(909) Thermochemistry Equations & Formulas - Lecture Review & Practice Problems

Thermochemistry Overview

  • Focus on thermochemistry concepts and equations crucial for mastering the chapter.

Internal Energy

  • Definition: Change in internal energy (ΔE) of a system.

  • Equation: ΔE = Q + W

    • Q: Heat energy that enters (+) or leaves (−) the system.

    • W: Work done on (+) or by (−) the system.

Heat Flow

  • Heat flows from hot to cold:

    • If system at 100°C and surroundings at 25°C:

      • Q (for system) = negative (exothermic process).

      • Surroundings absorb heat, making Q positive (endothermic).

  • Measurement Units:

    • Typically in joules (J).

    • 1 kJ = 1,000 J; 1 calorie = 4.184 J; 1 kcal = 1,000 calories.

Work (W)

  • Equation: W = PΔV

    • W > 0: Work done on the system.

    • W < 0: Work done by the system.

  • Expanding Gas: Change in volume (ΔV) positive, so W is negative.

  • Compressing Gas: Change in volume (ΔV) negative, so W is positive.

Example Problem

  • Given:

    • Q: 300 J (heat absorbed)

    • Expand from 2 L to 3 L at 5 atm.

  • Calculate ΔE:

    • W = PΔV = 5 atm * (3 L - 2 L) = 5 atm * 1 L = 5 L·atm = 5 * 101.3 J = -506.5 J (negative).

    • Final Calculation: ΔE = Q + W = 300 J - 506.5 J = -206.5 J (energy lost).

Heat Absorption and Release

Temperature Change

  • Equation: Q = mCΔT

    • m: Mass

    • C: Specific heat capacity (for water, 4.184 J/g°C).

    • ΔT: Change in temperature.

  • Example Calculation: Heating 50 g of water from 25°C to 75°C:

    • Q = 50 g * 4.184 J/g°C * (75°C - 25°C) = 10,460 J.

Phase Change

  • Equation: Q = m * ΔH

    • Usage of ΔH for fusion or vaporization.

  • Example: Melting 54 g of ice (heat of fusion = 6 kJ/mol).

    • Molar mass of water: 18 g/mol → 3 moles.

    • Required heat energy: 3 moles * 6 kJ/mole = 18 kJ.

Thermochemical Equations

Combustion Reaction Example

  • Combustion of propane forming CO2 and water:

    • Exothermic reaction releases 12,200 kJ of heat.

  • Calculating heat for 64 g of O2:

    • Molar mass of O2 = 32 g/mol → 2 moles of O2.

    • 2 moles ⇒ (12,200 kJ/5 moles * 2 moles) = 4,880 kJ released.

Back Conversion Example

  • If 3,600 kJ released, determining grams of CO2 produced:

    • Molar mass of CO2 = 44 g/mol.

    • Calculate moles based on kJ to moles:

      • Use ratio of 12,200 kJ for 3 moles of CO2.

      • Result: 396 g of CO2.

Hess's Law

  • Find enthalpy of reactions:

    • Example: 2A + 2B → D + E with known reactions.

  • Steps:

    1. Modify and add reactions to find the target reaction.

    2. Modify energies accordingly (reverse signs if reversing reactions).

    3. Sum enthalpy values to find total.

  • Final example yields: 800 kJ for overall reaction.

Conclusion

  • Summary of thermochemistry, emphasizing heat flow, energy changes, and calculations.