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Chapter 15

Chapter 15 . Thermochemistry. Energy. What is energy? Energy is the ability to do work or produce heat. The Law of Conservation of Energy: This law states that can not be created or destroyed only transferred. Two types of energy: Kinetic Potential. Heat.

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Chapter 15

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  1. Chapter 15 Thermochemistry

  2. Energy • What is energy? • Energy is the ability to do work or produce heat. • The Law of Conservation of Energy: • This law states that can not be created or destroyed only transferred. • Two types of energy: • Kinetic • Potential

  3. Heat • Heat is energy transferred from a warmer object to a cooler object. • Heat is represented mathematically as q • The amount of heat energy required to raise the temperature of one gram of pure water by one degree Celsius is called… • a calorie • The SI unit for energy is the Joule • 1 Joule = 0.2390 calories • 1 calorie = 4.184 Joules

  4. Problems • A breakfast of cereal, orange juice, and milk contains 230 Calories. Convert this amount of energy in to Joules. • 9.6 x 105 J

  5. Glucose is a simple sugar found in fruit. Burning 1.00 g of glucose releases 15.6 kJ of energy. How many Calories are released? • 3.73 Calories • An fruit and oatmeal bar contains 142 Calories. Convert this energy to Joules • 5.94 x 105 • A chemical reaction releases 86.5 kJ of heat. How many Calories are released? • 20.7 Calories

  6. Specific Heat • The specific heat of any substance is the amount of heat required to raise the temperature of one gram of that substance by 1 degree Celsius. • The specific heat of water is: • 1 cal/g-oC • 4.184 J/g-oC • The specific heat of concrete is 0.84 J/g-oC • This is why in the summer concrete gets hot and water says cool.

  7. Using Specific Heat • The specific heat of a substance can be used to calculate the heat energy absorbed or given off when that substance changes temperature. • q = (C)(m)(ΔT) • remember ΔT = Tfinal – Tinitial • Calculate the heat absorbed by a 5 x 103g block of concrete when its temperature is raised from 20 oC to 26oC. • 25,000 J or 25 kJ

  8. The temperature of a sample of iron with a mass of 10.0 g changed from 50.4 oC to 25.0oC and released 114 J. What is the specific heat of iron? • 0.449 J/g-oC • If the temperature of 34.4 g of ethanol increase from 25.0 oC to 78.8 oC, how much heat has been absorbed buy the ethenol. (C for ethanol is 2.44 J/g-oC) • 4.52 x 103 J

  9. A 155 g sample of an unknown substance was heated from 25.0 oC to 40.0 oC. In the process, the substance absorbed 5696 J of energy. What is the specific heat of the substance? • 2.45 J/g-oC • A 38.8 g piece of metal alloy absorbs 181 J as its temperature increases from 25 oC to 36 oC. What is the alloy’s specific heat? • 0.424 J/g-oC

  10. Thermochemistry • Thermochemistry is the study of heat changes during chemical reactions or phase changes. • When studying thermochemistry we look at two things: • System • The system is the specific part of the universe that we are studying. • Surroundings • The surroundings are everything else in the universe.

  11. Enthalpy and Enthalpy Change • Enthalpy is defined as the heat content of a system at constant pressure. • The change in enthalpy for a reaction is called the enthalpy (heat) of reaction. • ΔHrxn • ΔHrxn = Hproducts – Hreactnats

  12. Thermochemical Equations • A thermochemical equation is a balanced equation that includes the physical states of all reactants and products and the enthalpy change. • 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) ΔH = -1625 kJ • NH4NO3(s)  NH4+(aq) + NO3- ΔH = 27 kJ

  13. Hess’s Law • Hess’s Law states that if you can add two or more equations to produce a final equation for a reaction than the sum of the enthalpy changes of the individual reactions is the enthalpy change of the overall reaction.

  14. Calculate ΔH for the reaction • 2 H2O2(l)  2 H2O(l) + O2(g) • 2 H2(g) + O2(g)  2 H2O(l) ΔH = -572 kJ • H2(g) + O2(g)  H2O2(l) ΔH = -188 kJ

  15. Use equations (a) and (b) to determine ΔH for the following reaction: • 2 CO(g) + 2 NO(g)  2 CO2(g) + N2(g) • 2 CO(g) + O2(g)  2 CO2(g) ΔH = -566.0 kJ • N2(g) + O2(g)  2NO(g) ΔH = -180.6 kJ

  16. ΔH for the following reaction is -1789 kJ. Use equation (a) to determine ΔH for reaction (b). • 4 Al(s) + 3 MnO2(s)  2 Al2O3(s) + 3 Mn(s) • 4 Al(s) + 3 O2 2 Al2O3(s) ΔH = -3352 kJ • Mn(s) + O2(g)  MnO2 ΔH = ?

  17. Enthalpy of Formation • The enthalpy of formation for any reaction is defined as the heat change when all reactants are in their elemental form and only one mole of product is produced. • S(s) + 3 F2(g)  SF6ΔHof = -1220 kJ • Sometimes we need to use fractional coefficients. • H2(g) + F2(g)  HF ΔHof = -273 kJ

  18. Using Enthalpy of Formation • We can use the enthalpy of formation for components of a reaction to calculate the total enthalpy change of the reaction (ΔHrxn). • H2S(g) + 4 F2(g)  2 HF(g) + SF6(g) • ½ H2(g) + ½ F2(g)  HF ΔHof = -273 kJ • S(s) + 3 F2(g)  SF6ΔHof = -1220 kJ • H2(g) + S(s)  H2S(g) ΔHof = -21 kJ

  19. Determine ΔH for CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l) • Using: • ΔHof(CO2) = -394 kJ • ΔHof(H2O) = -286 kJ • ΔHof(CH4) = -75 kJ • ΔHof(O2) = 0 kJ

  20. Reaction Spontaneity • When things rust the reaction taking place is: • 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) ΔH = -1625 kJ • Any physical or chemical that occurs with no outside intervention is a spontaneous process. • If we reverse the above reaction: • 2 Fe2O3(s)  3 O2(g) + 4 Fe(s) ΔH = 1625 kJ • This process would be non-spontaneous

  21. Entropy • Entropy is a measure of the number of possible ways a system can be configured. • If we have a piece of paper cut into 8 different sections there would be 56 different ways we could arrange them. (8 x 7) • If we cut the paper into 16 different pieces there would be 240 different ways we could arrange them. (16 x 15) • We have increased the papers entropy.

  22. The Second Law Of Thermodynamics • The second law of thermodynamics states that a spontaneous reaction will always occur in such a way that entropy increases. • Remember that the change in enthalpy (ΔH) is defined as: • Hproducts–Hreactants • Similarly: • ΔS = Sproducts – Sreactants • If ΔS is positive the entropy of the system is increasing. • If ΔS is negative the entropy of the system is decreasing.

  23. Predicting Entropy Changes • Phase Changes: • When a phase change occurs from a more ordered state to a less ordered state ΔS will be positive. • Solid  Liquid ΔS > 0 • When a phase change occurs from a less ordered state to a more ordered state ΔS will be negative. • Gas  Liquid ΔS < 0 • Dissolving a gas into a solvent always results in a decrease in entropy.

  24. Assuming no change in physical state, entropy increases when the number of moles of products is greater than the number of moles of reactants. • 2 SO3(g)  2 SO2(g) + O2(g) • Entropy increases when a solute dissolves in a solvent. • NaCl(s) Na+(aq) + Cl-(aq) • Entropy increases as temperature increases.

  25. Predict the sign of ΔS for each of the following chemical of physical processes • ClF(g) + F2(g)  ClF3(g) ΔS = • NH3(g)  NH3(aq) ΔS = • Entropy has the units Joules/Kelvin

  26. Gibbs Free Energy • Named after physicist J. Willard Gibbs, free energy is the maximum amount of energy available during a chemical reaction. • Gibbs Free Energy Equation: • ΔG = ΔH – TΔS • When a reaction occurs at standard conditions (298 K and 1atm) • ΔGo = ΔHo - ΔSo

  27. ΔG = ΔH – TΔS • A reaction where ΔH is negative and ΔS is positive will always be spontaneous. • N2(g) + 3 H2(g)  2 NH3(g) • ΔH = -91.8 kJ ΔS = -197 J/k • ΔG = • ΔG = -33.1 kJ

  28. ΔG = ΔH - TΔS

  29. For a process ΔH is 145 kJ and ΔS is 322 J/K. Calculate ΔG for this reaction at 298 K. Is it spontaneous?

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