) DON T FORGET WHAT THIS REPRESENTS.

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1 5.J Thermo: Endo/Exo and ΔH I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS. 1. For each of the following laws of thermodynamics, what does each actually mean in terms of the behavior of energy and particles? 1 st Law of Thermodynamics? Energy is neither created nor destroyed in a chemical reaction AKA Law of Conservation of Energy 2 nd Law of Thermodynamics? Disorder/chaos of a system always increases in terms of particles/energy distribution AKA Probability states that disorder increased as order is less probable 3 rd Law of Thermodynamics? Disorder/chaos of a system becomes constant as temperature approches absolute zero AKA 0 K = 0 disorder = perfect Instructions: Classify each statement as talking about an [EXO]thermic or [ENDO]thermic reaction: 2. [EXO] surroundings get hot 6. [EXO] reactants have more energy 3. [EXO] dynamite explodes 7. [EXO] H is negative 4. [EXO] energy is a product 8. [ENDO] absorbing sunlight to make sugar 5. [ENDO] H is positive 9. [ENDO] surroundings get cold 10. [ENDO] products have more energy 11. [ENDO] energy is a reactant 12. [ENDO] conversion of sugar to fat 13. [EXO] wood logs on a fire Instructions: ON A SEPARATE SHEET OF PAPER, Preform the following thermo calcualtions that follow using Dimensional Analysis. Remember NW = NC. Note: I have to give you a balanced chemical equation or the H would be invalide. 14. How much heat is released when 9.22 grams of glucose C6H12O6 in your body reacts with according to the following equation? C6H12O6 + 6 O2 6 CO2 + 6 H2O kj 9.22 g C6H12O6 1 mol C6H12O kj = 143 kj g C6H12O6 1 mol C6H12O6 17. Calculate the heat released when 74.6 grams of SO2 reacts according to the following equation. 2 SO2 + O2 2 SO kj 74.6 g SO2 1 mol SO kj = 57.7 kj g SO2 2 mol SO2 15. How much heat is absorbed during photosynthesis when 9.22 grams of glucose C6H12O6 is produced? 2803 kj + 6 CO2 + 6 H2O C6H12O6 + 6O g C6H12O6 1 mol C6H12O kj = kj g C6H12O6 1 mol C6H12O6 16. How much heat is released when 147 grams of NO2(g) is dissolved in excess water? 3NO2 + H2O 2HNO3 + NO kj 147 g NO2 1 mol NO2 138 kj = 147 kj g NO2 3 mol NO2 18. Calculate the heat release when 266 grams of white phosphorus P4 burns in air according to the following equation. P4 + 5 O2 P4O kj 266 g P4 1 mol P kj = 6470 kj g P4 1 mol P4 19. Calculate the amount of heat released when 1.26 x 10 4 grams of ammonia NH3 are produced according to the following equation. N2 + 3 H2 2 NH kj 1.26 x 10 4 g NH3 1 mol NH3 92.6kJ = 3.42 x 10 4 kj g NH3 2 mol NH3 5.K Thermo: Entropy and Hess s Law I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS. 1. Identify which one of the following pairs of samples has the higher entropy? a. Br2(l) or Br2(g) Br2 (g) (g) has > S than (l) b. C2H6(g) or C3H8(g) C2H6 lower MM >S than higher MM c. MgO(s) or NaCl(aq) NaCl (aq) (aq) has > S than uniform (s) d. KOH(s) or KOH(aq) KOH(aq) (aq) has >S than uniform (s) 2. Predict the entropy change (ΔS) for the following processes: a. O2(g) 2O(g) + S 1 part. to 2 b. 2O3(g) 3O2(g) + S 2 part. to 3 c. CH4(s) + 2O2(g) CO2(g) + 2H2O(g) + S (s) to (g) d. NaCl(s) Na + (aq) + Cl - (aq) + S (s) to (aq) dissocation e. C2H5OH(l) C2H5OH(g) + S (l) to (g) f. Ag + (aq) + Cl - (aq) AgCl(s) S (aq) dissociation to (s) Instructions: Solve the following Hess s law problems. Remember NW = NC 3. Calculate the H for the following reaction: Sn + 2Cl2 SnCl4 Sn + Cl2 SnCl2 H = -325 kj Leave same SnCl2 + Cl2 SnCl4 H = -186 kj Leave same Sn + Cl2 SnCl2 SnCl2 + Cl2 SnCl4 Sn + 2Cl2 SnCl4 H = 325 kj H = 186 kj H = 511 kj 4. Calculate the H for the following reaction: NO + O NO2 You are given these three equations: O2 2O H = +495 kj As O is a react = FLIP 2O3 3O2 H = -427 kj FLIP to cancel O2 & O3 NO + O3 NO2 + O2 H = -199 kj x2 To Cancel 2O O2 H = 495 kj 3O2 2O3 H = kj 2NO + 2O3 2NO2 + 2O2 H = 398 kj 2 NO + 2 O 2NO2 H = 466 kj Have to simplify H = 233 kj

2 5. Calculate the H for the following reaction: 2H2O2 2H2O + O2 2H2 + O2 2H2O H = -572 kj H2 + O2 H2O2 H = -188 kj FLIP to have H2O2 as react and x2 to cancel H2 & O2 2H2 + O2 2H2O H = 572 kj 2 H2O2 2 H2 + 2 O2 H = kj 2H2O2 2H2O + O2 H = 196 kj 6. Calculate the H for the following reaction: 2CO + 2NO 2CO2 + N2 2 CO + O2 2CO2 H = kj N2 + O2 2NO H = kj FLIP to have N2 as a prod. / NO as a react. As well as cancel O2 2 CO + O2 2CO2 H = kj 2NO N2 + O2 H = kj 2CO + 2NO 2CO2 + N2 H = kj 7. Calculate the H for the following reaction: CS2 + 2H2O CO2 + 2H2S CS2 + 3O2 CO2 + 2SO2 H = kj H2S + 1.5O2 H2O + SO2 H = -563 kj FLIP to have H2O as react / H2S as prod and cancel O2 by x2 3H2O + 3SO2 2H2S + 3O2 H = kj CS2 + 3O2 CO2 + 2SO2 H = 1075 kj CS2 + 2H2O CO2 + 2H2S H = + 51 kj 8. Calculate the H for the following reaction: 4Al + 3MnO2 2Al2O3 + 3Mn 4Al + 3O2 2Al2O3 H = kj Mn + O2 MnO2 H = -521 kj FLIP to have MnO2 as react/ Mn & O2 as prod and cancel O2 by x3 4Al + 3O2 2Al2O3 3MnO2 3Mn + 3O2 4Al + 3MnO2 2Al2O3 + 3Mn H = 3352 kj H = kj H = 1789 kj 5.L Kinetics: Collision Theory I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS. 1. What conditions must be satisfied in order for a reaction to occur? (1) Particles must collide If they do not the other two conditions to noted do not matter. No collision = no interation = no reaction possible (2) When they collide, the must possess suffiecnt KE to start the reaction Vocabulary Term = Activaiton Energy = Ea (3) Must collide w/ proper orientation Bonding sites / lone pairs required for reaction must collide 2. What makes a collision effective? (1) Particles MUST COLLIDE. When they collide it MUST be w/ If they do not no other conditions matter. (a) Sufficent Energy Scientifically this should be Activation Energy. In order to be effective ALL conditions must be met (b) High frequency of collision If a reaction happens once every years it is not practical (c) Correct orientation for reaction Otherwise e - / e - repulsion cannot be overcome 3. What makes a collision ineffective? (1) Particles MUST COLLIDE. When they collide it CAN be w/ If they do not no other conditions matter. (a) Insufficent Energy Lacking Activation Energy. ANY TRUE and a reaction is classified as ineffective (b) Low frequency of collision If a reaction happens once every years it is not practical (c) Incorrect orientation for reaction Otherwise e - / e - repulsion results 4. Define reaction rate: How fast (or slow) reactant(s) are converted to product(s) The proportion of the # of effective collisions over time. 5. What is the difference between a reaction rate and the reaction mechanism? Mechanism = pathway (steps) take to go from reactants to products Rate = speed of rxn along path 6. What factors affect the rate of reaction? (1) Nature of Reactants (2) Temperature (3) Concentration (4) Particle size / Surface Area NOT CATALYSTS OR INHBITORS they effect the mechanism

3 7. What effect, and how, do the following factors have upon reaction rate? Nature of Reactants? Refers to complexity and # of bonds that must be broken and reformed in same manner in the course of the rxn. The more bonds that must be broken or formed, the more energy requried to break said bond, the more Ea required to do so Temperature? Higher the temp, higher rate of rxn. Higher the temp, the more KE, the more movement = higher probability of collisions and collision possessing Ea Concentration? More particles in smaller area incr the likelyhood of collisions. More collisions = more changes to collide w/proper orientation = higher rate of rxn Particle Size? Smaller the particle, the more surface area exposed/available for rxn. The more particles possible for reaction, the more that can collide w/proper orientation = higher rate of rxn Cayalyst? DOES NOT DIRECTLY CHANGE THE RXN RATE, CHANGES RXN MECHANISM Lowers the Ea, making more energy available for more rxn to occur. PRECIEVED as an incr in rxn rate, rather than shortening the mechanism. Inbitor? DOES NOT DIRECTLY CHANGE THE RXN RATE, CHANGES RXN MECHANISM Rasies the Ea, making less energy available for more rxn to occur. PRECIEVED as a decr in rxn rate, rather than extending / lengthening the mechanism. 5.M Kinetics: Reaction Pathway (Potential Energy) Diagrams I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS. Instructions: Use the reaction pathway diagram below to answer questions 1 9 for the forward reaction. 1. Which of the letters in the diagram represents the PE of the prods? e 2. Which letter indicates the potential energy of the activated complex? c 3. Which letter indicates the potential energy of the reactants? b 4. Which letter indicates the activation energy of the A + B C? a 5. Which letter indicates the heat of reaction? f 6. Is the reaction exothermic or endothermic? endothermic 7. Which letter indicates the activation energy of the reverse reaction? d 8. Which letter indicates the heat of reaction of the reverse reaction? f 9. Is the reverse reaction exothermic or endothermic? exothermic Instructions: Use the reaction pathway diagram on the right to answer questions Remember N 3 I have noted a margin of error for all diagram values in blue 10. The heat content of the reactants of the forward reaction is about 80 kj 11. The heat content of the products of the forward reaction is about 160 kj 12. The heat content of the activated complex of the forward reaction about 250 kj ±10 kj 13. The activation energy of the forward reaction is about 170 kj ±10 kj 14. The heat of reaction (ΔH) of the forward reaction is about +250 kj = need sign 15. The forward reaction is (endothermic/exothermic). 16. The heat content of the reactants of the reverse reaction is about 160 kj 17. The heat content of the products of the reverse reaction is about 80 kj 18. The heat content of the activated complex of the reverse reaction is about 250 kj ±10 kj 19. The activation energy of the reverse reaction is about 170 kj ±10 kj 20. The heat of reaction (ΔH) of the reverse reaction is about 250 kj = need sign 21. The reverse reaction is (endothermic/exothermic).

4 22. Chemical reactions occur when reactants collide. For what reasons may a collision fail to produce a chemical reaction? (1) Incorrect orientation Otherwise e - / e - repulsion cannot be overcome (2) Insufficent energy Lacking Activation Energy. 23. If every collision between reactants lead to a reaction, what determines the rate at which the reaction occurs? (1) Nature of Reactants The nature of bonds (strength and #) determines the amount of Ea required to react (2) and their possession of Ea Indicated by temperature. But temperature is an average, it not distributed equally. 24. What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction? Ea = Energy requeared to breakt eh bonds of reactant to enable new bonds to be formed in products. Activated Complex = Represents the total PE that must be present for the rxn to occur. Difference = The PE of the reactants and the activated complex is the Ea 25. What happens when a catalyst is used in a reaction? Alters the mechanism of the rxn, lowering the Ea as the mechanism is shorter More rxns w/same amount of energy 26. Name 4 things that will speed up or slow down a chemical reaction. (1) Nature of Reactants (2) Temperature (3) Concentration (4) Particle size / Surface Area NOT CATALYSTS OR INHBITORS they effect the mechanism 27. Sketch an energy diagram for a reaction. (Label the axes.) Potential energy of reactants = 350 kj Activation energy = 100 kj So PE of Activated Complex is 350 kj kj = 450 kj Potential energy of products = 250 kj 28. Is the reaction in #27 exothermic or endothermic? Explain. Exothermic, the PE of the products is less than the PE of the reactants as the PE is being REALEASED = exothermic 29. How could you lower the activation energy for the reaction in #27? Add a catalyst Will alter the mechanism and create a path that requires less Ea 5.N Equilibrium: Le Chatelier s Principle I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS. 1. What is the difference between a physical and chemical equilibrium? Physical = a closed constant temp/presurre in which rate of forward and reverse physical changes are equal NOT creating new substances Chem = a closed system w/constant temp/pressure/concentration in which rate of forward and reverse chemical change are equal Instructions: Complete the following chart by writing left, right, or none for equilibrium shift given the. Write increases, decreases, or remains the same for the concentration of reactants and products, and for the value of K (will need to come back and do later). N2 (g) + 3 H2 (g) 2 NH3 (g) kcal NaOH (s) Na + (aq) + OH - (aq) kcal Stress Equalibirum Shift [ N2 ] [ H2 ] [ NH3 ] K Stress Equalibirum Shift Amount of NaOH(s) [ Na + ] [OH - ] K add N2 right Decr. Decr. Incr. same Add NaOH right decr incr incr same add H2 right Decr Decr Incr same Add NaCl left incr decr decr same (adds Na + ) add NH3 left Incr Incr Decr same Add KOH left incr decr decr same (adds OH - ) remove N2 left Incr Incr Decr same Add H + right decr incr incr same (removes OH - ) remove H2 left Incr Incr Decr same incr. temp. left incr decr decr decr remove NH3 right Decr Decr Incr same decr. temp. right decr incr incr incr incr. temp. left Incr Incr Decr Decr incr. Press. none same same same same decr. temp. right Decr Decr Incr Incr decr. Press. none same same same same incr. Press. right Decr Decr Incr Incr Add Ca(OH)2 left incr decr decr decr decr. Press. left Incr Incr Decr Decr Add Na3PO4 left incr decr decr decr

5 Instructions: Consider the following equilibrium and answer the questions when these changes are made chart by writing left, right, or none for equilibrium shift given the. Write increases, decreases, or remains the same for the concentration of reactants and products. 5. Predict which way these equilibrium systems will shift when the total pressure is increased. Pressure only affects gassess and pressure will cause a shift to side with less moles in gas due to # of collisions. (A) N2 (g) + O2 (g) 2 NO (g) no shift (B) 2 SO2 (g) + O2 (g) 2 SO3 (g) right / product C) 4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g) left / react 6. Heat + CH4 (g) + 2 H2S (g) CS2 (g) + 4 H2 (g) and CH4 gas is added: (A) direction of equilibrium shift? right / product (B) [H2S]? decr (C) [CS2]? incr (D) [H2]? incr 7. Heat + CH4 (g) + 2 H2S (g) CS2 (g) + 4 H2 (g) and CS2 gas is removed: (A) direction of equilibrium shift? right / product (B) [CH4]? decr (C) [H2S]? incr (D) [H2]? incr 8. Heat + CH4 (g) + 2 H2S (g) CS2 (g) + 4 H2 (g) and H2 gas is added: (A) direction of equilibrium shift? left / reactant (B) [CH4]? decr (C) [CS2]? decr (D) [H2S]? incr 9. Heat + CH4 (g) + 2 H2S (g) CS2 (g) + 4 H2 (g) and temperature is increased: (A) direction of equilibrium shift? right / product (B) [H2S]? decr (C) [CS2]? incr (D) [H2]? incr (E) [CH4]? decr 5.O Equilibrium: Equilibrium Constant (K)) I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS. Instructions: ON A SEPARATE SHEET OF PAPER, write the Keq expressions for the reactions below. 1. N2 (g) + 3 H2 (g) 2 NH3 (g) 4. 2 CO(g) + O2 (g) 2 CO2 (g) Keq = [NH3] 2 Keq = [CO2] 2 [N2] [NH3] [CO] 2 [O2] 2. 2 KClO3 (s) 2 KCl (s) + 3 O2 (g) 5. Li2CO3 (s) 2 Li + (aq) + CO3 2- (aq) Keq = [O2] 3 Keq = [Li + ] 2 [CO3 2- ] 3. H2O(l) H + (aq) + OH - (aq) 6. CH4 (g) + 2 H2S (g) 4 CS2 (g) + H2 (g) Keq = [ H + ] [ OH - ] Keq = [H2] 4 [CS2] [H2S] 2 [CH4] 7. In considering the following equilibriums, which equilibrium favors products to the greatest extent? Reactants to the greatest extent? Notice that only 2 responses are req a) 2NO2 (g) N2O4 (g) Keq = 2.2 N/A Is not the largest or the smallest b) Cu 2+ (aq) + 2Ag(s) Cu(s) + 2Ag + (aq) Keq = 1 x React to greatest extent Denominator greater c) Pb 2+ (aq) + 2 Cl - (aq) PbCl2(s) Keq = 6.3 x 10 4 Prod to greatest extent Numerator greater d) SO2(g) + O2 (g) SO3 (g) Keq = 110 N/A Is not the largest or the smallest 8. What is the only way to change the value of the Keq? Changing the heat content of the rxn (temperature) or pressure of the system (for gasses only) 9. In the reaction: A + B C + D + 100kJ, what happens to the value of Keq if we increase the temperature? Will incr Denominator incr / numerator decr. 10. If the value of Keq decreases when we decrease the temperature, is the reaction exothermic or endothermic? Endo Denominator decr / numerator incr. 11. In the reaction; W + X + 100kJ Y + Z, what happens to the value of Keq if we increase the [X]? Explain your answer. Keq constant The shift caused by [X] is recipricated by both products and reactants. Keq is only affected by temperature (pressure too for gasses) 12. If the value of Keq increases when we decrease the temperature, is the reaction exothermic or endothermic? Exo Denominator incr / numerator decr. 13. Predict whether reactants of products are favored in the following equilibrium systems (a) CH3COOH(aq) H + (aq) + CH3COO - (aq) Keq = 1.8 x 10-5 Reacts K < 1, numerator greater = reacts (b) H2O2(aq) H + (aq) + HO2(aq) Keq = 2.6 x Reacts K < 1, numerator greater = reacts (c) CuSO4(aq) + Zn(s) Cu(s) + ZnSO4(aq) Keq = Prods K > 1, denominator greater = prods

6 Instructions: ON A SEPARATE SHEET OF PAPER, use the information below to calculate the equilibrium constant (Keq) for the following reactions and tell whether equilibrium lies favors the reactants or the products.. NW = NC. Hint: Common mistakes will be made here H2 (g) + Cl2 (g) 2 HCl (g) At equilibrium [H2] = 0.42 M, [Cl2] = M, and [HCl] = 0.95 M Keq = [0.95 M] 2 = 29, prod favored [0.42 M] [0.75 M] 15. N2 (g) + 3 H2 (g) 2 NH3 (g) At equilibrium [N2] = 0.34 M, [H2] = 0.13 M, and [NH3] = 0.19 M. Keq = [0.19 M] 2 = 48, prod favored [0.13 M] 3 [0.34 M] NO (g) + O2 (g) 2 NO2 (g) At equilibrium [NO] = M, [O2] = M, and [NO2] = 0.95 M. Keq = [0.95 M] 2 = 1.1x10 9, prod favored [2.4x10-3 M] 2 [1.4x10-4 M] 17. C (s) + CO2 (g) 2 CO (g) At equilibrium [CO2] = M, [CO] = M Keq = [5.4x10-5 M ] 2 = 3.5x10-4, react favored [8.3x10-6 M] 18. 2NO (g) + Br2 (g) 2 NOBr (g) At equilibrium [NO] = 0.5 M, [Br2] = 0.25 M, and [NOBr] = 3.5 M. Keq = [3.5 M] 2 = 200, prod favored [0.5 M] 3 [0.25 M] Fe (s) + 3H2O (g) Fe2O3 (s) + 3 H2 (g) At equilibrium [H2O] = 1.0 M, and [H2] = 4.5 M. Keq = [4.5 M] 3 = 91, prod favored [1.0 M] CaCO3 (s) CaO (s) + CO2 (g) At equilibrium [CO2] = M. Keq = [4.0x10-3 M] = 4.0x10-3, react favored 21. PCl5 (g) PCl3 (g) + Cl2 (g) Equilibrium concentrations are: [PCl5] = 0.25 M, [PCl3] = M, and [Cl2] = M. Keq = [3.2x10-3 M] [9.7x10-4 M] = 1.2x10-5, react favored [0.25 M]