example a gas pushing back the air. Work is usually small unless a lot of gas is made or used up.
Why is "cold flows in" an incorrect description of cooling?
Only energy moves. Cooling means energy left the surroundings. Also, a temperature change smaller than the thermometer's precision is not evidence of a process.

6.2 Energy Diagrams

What does an energy diagram show, and how are its axes labelled?
The energy of the system at each stage of a process (the surroundings are not drawn). Vertical axis: "Energy (kJ mol⁻¹)", not temperature. Horizontal axis: progress from start to finish, with no scale or unit. Zero can go anywhere; only differences matter.
How do you read ΔH from an energy diagram?
ΔH = E(final) − E(start); the arrow runs from start level to end level. Reactants at 20 kJ mol⁻¹ and products at 85 kJ mol⁻¹ give ΔH = +65 kJ mol⁻¹, endothermic.
How do exothermic and endothermic diagrams differ?
Exothermic: final level lower, arrow down, ΔH negative, surroundings warm. Endothermic: final level higher, arrow up, ΔH > 0, surroundings cool.
What happens to ΔH for the reverse process?
The same two levels with the arrow reversed: ΔH changes sign. If ΔH = −184 kJ mol⁻¹, the reverse is +184 kJ mol⁻¹ and its arrow points up.
How are the phases of one substance arranged on an energy diagram?
Gas highest, then liquid, then solid lowest, because energy is needed to overcome attractions on melting and boiling.
How is dissolving shown as a step diagram?
Step 1 (separate solute particles) up, step 2 (separate solvent particles) up, step 3 (solute–solvent attractions form) down. The overall ΔH is the sum of the three. This is a model; the real process does not happen in separate steps.
Does adding an activation-energy hump or a catalyst change ΔH on the diagram?
No. A hump can be added, and a catalyst lowers it, but the start and end levels, and so ΔH, stay the same.

6.3 Heat Transfer and Thermal Equilibrium

What does temperature measure at the particle level?
The average kinetic energy of the particles. Particles in a warmer body have greater average kinetic energy than those in a cooler body, though at any moment they have a spread of energies.
How does heat transfer happen between two bodies in contact?
Their particles collide. On average energy passes from faster particles to slower ones, so energy moves from the warmer body to the cooler body. The net transfer slows as the temperature gap closes.
What is thermal equilibrium?
When the average kinetic energies, and so the temperatures, are equal. Particles still move and collide, but energy passes equally each way, so there is no net transfer. Two blocks both at 30 °C have no net heat transfer.
Where does the final temperature lie when a warm and cool body are mixed in an insulated container?
Between the two starting temperatures, because energy lost by the warmer body = energy gained by the cooler body. It is halfway only if the two heat capacities are equal.
How does doubling the Kelvin temperature affect average kinetic energy?
Average KE ∝ T in kelvin: KE₂/KE₁ = T₂/T₁. 200 K → 400 K doubles it. 100 °C → 200 °C does not: 373.15 K → 473.15 K is a factor of 1.268. Convert to kelvin before any ratio.
What is the difference between heat and temperature?
Temperature is a property of a body (average kinetic energy of its particles). Heat is energy in transit because of a temperature difference. Same temperature does not mean same total energy: a bigger body has more particles.
Do particles at the same temperature have the same speed?
No. They have the same average kinetic energy, KE = ½mv², so lighter particles move faster on average.

6.4 Heat Capacity and Calorimetry

What does the sign of q mean?
q is energy transferred as heat (J or kJ). Positive q: the system absorbs heat and its energy rises. Negative q: it releases heat and its energy falls. ΔT = T(final) − T(initial) gives q its sign.
What is specific heat capacity, and what is the value for water?
The energy to warm 1 g of a substance by 1 °C, c, in J g⁻¹ °C⁻¹. Water: 4.18 J g⁻¹ °C⁻¹, unusually large. A substance with a high c heats slowly; for equal masses and energy, the smaller c gives the larger ΔT.
How do you calculate heat from a temperature change?
q = mcΔT (q in J, m in g, c in J g⁻¹ °C⁻¹, ΔT in °C or K). Warming 10.0 g of water by 5.0 °C needs 10.0 × 4.18 × 5.0 = 209 J. Use the mass whose temperature changes; for a solution, water plus solute.
What are molar heat capacity and heat capacity of an object?
Molar heat capacity C_m = c × M (J mol⁻¹ °C⁻¹), used as q = nC_mΔT. Heat capacity of an object or calorimeter C = mc (J °C⁻¹), used as q = CΔT. Match c with grams and C_m with moles.
How do you find q for a process happening in a calorimeter solution?
q(process) = −q(solution). If the solution's q is −850 J (it cooled), the process has q = +850 J and is endothermic. In any insulated calorimeter q(lost) + q(gained) = 0.
What assumptions does coffee-cup calorimetry make?
No energy is exchanged with the air or bench; the calorimeter absorbs negligible energy unless its heat capacity is given; a dilute solution has c = 4.18 J g⁻¹ °C⁻¹; the final reading is the equilibrium temperature after stirring.
100.0 g of water and 100.0 g of copper each absorb 2.00 kJ. Which warms more?
Copper, by about 52 °C, against 4.78 °C for water, because copper's specific heat capacity is much smaller than water's.
How many significant figures should a calorimetry answer have?
ΔT is found by subtraction, so it often has fewer significant figures than the readings themselves. Let ΔT limit the answer.

6.5 Energy of Phase Changes

Why does temperature stay constant during a phase change?
The energy changes the particles' potential energy (overcoming or forming attractions), not their average kinetic energy. So particles do not speed up while melting.
How do you calculate the energy for a phase change?
q = n × ΔH, with n in mol (n = m / M first; do not multiply grams by ΔH). Melting 2.00 mol of ice at 0 °C: 2.00 × 6.01 = +12.0 kJ.
What are the values of ΔH_fus and ΔH_vap for water?
ΔH_fus = 6.01 kJ mol⁻¹ at 0 °C; ΔH_vap = 40.7 kJ mol⁻¹ at 100 °C.
Why is ΔH_vap much larger than ΔH_fus?
Boiling separates the particles completely, while melting only loosens them. Stronger intermolecular forces mean a larger ΔH_vap.
How are the enthalpies of freezing and condensation related to fusion and vaporization?
Same size, opposite sign: ΔH_freezing = −ΔH_fus and ΔH_cond = −ΔH_vap. Condensing 0.500 mol of steam: −20.4 kJ (released). Freezing 9.01 g of water: −3.01 kJ.
How do you read a heating curve?
Flat sections are phase changes: the temperature is the melting or boiling point, and the length is the energy needed. Sloping sections are one phase warming, using q = mcΔT; a steeper slope means a smaller heat capacity.
How do you solve a multi-step heating problem (e.g. ice to steam)?
Split at every melting and boiling point. Use q = mcΔT for each phase, with its own c (ice, liquid water and steam differ), and q = nΔH for each phase change. Convert every step to the same unit (J or kJ) before adding.

6.6 Introduction to Enthalpy of Reaction

What is the enthalpy change of reaction, ΔH?
The heat transferred by a reaction at constant pressure. Negative: heat released (exothermic). Positive: heat absorbed (endothermic). It is quoted per mole of reaction, for the amounts in the balanced equation with state symbols.
How do you calculate the heat for a given amount of reactant?
Moles of reaction = n ÷ coefficient (using the limiting reactant), then q = moles of reaction × ΔH. CH₄ + 2O₂ → CO₂ + 2H₂O(l), ΔH = −890 kJ mol⁻¹: 2.00 mol CH₄ gives q = −1780 kJ.
How does ΔH change when an equation is multiplied or reversed?
Multiply the equation by k → multiply ΔH by k. Reverse it → change the sign. N₂ + 3H₂ → 2NH₃ has ΔH = −92 kJ mol⁻¹, so 6NH₃ → 3N₂ + 9H₂ has +276 kJ mol⁻¹.
Why do products of an exothermic reaction briefly get hotter than the surroundings?
Bond changes lower the chemical potential energy, so the products' particles move faster. Energy then flows to the surroundings until thermal equilibrium. An endothermic reaction leaves the products colder, and energy flows in.
A reaction in water makes the temperature fall. Is ΔH positive or negative?
Positive. The system absorbed energy from the water: q_rxn = −q_soln = −mcΔT, and ΔT is negative.
What is a common error when using coefficients with ΔH?
Forgetting to divide by the coefficient. If the equation has 2NH₃, divide moles of NH₃ by 2 to get moles of reaction. Also use the limiting reactant, not the excess one.

6.7 Bond Enthalpies

What is a bond enthalpy?
The energy to break one mole of a bond in gaseous molecules. It is always positive. Forming the same bond releases the same amount. Tables give averages over many molecules, so answers are estimates.
How do you estimate ΔH from bond enthalpies?
ΔH ≈ Σ(bonds broken) − Σ(bonds formed). Broken minus formed, not the reverse. H₂ + F₂ → 2HF with H–H 432, F–F 154, H–F 565: 586 − 1130 = −544 kJ mol⁻¹.
What makes a reaction exothermic in terms of bond enthalpies?
More energy is released forming bonds in the products than is needed to break bonds in the reactants, so ΔH negative. If more is needed than released, ΔH > 0.
How do you count bonds correctly for a bond-enthalpy calculation?
Use the Lewis structure: bonds per molecule × coefficient. 2NH₃ has 6 N–H bonds; 2 mol CH₄ contain 8 mol C–H. Use the right bond order: O₂ is O=O, N₂ is N≡N, CO₂ has two C=O. Use the C=C value itself, not two C–C.
Why can bonds that are unchanged be left out of a bond-enthalpy calculation?
A bond broken and then formed again contributes equal and opposite amounts, so it cancels and the answer does not change.
Why does a bond-enthalpy estimate differ from the measured ΔH?
Table values are averages over many molecules, not the exact value in each one; and the method assumes all species are gases, so it ignores intermolecular forces in liquids or solids.

6.8 Enthalpy of Formation

What is a standard enthalpy of formation?
ΔH°f is the enthalpy change when 1 mol of a substance forms from its elements in their standard states under standard conditions (1 bar, data usually at 298 K). For CO₂(g): C(graphite) + O₂(g) → CO₂(g). Fractions such as ½N₂ are allowed.
Which substances have ΔH°f = 0?
Elements in their standard states (most stable form): O₂(g), N₂(g), H₂(g), C(graphite), Br₂(l), Fe(s). Other forms are not zero: O₃(g) and C(diamond) have positive values.
How do you calculate ΔH° for a reaction from enthalpies of formation?
ΔH°rxn = ΣnΔH°f(products) − ΣnΔH°f(reactants), multiplying each by its coefficient. 2NO + O₂ → 2NO₂ with NO +90.3, NO₂ +33.2: 66.4 − 180.6 = −114.2 kJ mol⁻¹. Use brackets when subtracting negative totals.
Why does the state of a substance matter when using ΔH°f values?
Different states have different values: H₂O(l) −285.8 kJ mol⁻¹, H₂O(g) −241.8 kJ mol⁻¹. H₂O(l) → H₂O(g) has ΔH° = +44.0 kJ mol⁻¹. Use the table value for the state shown.
How is the formation-enthalpy formula different from the bond-enthalpy formula?
Formation: products − reactants. Bond enthalpies: broken − formed, so reactants come first. Mixing them up gives the wrong sign.
How do you find an unknown ΔH°f from a reaction's ΔH°?
Write ΔH°rxn = ΣnΔH°f(products) − ΣnΔH°f(reactants), substitute the known values (zero for standard-state elements), and solve for the missing one.

6.9 Hess’s Law

What is Hess's law?
The ΔH of a process equals the sum of the ΔH values of any series of steps that start and end with the same substances. The steps can be imaginary. ΔH depends only on the start and end.
Why does Hess's law work?
Energy is conserved (first law). Each step transfers energy to or from the surroundings, so the net transfer is the sum. If two routes gave different totals, you could go out by one and back by the other and create energy, which is impossible.
What happens to ΔH when you reverse, multiply or add equations?
Reverse: same size, opposite sign (−96 → +96). Multiply by c: multiply ΔH by c (½ × −312 = −156). Add equations: add ΔH values (+57 + −120 = −63). Reverse and double J → L (−96): 2L → 2J is +192 kJ mol⁻¹.
What is the method for combining equations to reach a target equation?
For each target substance, find the given equation containing it (start with substances that appear once), reverse it if it is on the wrong side, multiply to match the coefficient, add the equations, cancel what appears on both sides, check the sum is exactly the target, then add the adjusted ΔH values.
Why do state symbols matter in Hess's law?
H₂O(l) and H₂O(g) are different substances. A liquid and a gas of the same compound do not cancel; the phase change is a step with its own ΔH.
How is the enthalpy-of-formation formula related to Hess's law?
It is Hess's law with a route through the elements: break reactants into elements (reverse their formations), then form the products from the elements.
What cancelling error is common in Hess's law problems?
Cancelling the whole of a substance when the amounts on each side differ. Cancel only the smaller amount, and keep the remainder on its side.

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