THERMOCHEMISTRY

1. THERMOCHEMISTRY SC5. Obtain, evaluate, and communicate information about the Kinetic Molecular Theory to model atomic and molecular motion in chemical and physical processes. a. Plan and carry out an investigation to calculate the amount of heat absorbed or released by chemical or physical processes. (Clarification statement: Calculation of the enthalpy, heat change, and Hess’s Law are addressed in this element.) b. Construct an explanation using a heating curve as evidence of the effects of energy and intermolecular forces on phase changes. c. Develop and use models to quantitatively, conceptually, and graphically represent the relationships between pressure, volume, temperature, and number of moles of a gas.

2. Thermochemistry… • is the study of the energy and heat associated with chemical reactions and/or physical transformations • focuses on these energy changes, particularly on the system's energy exchange with its surroundings

3. Energy, Heat & Temperature • Temperature (ΔT) is the quantitative measurement of average thermal energy. • It is NOT affected by mass • This should make sense – Can’t two different items have the same temperature? • It is usually expressed as and is measured in Celsius or Kelvin • Recall K = 273 + °C

4. Energy, Heat & Temperature • Thermal energy is the portion of the internal energy that is responsible for a system's temperature • It is the measure of the total amount of kinetic energy(energy of motion) in the system • It increases with mass

5. Energy, Heat & Temperature • Heat is the thermal energy that is transferred between systems at different temperatures. • It flows spontaneously from systems at high T to systems at low T. • Once the temperatures of the systems are the same, heat no longer flows (thermal equilibrium) • It is usually expressed as q or Q and is measured in Joules or calories • 1 calorie = 4.184 Joules • Food calories = kilocalories (kcal)

6. Entropy • Entropy (ΔS) is measure of disorder in a system. • In general, the universe (and everything in it) naturally becomes MORE disorderly. • Changes are more favorable (spontaneous) when the system has an increase in entropy. • Decomposition reactions have +ΔS • Synthesis reactions have -ΔS

7. Enthalpy • Enthalpy(ΔH) is the amount of heat content used or released in a system at constant pressure. • Changes are more favorable (spontaneous) when the system has a decrease in enthalpy. • ENDOTHERMIC processes have +ΔH • EXOTHERMIC processes have -ΔH

8. Enthalpy, Entropy, & Spontaneity • Spontaneous changes have • POSITIVE entropy and • NEGATIVEenthalpy • For example – • Burning is a spontaneous process because • The process is EXOTHERMIC and • The particles are INCREASING IN ENTROPY as they vaporize

9. Gibbs Free Energy • So what happens if one of the driving forces is favorable while the other is not (i.e. -ΔS while -ΔH)? • ΔG = ΔH - TΔS • Spontaneous reactions have -ΔG

10. More about enthalpy… • The energy used to form a compound is called enthalpy (or latent heat) of formation (ΔHf) • ΔHf C6H12O6 = -1273 kJ/mol • The energy released when a compound is burned in oxygen is called enthalpy (or latent heat) of combustion (ΔHc) • ΔHc C6H12O6 = -2801 kJ/mol

11. Enthalpy (Heat) of Reaction • Heat of reaction (ΔHrxn)is the amount of heat absorbed or released in the transformation of reactants into products under the same conditions. • 6 CO2 + 6 H2O  C6H12O6 + 6 O2 ΔHrxn = +2801 kJ/mol • CH4(g) + 2 O2(g)  CO2 + 2 H2O ΔHrxn = -891 kJ/mol

12. Hess’s Law (single rxn) • ΔHrxn=ΣΔHfproducts - ΣΔHf reactants • Calculate the ΔHrxn for the following: C3H6O(ℓ) + 4O2(g) ---> 3CO2(g) + 3H2O(ℓ) ΔH°f, C3H6O: -248 kJ/molΔH°f, O2: 0 kJ/mol ΔH°f, CO2: -393.5 kJ/molΔH°f, H2O: -285.83 kJ/mol • ΔHrxn = [3 (-393.5) + 3 (-285.83)] - [ (-248) + (4) (0) ] = -1790 kJ

13. Hess’s Law (multiple rxns) During discharge of a lead-acid storage battery, the following chemical reaction takes place: Pb+ PbO2 + 2H2SO42PbSO4 + 2H2O Using the following two reactions, (A) Pb + PbO2 + 2SO3 2PbSO4 ΔH° = -775 kJ (B) SO3 + H2O  H2SO4 ΔH° = -113 kJ determine the enthalpy of reaction for the discharge reaction above.

14. Hess’s Law (multiple rxns) Solution: 1) Multiply chemical equation (B) by 2: 2SO3 + 2H2O 2H2SO4 ΔH = -226 kJ 2) Reverse the reaction in chemical equation (B). Because of that, the sign of the change in enthalpy becomes positive. Let's identify the following chemical equation as (C): (C) 2H2SO42SO3 + 2H2O ΔH° = 226 kJ 3) Add chemical equations (A) and (C) together: Pb+ PbO2 + 2SO3 + 2H2SO42PbSO4 + 2SO3 + 2H2O 4) Then, add the enthalpy changes of equations (A) and (C): ΔH° = -775 kJ + 226 kJ = -549 kJ Remember to cancel out 2SO3 because it appears on both the reactant and product side, leaving you with the desired chemical reaction.

15. Hess’s Law (multiple rxns) The following two reactions are known: Determine the ΔH value for the reaction below: Fe2O3(s) + CO(g) ---> 2FeO(s) + CO2(g)

16. Phase Changes, Revisited • Remember that as a substance is undergoing a phase change, its temperature does NOT change. Why not? • Energy is being used to separate the particles and is being stored as potential energy, so there is no heat available to be measured

17. Even more about enthalpy… • The heat used to melt a substance is called enthalpy (or heat) of fusion (ΔHfus) • ΔHfus ice = 334 J/g OR 79.7 cal/g • The heat used to vaporize a substance is called enthalpy (or heat) of vaporization (ΔHvap) • ΔHvap water = 2258 J/g OR533 cal/g

18. Latent Heat • Q=mΔH or Q=mL is used to calculate the heat absorbed or released during phase changes. Q = heat (in joules), m = mass, ΔH or L = ΔHfus or ΔHvap depending on situation • Notice that temperature does NOT factor in this equation.

19. Latent Heat Calculations • How much energy is required to vaporize a 25 g block of ice?

20. More about phase changes… • So how do we figure out the heat absorbed or released in between the phase changes, where temperature IS a factor?

21. Specific Heat Q = mcpΔT m = mass cp = specific heat capacity ΔT = change in temperature (Tfinal – Tinitial)

22. Specific Heat Capacity… • Is the amount of heat required to raise the temperature of 1 gram of a substance by 1°C or 1 K • It varies among substances, and it can vary with temperature

23. Examples of cp Values Substance Specific heat capacity J/(g.K) Hydrogen (g) 14.3 Helium (g) 5.19 Water (l) 4.18 Ethanol (l) 2.44 Water (s) 2.11 Water (g) 2.08 Air (g) 1.01 Asphalt (s) 0.920 Aluminum (s) 0.897 Concrete (s) 0.880 Graphite (s) 0.710 Diamond (s) 0.509 Iron (s) 0.449 Copper (s) 0.385 Gold (s) 0.129 Lead (s) 0.129

24. Heating Curve for Water F 140 120 100 80 60 40 20 0 -20 -40 -60 -80 -100 Heat= m x Hvap Hv = 2258 J/g E D BP Heat= m x Hfus Hf = 334 J/g Heat= m x Cp, gas x ΔT Cp (steam) = 2.08 J/g°C Heat= m x Cp, liquid x ΔT Temperature (oC) Cp = 4.18 J/g°C B MP C A  B warm ice B  C melt ice (solid  liquid) C  D warm water D  E boil water (liquid  gas) E  F superheat steam Heat= m x Cp, solid x ΔT Cp (ice) = 2.11 J/g°C A Heat

25. Specific Heat Calculations How much heat is released when an 869 g iron bar cools from 94°C to 5°C?

26. Heat Calculations How much heat is absorbed by 50.0 grams of water as it changes from 0°C ice into 100°C steam?

27. Calorimetry… • is the technique of measuring the heat of chemical reactions or physical changes • The device used is called a calorimeter

28. Calorimetry “Coffee cup calorimeter” • Typically the cup is filled with water • Another substance (either hotter or colder) is added to the water • The temperature change of the water is measured

29. Calorimetry Calculations • Because the system is insulated, the heat remains within the system. So… Heat gained by water = Heat lost by the substance +Q = -Q mcp,waterΔT = -[mcp,substanceΔT] • Now, you just substitute the values you know, and solve for the value you don’t!

30. And now, a calorimetry problem! 240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC). When thermal equilibrium is reached, the system has a temperature of 42oC. Find the mass of the iron.

31. Gas Laws • Gas Laws proportionally relate various measurable quantities relating to gases under certain conditions: Pressure Volume Temperature Number of moles Boyle’s Law Gay-Lussac’s Law Charles’s Law Avogadro’s Law

32. Unit Standards • Pressure should be in atmospheres 1 atm = 760 mmHg = 760 torr = 101.3 kPa • Volume should be in liters 1 L = 1000 mL = 1000cm3 • Temperature should be in Kelvin K = 273 + °C • Number of moles should be in moles mol = mass ÷ molar mass

33. Boyle’s and Charles’s Laws • Boyle’s Law (pressure and volume are inversely proportional, given constant temperature and number of moles) • P1V1 = P2V2 • Charles’s Law (volume and temperature are directly proportional, given constant pressure and number of moles) • V1/T1 = V2/T2

34. Combined Gas Law • The combined gas law is simply a merging of Boyle’s and Charles’s Laws • P1V1/T1 = P2V2/T2

35. Gay-Lussac’s and Avogadro’s Laws • Gay-Lussac’s Law (pressure and temperature are directly proportional, given constant volume and number of moles) • P1/T1 = P2/T2 • Avogadro’s Law (volume and number of moles are directly proportional, given constant pressure and temperature) • V1/n1 = V2/n2

36. The Ideal Gas Law • This law works for all ideal gases, even ones NOT under constant conditions • PV=nRT • R = 0.0821 L·atm/mol·K

37. Gas Law Practice Problems A sample of oxygen gas has a volume of 36.7 L at 145 kPa and 28.0˚C. What volume will the sample have at STP (0°C and 100 kPa)?

38. Gas Law Practice Problems If 1.9 moles of gas is held at a pressure of 5.0 atm and in a container with a volume of 34 liters, what is the temperature of the gas?