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Chapter 16 Reaction Energy & Kinetics

Chapter 16 Reaction Energy & Kinetics. 16-1 Thermochemistry. Thermochemistry. The study of the transfers of energy as heat that accompany chemical reactions and physical changes This heat can be measured in a calorimeter. Units.

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Chapter 16 Reaction Energy & Kinetics

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  1. Chapter 16Reaction Energy & Kinetics 16-1 Thermochemistry

  2. Thermochemistry • The study of the transfers of energy as heat that accompany chemical reactions and physical changes • This heat can be measured in a calorimeter

  3. Units • Temperature units may be in Kelvin (K) or degrees Celsius (°C) • Energy units are the joule (J) which is the SI unit for energy. The kJ maybe used as well.

  4. Heat Transfer • Think of heat transfer as the difference between two objects’ temperatures. The energy will flow from areas of high temperature to areas of low temperature.

  5. Heat Capacity & Specific Heat • Different materials can absorb energy differently. • Specific Heat is the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius or Kelvin. • See page 533 Table 16-1

  6. Heat Capacity & Specific Heat • Cp = q / (m x ∆T) or q = Cp x m x ∆T Where: Cp = specific heat q = energy lost or gained m = mass of sample ∆T = difference between initial & final temperatures

  7. Heat of Reaction • The quantity of heat released or absorbed during a chemical rxn. • When a rxn includes this information, we call it a thermochemical equation: 2H2(g) + O2(g)→ 2H2O (g) + 483.6 kJ If you double the amount of reactants, you double the amount of energy. Sometimes fractional coefficients are used.

  8. Thermochemistry • Enthalpy – The heat energy of reaction ∆H Enthalpy can only be measured by a change, not directly. ∆H is the amount of energy absorbed or lost by a rxn at constant pressure

  9. ∆H will be negative since energy has left the system Reaction Pathways The products have less energy than the reactants. The rxn released energy (heat) = exothermic

  10. ∆H will be positive since energy has been added to the system Reaction Pathways The products have more energy than the reactants. The rxn absorbed energy (heat) = endothermic

  11. Tips • Coefficients = moles • Physical states must be indicated as ΔH changes for different states of matter • ΔH is directly proportional to the molar coefficients, as they change, ΔH changes • Temperature usually does not effect ΔH

  12. Heat of Formation, ΔHf • The energy released or absorbed when one mole of a substance is formed from its elements (in terms of product) • Standard Heats of Formation, ΔHf0are values when the reactants and products are at their standard states of matter, at atmospheric pressure and room temperature (298 K)

  13. Heat of Formation, ΔHf • Compounds with a high negative heat of formation, largely exothermic, are very stable and occur readily. • An element’s ΔHf0 = zero • Compounds with a positive, or small negative heat of formation values are unstable and will decompose into their constituents.

  14. Heat of Combustion, ΔHc • The energy released when one mole of substance undergoes complete combustion (in terms of reactant) • All reactants are in their standard state. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g) ΔHc = -2219.2

  15. Calculating Heats of Reactions: Hess’s Law • The overall enthalpy change in a rxn is equal to the sum of enthalpy changes for the individual steps in the process. • Rules: • If a rxn is reversed, the sign on ΔH is reversed • Multiply the coefficients, if necessary, on rxns so the end rxn has the desired quantities

  16. These values are only for the formation of the compound Heat of Formation, ΔHf NaOH + HC2H3O2 NaC2H3O2 + H2O Na + O2 + H2  NaOH ∆Hf = - 469.15 kJ O2 + H2 + C  HC2H3O2∆Hf = - 484.5 kJ Na + O2 + H2+ C  NaC2H3O2∆Hf = 298.15 O2 + H2 H2O ∆Hf = -285.83 kJ

  17. NaOH + HC2H3O2 NaC2H3O2 + H2O Reverse these signs These are not being formed, rather they are being used Na + O2 + H2  NaOH ∆Hf = - 469.15 kJ O2 + H2 + C  HC2H3O2∆Hf = - 484.5 kJ Na + O2 + H2+ C  NaC2H3O2∆Hf = 298.15 O2 + H2 H2O ∆Hf = - 285.83 kJ Keep these the same since these compounds are being formed in the rxn

  18. NaOH + HC2H3O2 NaC2H3O2 + H2O NaOH  Na + ½O2 + ½H2 ∆Hf = 469.15 kJ HC2H3O2 O2 + 2H2 + 2C ∆Hf = 484.5 kJ Add them Na + O2 + 3/2H2+ 2C  NaC2H3O2∆Hf = 298.15 kJ ½O2 + H2 H2O ∆Hf = -285.83 kJ The value is positive, so the rxn is endothermic ∆H = 965.97 kJ Hess’s Law

  19. Practice Problem! Calculate ∆H for the following reaction: 2N2 (g) + 5O2 (g) 2N2O5 (g) Using the following data: ½O2 (g) + H2 (g) H2O (l) ∆Hf° = -285.8 kJ N2O5 (g) + H2O (l)  2HNO3 (l) ∆H° = -76.6 kJ 3/2O2(g)+½H2(g)+½N2(g)HNO3 (l) ∆Hf° = -174.1 kJ ∆ H = 28.4 kJ

  20. Chapter 16Reaction Energy & Kinetics 16-2 Driving Force of Reactions

  21. Enthalpy & Reaction Tendency • Most rxns tend to favor production of more stable products from less stable reactants. • That means exothermic rxns are usually favored in nature. • However, some endothermic rxns do occur spontaneously.

  22. Entropy, S • A measure of the degree of randomness of the particles in a system; tendency toward disorder. • There is a tendency for systems in nature to favor the disordered state. Gas Solid Liquid Increase in Entropy

  23. Entropy • A positive value for ∆S means an increase in entropy (disorder) • A negative value for ∆S means a decrease in entropy (disorder) • Rxns tend to favor high entropy and low enthalpy

  24. Free Energy, G • A combined enthalpy-entropy function • Rxns tend to favor a lower free energy system • Free Energy cannot be measured, only the change can be measured, ∆G ∆G0 = ∆H0 - T∆S0 @ Standard States

  25. Free Energy • If ∆G is positive, the reaction is nonspontaneous and will not readily occur. • If ∆G is negative, the reaction is spontaneous and will occur.

  26. Relationship between Enthalpy, Entropy & Free Energy Exothermic Disordering Always Ordering Exothermic @ low temps Endothermic Disordering @ high temps Endothermic Ordering Always

  27. Practice • Predict whether the value of ∆S will be greater than, less than or equal to zero: • 3H2 (g) + N2 (g) 2NH3 (g) • 2Mg (s) + O2 (g) 2MgO (s) • C6H12O6 (s) + 6O2 (g) 6CO2 (g)+ 6H2O (g) • KNO3 (s) K+ (aq) + NO3-(aq) ∆ S < 0 ∆ S < 0 ∆ S > 0 ∆ S > 0

  28. Practice • Calculate the value of ∆G0 for the reaction below. Will the reaction be spontaneous at 298 K? Cu2S (s) + S(s) 2CuS(s) ∆H0 = -26.7 kJ ∆S0 = -19.7 J/(mol∙K) ∆ G0 = -20.8 kJ/mol Yes, it is spontaneous

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