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Honors Chemistry

Honors Chemistry. Unit 10 – Chapter 16 ( p.531-559), Chapter 17 (p.560-587), and Chapter 18 (p.589-629) Thermochemistry, Kinetics and Equilibrium. Energy Transformations. “ Thermochemistry ” - concerned with heat changes that occur during chemical reactions

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Honors Chemistry

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  1. HonorsChemistry Unit 10 – Chapter 16 (p.531-559), Chapter 17 (p.560-587), and Chapter 18 (p.589-629) Thermochemistry, Kinetics and Equilibrium

  2. Energy Transformations • “Thermochemistry” - concerned with heat changes that occur during chemical reactions • Energy - capacity for doing work or supplying heat • weightless, odorless, tasteless • if within the chemical substances- called chemical potential energy

  3. Generic Heating Curve

  4. Heating Curve of Water

  5. Heating Curve of Water Heating curve observations (from left to right as heat is added): • Below 0C water is a solid • At 0C there is no change in temperature. Only a phase change is occurring (solid to liquid) • From 0C up to 100C water is a liquid • At 100C there is no change in temperature. Only a phase change is occurring (liquid to gas) • Above 100C water is a vapor (steam)

  6. Q = mcT Calculating how much energy is required to go from one phase to another • the diagonal parts of the curve • temperature change Q = mcT Q = heat energy m = mass T = Tfinal – Tinitial c = specific heat -for the phase and substance

  7. Q = nΔH The flat regions of the graph are not experiencing any temperature change. Q = nΔH n = number of moles ΔH = molar heat - depends phase change Molar heat of fusion ( DHf )- energy required to melt or freeze 1 mole of a substance Molar heat of vaporization ( DHv )- energy required to vaporize or condense 1 mole of a substance

  8. Molar Heats Molar heat of fusion ( DHf )- energy required to melt or freeze 1 mole of a substance (segment B) Ex. water’s molar heat of fusion = 6.02 kJ/mol Molar heat of vaporization ( DHv)- energy required to vaporize or condense 1 mole of a substance (segment D) Ex. water’s molar heat of vaporization = 40.6 kJ/mol

  9. ΔH and c The following are common constants you will use for water: ΔH fusion = 6.02 kJc liquid = 4.18 J molg°C ΔH vaporization = 40.6 kJc vapor = 2.03 J molg°C ΔH will be negative if cooling c solid = 2.06 J g°C

  10. Example • The following is an example of a problem involving a heating curve: How much energy is needed to melt 10. g of water at 0C?

  11. 10. g H2O x 1 mol H2O = 0.55 molH2O 1 18.02 g H2O Q = 0.55 mol H2O x 6.02 kJ = 3.3 kJ mol

  12. Example How much energy is needed to raise 10. g of water from 0C to 20. C?

  13. Q = mcT T = Tfinal – Tinitial = 20. C - 0C = 20. C Q = 10. g x 4.18 J x 20. C gC Q = 10. g x 4.18 J x 20. C = 836 J = 840 J gC

  14. Harder Example How much energy is needed to completely melt 10. g of water at 0C and then raise the water to 20. C?

  15. Sketch

  16. Combination of last two problems Flat region: Q= 3.3 kJ Diagonal region: Q = 840 J 840 J x 1kJ = 0.84 kJ 1 1000 J 0.84 kJ + 3.3 kJ = 4.14 kJ = 4.1 kJ

  17. Try it How much energy needs to be removed to condense 10.g of steam at 100. C and cool to 40. C?

  18. Sketch

  19. Flat region: Q1= nΔHv Q1= (_0.55mol_)(_-40.6KJ_) = -22KJ 1 mol Diagonal region: Q2= mcT. T = (40ºC) – (100ºC) = -60ºC Q2 = (_10.g_)(_4.18J_)(-60ºC) = -2500J 1 g ºC QT = ( -22KJ ) + ( -2.5KJ ) = -24.5KJ

  20. The graph below is a comparison of the activation energies for an uncatalyzed reaction and for the same reaction with a catalyst present. This is a generic chart without energy values on the vertical axis, but the potential energy of the reactants and the potential energy of the products can be seen.

  21. Endo vsExo Endothermic Reaction • Within the bonds of the products is stored some energy that came out of the collisions; i.e. energy was “put into” the reaction

  22. Endo vsExo Exothermic Reaction • The energy of the products is lower than the energy of the reactants; i.e. energy is released from the bonds of the reactants and is “given off” from the reaction

  23. Heat of Reaction Enthalpy - Heat of reaction– ( H) – The difference between the energy of the products and the energy of the reactants (enthalpy change) “” = final minus initial • If H is positive, energy was put into the reaction and it is endothermic. • If H is negative, energy was given off from the reaction and it is exothermic.

  24. Entropy • Entropy (S ) – a measure of the disorder of a system • Ex. when playing cards are ordered by number and suit, they have a low entropy • When a deck of cards is thrown into the air, they have a high entropy • + S = increasing entropy/disorder • - S = decreasing entropy/disorder • Law of Disorder – Processes move in the direction of increasing entropy

  25. Spontaneous Processes

  26. Hess Law • The overall enthalpy changes in a reaction is equal to the sum of the enthalpy changes for the individual steps. • A way to calculate H values

  27. Determine the heat of reaction for the reaction: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ Hint: The three reactions must be algebraically manipulated to sum up to the desired reaction. and.. the H values must be treated accordingly.

  28. Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ 4NH3 2N2 + 6H2 H =+183.6 kJ NH3: Reverse and x 2 O2 : Found in more than one place, SKIP IT (its hard). NO: x2 2N2 + 2O24NO H = 361.2 kJ H2O: x3 6H2 + 3O26H2O H = -1451.1 kJ

  29. Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) 4NH3  2N2 + 6H2 H =+183.6 kJ NH3: Reverse and x2 O2 : Found in more than one place, SKIP IT. NO: x2 2N2 + 2O24NO H = 361.2 kJ H2O: x3 6H2 + 3O26H2O H = -1451.1 kJ Cancel terms and take sum. H = -906.3 kJ + 5O2  + 6H2O 4NH3 4NO The reaction is Exothermic

  30. Try It Determine the heat of reaction for the reaction: C2H4(g) + H2(g)  C2H6(g) Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + ½ O2(g)  H2O(l) H = -286 kJ

  31. Determine the heat of reaction for the reaction: Goal: C2H4(g) + H2(g)  C2H6(g) H = ? Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H4(g) :use 1 as is C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ H2(g) :# 3 as is H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H6(g) : rev #2 2CO2(g) + 3H2O(l) C2H6(g) + 7/2O2(g) H = +1550 kJ C2H4(g) + H2(g) C2H6(g) H = -137 kJ

  32. Try It H2O(g) + C(s) → CO(g) + H2(g) Use these equations to calculate the molar enthalpy change which produces hydrogen gas. C(s) + ½ O2(g) → CO(g) ∆H = -110.5kJ H2(g) + ½ O2(g) → H2O(g) ∆H = -241.8kJ

  33. H2O(g) + C(s) → CO(g) + H2(g) Use these equations to calculate the molar enthalpy change which produces hydrogen gas. C(s) + ½ O2(g) → CO(g) ∆H = -110.5kJ H2O(g) → H2(g) + ½ O2(g) ∆H = +241.8kJ ___________________________________ C(s) + H2O(g) → H2(g) + CO(g) ∆H=+131.3kJ

  34. Try It Again Determine the heat of reaction for the reaction: C2H4(g) + H2(g)  C2H6(g) Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ Consult your neighbor if necessary.

  35. Determine the heat of reaction for the reaction: Goal: C2H4(g) + H2(g)  C2H6(g) H = ? Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H4(g) : use 1 as is C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ H2(g) : # 3 as is H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H6(g) : rev #2 2CO2(g) + 3H2O(l) C2H6(g) + 7/2O2(g) H = +1550 kJ C2H4(g) + H2(g) C2H6(g) H = -137 kJ

  36. Potential Energy (PE) Diagrams

  37. Reversible reaction Many reactions do not continue until all of the reactants have been converted into products. This is because many reactions are reversible. Reversible reaction – A reaction where reactants form products and products form reactants – indicated by the sign

  38. Reversible reaction • In a reversible reaction, equilibrium is reached when the forward reaction and the reverse reaction occur at the same rate. • Let’s look at the following equation: • Equilibrium Point: where the rate of the forward reaction (reaction of H2O with Cl2) is equal to the rate of the reverse reaction (reaction of HCl with O2).

  39. Dynamic equilibrium • Dynamic equilibrium – for every collision that produced product, there is a collision that produces reactant (occurs in a closed system) Equilibrium constant (K)= [product]coefficient from equation x [product]coefficient from equation [reactant]coefficient from equationx [reactant]coefficient from equation

  40. Equilibrium Expressions The equilibrium constant (K) remains the same as long as temperature is constant. The concentrations of the products and reactants may change, but as long as temperature is constant the value of K for a reaction will remain unchanged. As long as you have a balanced equation, you can write the K expression for any reaction.

  41. K Expression Examples The K expression for the reaction 2H2O (g) + 2Cl2(g)  4HCl (g) + O2(g) would be: K = [HCl]4 [O2] [H2O]2 [Cl2]2

  42. Example – try it Ex. Write the K expression for the following reaction: 4NH3 (g) + 7O2 (g)  4NO2 (g) + 6H2O (g)

  43. Example 2 - Try It Ex. Write the K expression for the following reaction: 2CO2 (g)  2CO (g) + O2(g)

  44. IMPORTANT !!!!! In the prior examples, all reactions were in gaseous form. This is called a homogenous equilibrium since all the reactants and products were in the same state of matter. With K expressions, we include concentrations that change during a reaction. The concentration of pure solids and pure liquids do not change very much in a reaction, so we ignore those terms when writing a K expression for heterogeneous equilibrium (where reactants and products are in different states of matter).

  45. Example Write the K expression for the following reaction: Fe2O3 (s) + 3H2 (g)  3H2O (g) + 2Fe (s) K = [product]coefficient from equation x [product]coefficient from equation [reactant]coefficient from equationx [reactant]coefficient from equation Since the concentrations of pure solids and liquids do not change very much, we exclude them from the K expression! K = [H2O]3 [H2]3

  46. Example Write the K expression for the following reaction: AgCl (s) Ag+(aq) + Cl- (aq) K = [Ag+] [Cl-] 1 (make sure you include the charges of the ions) (Since the reactant is a solid, you do not include it. You don’t really have to write the “1” in the denominator, but I wanted you to see where the reactant term went.)

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