Unit 9: Thermochemistry

System & Surroundings · q=mcΔT · Calorimetry · Bond Energy · Heating Curves · Thermal Stoichiometry

🔥 Concept Overview

9.1–9.4 — Energy Fundamentals

  • System: the part of the universe being studied (e.g., the reaction). Surroundings: everything else.
  • 1st Law of Thermodynamics: Energy cannot be created or destroyed; it can only change form. q_system + q_surroundings = 0
  • Heat (q) is energy transferred between system and surroundings. Temperature is the average kinetic energy of particles.
  • Heat transfer mechanisms: conduction (contact), convection (fluid flow), radiation (electromagnetic waves)

9.5–9.8 — Enthalpy & ΔH

  • Enthalpy (H): heat content of a system. ΔH = H_products − H_reactants
  • Exothermic (ΔH < 0): heat released to surroundings; reactants have more energy than products. Surroundings get warmer.
  • Endothermic (ΔH > 0): heat absorbed from surroundings; products have more energy than reactants. Surroundings get cooler.
  • Thermal equation notation: write ΔH as a value on the right side, OR write heat as a reactant (endothermic) or product (exothermic)

9.9–9.10 — Specific Heat & q = mcΔT

q = m × c × ΔT  |  q_system + q_surroundings = 0
  • q = heat (J or kJ); m = mass (g); c = specific heat (J/g·°C); ΔT = T_final − T_initial
  • Water's specific heat: c = 4.184 J/(g·°C) — one of the highest known (why oceans moderate climate)
  • Positive q = heat absorbed (endothermic); Negative q = heat released (exothermic)

9.11–9.12 — Calorimetry

  • A calorimeter is an insulated container used to measure heat exchange in reactions
  • Assumption: no heat escapes → q_reaction = −q_solution
  • Heat of solution/combustion: q = m_solution × c_water × ΔT, then ΔH = q / moles of reactant

9.14 — Bond Energy

ΔH_rxn = Σ(bonds broken) − Σ(bonds formed)
  • Breaking bonds requires energy (endothermic); Forming bonds releases energy (exothermic)
  • Bond energies are always positive (energy required to break); use them to calculate theoretical ΔH
  • If ΔH is negative → exothermic (more energy released forming bonds than spent breaking them)

9.15 — Thermal Stoichiometry

Use ΔH from a thermochemical equation as a conversion factor: kJ per mole of substance in the reaction.

mol substance × |ΔH| kJ/mol = kJ released or absorbed

9.16–9.17 — Heating/Cooling Curves

Temperature
___________
/ \ ← gas (slope = q=mcΔT using c_gas)
____/ \
←plateau: boiling \___________
(ΔH_vap, T constant) plateau: condensing

Slopes = temperature change → q = mcΔT (KE changes)
Plateaus = phase change → q = m × ΔH_fus or m × ΔH_vap (PE changes, T constant)
  • Slopes: temperature is changing → KE of particles changes → q = mcΔT
  • Plateaus: phase change occurring → temperature is constant → bonds between molecules breaking/forming → PE changes → q = m × ΔH (fusion or vaporization)
  • At plateaus: both phases coexist; adding heat breaks intermolecular bonds, not raising temperature

🃏 Flashcards

Click any card to flip it.

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System vs. Surroundings
System: the reaction/substance being studied. Surroundings: everything else. Heat flows from hot → cold. q_system + q_surroundings = 0.
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Exothermic
ΔH < 0. Heat is RELEASED to surroundings. Reactants have more energy than products. Surroundings warm up. Example: combustion, hand warmers.
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Endothermic
ΔH > 0. Heat is ABSORBED from surroundings. Products have more energy than reactants. Surroundings cool down. Example: melting ice, photosynthesis.
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Specific Heat (c)
Energy required to raise 1 g of a substance by 1°C. Units: J/(g·°C). Water = 4.184 J/(g·°C). High specific heat = substance resists temperature change.
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q = mcΔT
q = heat (J); m = mass (g); c = specific heat J/(g·°C); ΔT = T_final − T_initial. Positive q = absorbed; negative q = released.
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Calorimetry
Measuring heat exchange in a reaction using an insulated container (calorimeter). Assumption: q_reaction = −q_solution. ΔH = q ÷ moles of limiting reactant.
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Bond Energy Rule
ΔH = Σ(bonds broken) − Σ(bonds formed). Breaking bonds = endothermic (positive). Forming bonds = exothermic (negative). Net negative ΔH = exothermic reaction.
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Heating Curve: Slope
Temperature is increasing. Kinetic energy of particles increases. Use q = mcΔT to calculate energy. Different slopes = different specific heats for each phase.
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Heating Curve: Plateau
Phase change is occurring. Temperature stays constant. Potential energy changes (intermolecular bonds breaking/forming). Use q = m × ΔH_fus or m × ΔH_vap.
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Thermal Stoichiometry
Use ΔH as a conversion factor from a thermochemical equation. Example: if ΔH = −890 kJ/mol CH₄, then burning 2 mol CH₄ releases 1780 kJ.
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1st Law of Thermodynamics
Energy cannot be created or destroyed, only transferred or converted. q_system + q_surroundings = 0. This is why calorimetry works — all heat from reaction goes to solution.
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Heat vs. Temperature
Heat (q): total energy transferred — depends on mass (extensive). Temperature: average kinetic energy of particles — independent of mass (intensive). A large cold lake has more heat than a small hot flame.

⚡ Quick Review

📝 Practice Quiz

20 multiple choice + 3 short answer.

1. A 50.0 g sample of water absorbs 4,184 J. What is the temperature change? (c = 4.184 J/g·°C)

2. A 100.0 g sample of iron (c = 0.449 J/g·°C) is heated from 20.0°C to 70.0°C. How much heat was absorbed?

3. A reaction releases heat to its surroundings. This reaction is:

4. In a calorimetry experiment, 100.0 mL of 1.00 M HCl mixes with 100.0 mL of 1.00 M NaOH. Temperature rises from 22.0°C to 28.9°C. (density = 1.00 g/mL; c = 4.184 J/g·°C). What is q_solution?

5. Using bond energies: H₂ + Cl₂ → 2HCl. Bond energies: H–H = 436, Cl–Cl = 243, H–Cl = 431 kJ/mol. What is ΔH?

6. Which statement about a heating curve plateau is correct?

7. For the thermochemical equation: CH₄ + 2O₂ → CO₂ + 2H₂O ΔH = −890 kJ. How much heat is released when 2.00 mol of CH₄ burns?

8. Heat is different from temperature because:

9. According to the 1st Law of Thermodynamics, if q_system = −850 J, then q_surroundings =

10. Which process is exothermic?

11. A student burns a Cheeto and uses it to heat 100. g of water. Temperature rises from 20.0°C to 35.0°C. How much heat was transferred to the water?

12. For the reaction N₂ + 3H₂ → 2NH₃, ΔH = −92 kJ. How many kJ are released when 34.0 g of NH₃ (MM = 17.03 g/mol) are produced?

13. During the melting of ice at 0°C, which energy type is changing?

14. Which substance has the highest specific heat and thus best resists temperature changes?

15. A reaction has products at higher energy than reactants. A PE diagram of this reaction shows:

16. A 200. g sample of aluminum (c = 0.900 J/g·°C) cools from 80.0°C to 20.0°C. How much heat did it release?

17. In a bond energy calculation, ΔH = +50 kJ. This means:

18. When 5.00 g of NaOH (MM = 40.00 g/mol) dissolves in 100. g of water and the temperature rises from 22.0°C to 28.6°C, the q_solution is approximately:

19. On a heating curve, as a liquid is heated from 30°C to 80°C (no phase change), which formula applies?

20. The specific heat of water (4.184 J/g·°C) is much higher than metals. This means that compared to metals, water:

Short Answer Questions

SA1. In a coffee-cup calorimeter, 100.0 mL of 1.00 M HCl is mixed with 100.0 mL of 1.00 M NaOH. Temperature rises from 22.0°C to 28.9°C. (density = 1.00 g/mL; c = 4.184 J/g·°C) (a) Calculate q_solution. (b) Calculate q_reaction. (c) Calculate ΔH in kJ/mol of water formed. (d) Is the reaction exothermic or endothermic?

(a) m_solution = 200. mL × 1.00 g/mL = 200. g
q_solution = 200. g × 4.184 J/g·°C × (28.9 − 22.0)°C = 200 × 4.184 × 6.9 = +5774 J

(b) q_reaction = −q_solution = −5774 J

(c) mol H₂O = 0.100 L × 1.00 mol/L = 0.100 mol
ΔH = −5774 J / 0.100 mol = −57,740 J/mol = −57.7 kJ/mol

(d) Exothermic — ΔH is negative; heat flows from reaction to the solution (solution warms up).

SA2. Use bond energies to calculate ΔH for: N₂ + 3H₂ → 2NH₃. Bond energies: N≡N = 945 kJ/mol, H–H = 436 kJ/mol, N–H = 391 kJ/mol. (a) List all bonds broken and formed. (b) Calculate ΔH. (c) Is this reaction exothermic or endothermic?

(a) Bonds broken: 1 × N≡N = 945 kJ; 3 × H–H = 3 × 436 = 1308 kJ. Total broken = 2253 kJ
Bonds formed: 2 × NH₃, each has 3 N–H bonds = 6 × N–H = 6 × 391 = 2346 kJ

(b) ΔH = 2253 − 2346 = −93 kJ

(c) Exothermic (ΔH < 0) — more energy released forming bonds than used breaking bonds.

SA3. A 100. g sample of water is heated from 10.0°C to 100.0°C, then completely vaporized at 100.0°C. (a) How much energy is needed to raise the temperature to 100.0°C? (c_water = 4.184 J/g·°C) (b) How much additional energy is needed to vaporize the water? (ΔH_vap = 2260 J/g) (c) What is the total energy needed?

(a) q = mcΔT = 100. g × 4.184 J/g·°C × (100.0 − 10.0)°C = 100 × 4.184 × 90.0 = 37,656 J = 37.7 kJ

(b) q_vap = m × ΔH_vap = 100. g × 2260 J/g = 226,000 J = 226 kJ

(c) Total = 37,656 + 226,000 = 263,656 J ≈ 264 kJ

Notice: vaporization requires ~6× more energy than heating from 10–100°C!