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Reaction Energy and Kinetics Equilibrium

Reaction Energy and Kinetics Equilibrium. Chapter 17 and 18. I. Energy. A. Definitions 1. Energy is the ability to do work or produce heat. 2. Potential energy- energy due to the composition or position of an object . 3. Kinetic energy – energy of motion.

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Reaction Energy and Kinetics Equilibrium

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  1. Reaction Energy and KineticsEquilibrium Chapter 17 and 18

  2. I. Energy A. Definitions 1. Energy is the ability to do work or produce heat. 2. Potential energy- energy due to the composition or position of an object. 3. Kinetic energy – energy of motion

  3. 4. Law of conservation of energy – in any chemical reaction energy can be converted from one form to another, but it is not created or destroyed. 5. Chemical potential energy – energy stored in a substance because of its composition.

  4. B. Heat 1. Definition – energy that is in the process of flowing from a warmer object to a cooler object. 2. Symbol – q 3. Measuring heat a. calorie – amount of heat required to raise the temperature of 1 gram of pure water by 1 degree Celsius

  5. b. Joule – SI unit of heat and energy 1 J = 0.2390 calories 1 cal = 4.184 J

  6. C. Specific Heat 1. Definition – 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.

  7. 2. Specific Heat Table Substance Specific Heat J/gK Water(l) 4.18 Aluminum 0.897 Carbon 0.709 Copper 0.385 Iron 0.449 Gold 0.129 Which substance would be the best conductor of heat? Which substance would be the best insulator of heat?

  8. 3. Calculating Heat Evolved & Absorbed a. The heat absorbed or evolved in a rxn depends on the specific heat, mass of substance, and amount of which the temp changes.

  9. b. Equation q = c x m x ∆T q = heat absorbed or evolved J c = specific heat of substance J/(g ·°C) m = mass of the sample in grams ∆T= change in temp °C (Tfinal – Tinitial)

  10. Problem 1. In the construction of bridges and skyscrapers, gaps must be left between adjoining steel beams to allow for the expansion and contraction of the metal due to heating and cooling. The temperature of a sample of iron with a mass of 10.0 g changed from 50.4C to 25C with the release of 114 J heat. What is the specific heat of iron?

  11. 2. Determine the specific heat of a material if a 35 g sample absorbed 48J as it was heated from 293K to 313K. 3. If 980 kJ of energy are added to 6.2L of water at 291K, what will the final temperature of the water be?

  12. II. Heat in Chemical Reactions A. Measuring Heat 1. Calorimeter – insulated device used for measuring the amount of heat absorbed or released during a chemical rxn. a. A known mass of water is placed in the calorimeter. b. The water absorbs the energy released from the reacting system or provide the energy absorbed by the system. (energy flows from hot to cold)

  13. 2. Determining specific heat using calorimeter Example: Suppose you put 125 g of water into a foam cup calorimeter and find that its temperature is 25.6 C. Then, you heat a 50 g sample of metal to a temperature of 115C and put the metal into the water. The final temperature of the water is 29.3C. What is the specific heat of the metal?

  14. 2. The temperature of a sample of water increases from 20C to 46.6C as it absorbs 5650 J of heat. What is the mass of the sample?

  15. III. Thermochemistry • Definitions 1. Thermochemistry – study of heat changes that accompany chemical rxns and phase changes. 2. System – specific part of the universe that contains the reaction or process you wish to study

  16. 3. Surroundings – everything in the universe other than the system 4. Universe – system plus surroundings

  17. B. Enthalpy 1. Enthalpy, H, – heat content of a system at constant pressure 2. ∆Hrxn(change in heat of system)= heat absorbed or released in a chemical rxn ∆ Hrxn = Hproducts – Hreactants

  18. 3. Negative ∆ H – Exothermic Rxn 4Fe (s) + 3O2(g) 2Fe2O3(s) + 1625 kJ or 4Fe (s) + 3O2(g) 2Fe2O3(s) ∆H =-1625 kJ Both equations indicate 1625kJ released after reaction. Therefore, (-) sign means products have 1625 kJ less energy than reactants (lost to surroundings) Energy Diagram:

  19. 4. Positive ∆ H- Endothermic Rxn 27kJ + NH4NO3(s)  NH4+1(aq)+ NO3-1(aq) or NH4NO3(s)  NH4+1(aq)+ NO3-1(aq) ∆H =27 kJ Both equations indicate 27 kJ absorbed: endothermic rxn: ∆H (+) sign; gain of energy by products - ∆Hreactants<∆Hproducts Energy Diagram:

  20. 5. Calculating ∆H°rxn a. ∆H°rxn = ∆H°products - ∆H°reactants Equation must balanced and you must multiply by the number of moles of products and reactants.

  21. b. Example: Using the standard heats of formation given, calculate the ∆H°rxn for: CH4(g) + 2O2 CO2(g) + 2H2O(g) Substance ∆H°rxn (kJ/mol) CH4 (g) -74.81 CO2(g) -393.5 H2O(g) -241.8

  22. IV. Thermochemical Equations A. Writing Thermochemical Equations 1. Thermochemical Equation – balanced rxn that includes the states of all reactants and products and the energy change 2. The nature of the rxn or process is written as a subscript.

  23. 3. A zero is written as a superscript if the reaction is carried out at standard conditions (1 atm and 298K). 4. Enthalpy(heat) of combustion-∆H°comb– enthalpy change for the complete burning of 1 mole of a substance.

  24. B. Changes of State 1. Definitions a. Molar enthalpy (heat) of vaporization-∆Hvap-Heat required to vaporize one mole of the a liquid. b. Molar enthalpy(heat) of fusion – ∆Hfusheat required to melt one mole of a solid

  25. 2. Thermochemical Equation for Changes of State a. The vaporization of water and melting of ice equations: H2O(l)  H2O(g) ∆Hvap = 40.7 kJ H2O(s)  H2O (l) ∆Hfus = 6.01 kJ b. The reverse processes have the same numerical values but opposite signs. ∆Hfus = -∆Hsolid ∆Hvap = -∆Hcond

  26. c. Table of Standard Enthalpies

  27. 3. Calculating Energy in a Reaction Problem 1: A bomb calorimeter is useful for measuring the energy released in combustion reactions. The reaction is carried out in a constant volume bomb with a high pressure of oxygen. How much heat is evolved when 54.0g glucose(C6H12O6) is burned according to this rxn? C6H12O6 (g)+ 6O2 (g) 6CO2(g) + 6H2O(l) ∆Hcomb = -2808 kJ

  28. Problem 2 Calculate the heat required to melt 25.7g of solid methanol at its melting point. Problem 3 What mass of methane must be burned in order to liberate 12880 kJ of heat? Hcomb = -891 kJ/mol

  29. V. Shifting Equilibrium A. Definitions 1. Equilibrium – a dynamic condition in which two opposing changes occur at equal rates in a closed system. 2. Le Chatelier’s Principle – states that if a system is subjected to a stress, the equilibrium is shifted in the direction that tends to relieve the stress.

  30. B. Le Chatelier’s Principle 1. Provides a means to predict the influence of stress factors on equilibrium systems. 2. In a reversible chemical reaction: the rxn can move forward or backwards. A + B = C + D

  31. C. Stress Factors and Equilibrium 1. Changes in Pressure a. Increase in pressure – shift to reduce # of gas molecules Note: pressure doesn’t affect solids or liquids Example: b. Decrease in pressure – shift to increase # gas molecules. Example:

  32. 2. Changes in Concentration A + B  C+D a. Increase in concentration of reactants – shift to producing more product (to use up reactant) Example: b. Decrease in concentration of reactants – shift to reverse reaction to make more of reactant Example

  33. 3. Changes in Temperature A + B C+D a. Increase in temp. – equil shifts to use up heat (energy) Example: b. Decrease in temp – equil shifts to form more energy Example:

  34. C. Equilibrium Constant 1. Definition a. Equilibrium Expression, Keq - numerical value of ratio of product concentrations to reactant concentrations. Keq = [C]c[D]d [A]a[B]b aA + bB  cC + dD *Don’t include solids and liquids in expression

  35. 2. Examples - Write, Keq , equilibrium constants a. N2 (g) + 3H2 2NH3(g) b. 2NaHCO3(s)  Na2CO3(s) +CO2(g) + H2O(g)

  36. 3. Calculating Keq – when concentrations are given. Example: Calculate the Keq for part a above [NH3] = .933 mol/L [N2] = .533 mol/L [H2] = 1.600 mol/L

  37. a. The Keq depends on the temperature of reactants and products. b. Keq is constant at a given temperature. c. A large value of Keq means that at equilibrium, the product is present in a larger amt than the reactants.(forward rxn favored.)

  38. d. A small value of Keq means that at equilibrium, the product is present in smaller amounts than the reactants. (reverse rxn favored) e. When the Keq is 1, the forward or reverse rxn is not favored.

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