Kinetics & Equilibrium. Le Châtelier's Principle. reaction rates. + Packet 9: Daily Assignment Sheet '19 Name: Per

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1 Daily Assignment Sheet '19 Name: Per (check them off as you complete them) Due Date Assignment Thur 2/21 Do clock reaction lab Fri 2/22 Do Not Write In That Box Below Mon 2/25 Do WS 9.1 Tue 2/26 Do WS (side 1) Wed 2/27 Do WS (side 2) Thur 2/28 Do WS 9.3 Fri 3/1 Do WS 9.4 -mustard day - play video games today Mon 3/4 Do WS 9.5 QUIZ TODAY PACKETS DUE TODAY Come to class with packets ready to be turned in, with the above underlined assignments in proper order, in your pocket folder (1/2 pt), with this page as the cover page & grade report stapled inside (1/2 pt) For 1/2 point, do not turn in old packets. -1/2 for no name on top. + Packet 9: Kinetics & Equilibrium Le Châtelier's Principle FOR OFFICE USE ONLY do not write in this space all claims filed must be submitted by the end of the fiscal year on form #G-55. reaction rates

2 + WS 9.1 Reaction Kinetics & Rate Laws 1. According to collision theory, in order for a chemical reaction to proceed, the reactants must collide with proper and with sufficient. 2. Draw a graph of energy vs reaction progress as was shown in class. 3. Factors which can change the speed of a reaction are: 1) 2) 3) 4) Rate Laws After timing the clock reaction at various concentrations, you have probably come to appreciate that the rate of a reaction depends (in part) on the of the. A rate law is an equation that relates the rate of a reaction to the concentration of reactants. 2 NO2(g) + F2(g) ---> 2 NO2F(g) rate = k[no2] a [F2] b k is called the rate constant. It is a proportionality constant in the relationship between rate and concentrations. In general, when k is small, the reaction is. When k is large, the reaction is. The exponents "a" & "b" must be determined by analyzing the experimental data, like below: [NO2] (mol/l) [F2] (mol/l) initial rate (mol/l s) experiment x 10-5 experiment x 10-5 experiment x 10-5 Compare experiment 1 & 2. Doubling the [NO2] causes the rate to double. Solve for the exponent like this: 2 a = 2 a = Compare experiment 1 & 3. Doubling the [F2] causes the rate to remain the same. Solve for the exponent like this: 2 b = 1 b = The rate law for this reaction is: These exponents refer to the reaction order. It is the power to which the concentrations of the reactant must be raised to give the observed relationship between concentration & rate. The above reaction is order with respect to [NO2] and order with respect to [F2]. The sum of these orders is, which is the overall order of the reaction. Note: It is possible for reaction orders to be fractional or negative. For more info, go ask someone else!

3 WS 9.1 (side 2) Determine the reaction order: - In zeroth order reactions, doubling the concentration of a reactant will have no effect on the rate. 2 0 = 1 - In first order reactions, doubling the concentration of a reactant will double the rate. 2 1 = 2 - In second order reactions, doubling the concentration of a reactant will quadruple the rate. 2 2 = 4, etc... 2 N2O5(g) ---> 4 NO2(g) + O2(g) example 1: initial [N2O5] (mol/l) initial rate (mol/l s) experiment x 10-6 experiment x ) Write the rate law: 2) Solve for the rate law constant: example 2: aa ---> B initial [A] (mol/l) initial rate (mol/l s) experiment x 10-4 experiment x 10-4 experiment x ) Write the rate law: 2) Solve for the rate law constant: Problem 1: Br2 + C3H6O ---> C3H5OBr + HBr [Br2] (mol/l) [C3H6O] (mol/l) initial rate (mol/l s) experiment x 10-5 experiment x 10-5 experiment x 10-5 Find the reaction order with respect to Br2 & C3H6O, the overall order, the rate law, and k. Problem 2: NO2 - + NH > N2 + 2 H2O [NO2 - ] (mol/l) [NH4 + ] (mol/l) initial rate (mol/l s) experiment x 10-7 experiment x 10-7 experiment x 10-7 experiment x 10-7 experiment x 10-7 experiment x 10-7 Find the reaction order with respect to NO2 - & NH4 +, the overall order, the rate law, and k.

4 + WS Equilibrium (side 1) 1. When a reaction has reached equilibrium, the and rates of reaction are. During dynamic equilibrium, molecules of and are still being formed, however their overall rates of are the same. At, it is not necessary to have concentrations of and. The equilibrium constant, known as, is specific for a given. It is calculated as the ratio of the concentrations of to reactants, with each concentration raised to a equal to the of that substance. If Keq is greater than 1, then are favored. If Keq is less than one, then reactants are. When calculating Keq, only substances which are or are considered. Solids and liquids are since their concentrations generally won't. Ans (IAO+1): aqueous change equal equal equilibrium favored formation forward gases ignored Keq marshmallows moles power products products products products reactants reactants reverse temperature 2. In a 4.0 L vessel, mol N2, mol O2, and mol NO have reached equilibrium at 890 K. Calculate keq for this reaction: N2(g) + O2(g) 2 NO(g) 3. For the following reaction at equilibrium, [HCl] = M, [O2] = M, [Cl2] = M, and Keq = 214 at 560 K. Calculate the concentration of the water at equilibrium. 4 HCl(g) + O2(g) 2 Cl2(g) + 2 H2O(g) 4. In a 1.0 L tank, 0.95 mol SO3, 8.95 mol H2O, and 1.35 mol H2SO4 are at equilibrium at 25 C. Calculate keq. SO3(g) + H2O(l) H2SO4(aq) 5. A 15.0 L tank at equilibrium is found to contain 4.50 mol of N2 and 2.11 mol ammonia (NH3). Given that Keq = 3.2 x 10-4, calculate how many moles of hydrogen are present. N2(g) + 3 H2(g) 2 NH3(g)

5 WS Equilibrium (side 2) 6. When 0.10 mol H2S was placed into a 10.0 L vessel at 1132 C, it resulted in an equilibrium mixture containing mol H2. What is keq? 2 H2S(g) 2 H2(g) + S2(g) 7. Keq for the following 150 C is 1.2 x What are the concentrations of each component at equilibrium if 5.00 mol of I2 and 8.00 mol of Br2 are reacted in a 10.0 L vessel? I2(g) + Br2(g) 2 IBr(g) Ans (IRO+1): 2.3 x Units: M M M M mol

6 + WS Le Châtelier's Principle Use arrows (up / down) to indicate the affect of each of these disturbances (stresses) on the concentration of the reactants and products in this equilibrium: 9 KJ + 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) disturbance affect on affect on affect on [SO 2 ] [O 2 ] [SO 3 ] [SO2] [O2] [SO3] pressure temp N 2 (g) + 3 H 2 (g) 2 NH 3 (g) KJ disturbance affect on affect on affect on [N 2 ] [H 2 ] [NH 3 ] [N 2 ] [H2] [NH3] pressure temp

7 + WS C 6 H 6 (g) + 3 H 2 (g) C 6 H 12 (g) + 80 KJ disturbance affect on affect on affect on [C 6 H 6 ] [H 2 ] [C 6 H 12 ] [C 6 H 6 ] [H2] [C 6 H 12 ] pressure temp 95 KJ + 2 KClO 3 (s) 2 KCl(s) + 3 O 2 (g) disturbance affect on affect on affect on [KClO 3 ] [KCl] [O 2 ] [KClO 3 ] [KCl] [O2] pressure temp In general, Le Châtelier's principle states that if a system at is subjected to a, the equilibrium is in the direction that tends to the stress. If there is a change in pressure, this only affects equilibriums in which are involved. If the total # of of gases differs from reactants to, then under pressure, the equilibrium will in the direction to produce gases. The effect of changing the temperature of a reaction depends on which of the opposing reactions is and which is. According to the principle, the addition of heat the equilibrium so that the heat is absorbed (endothermic reaction ). The removal of heat would favor reactions. ANS (IAO+2): endothermic, equilibrium, exothermic, exothermic, favored, gases, heavier, Le Châtelier, less, moles, products, relieve, reversible, shift, shifted, shifts, solubility, stress

8 + WS Solubility Product K sp is called the solubility product constant. It is the product of the concentrations of a substances' ions in a saturated solution, each raised to the power of the coefficient. CaF 2(s) Ca +2 (aq) + 2 F - (aq) K sp = [Ca +2 ] [F - ] 2 Remember, pure solids and liquids are not included in equilibrum expressions. The numerical value for K sp can be determined from solubility data (see reference chart). Calculate the value for Ksp for the following: 1. Mg(OH)2 at 100 C 2. AgC2H3O2 at 0 C 3. PbCl 2 at 0 C 4. PbCl 2 at 20 C (answers on side 2...)

9 + WS (side 2) K sp values can be looked up in charts (see reference chart) and be used ot determine solubility data. Calculate the solubility of each 25 C, in g/l: 5. BaCO3 6. ZnS 7. Ag2S 8. Fe(OH)2 Ans (IRO+2): 0.014, 1.9E-3, 1.9E-4, 5.2E-4, 1.4E-5, 4.6E-5, 5.6E-5, 2.5E-8, 1.3E-9, 8.4E-9, 1.3E-10, 3.2E-10, 6.2E-15 Units: g/l g/l g/l g/l

10 + WS 9.5 Review Sheet - Kinetics and Equilibrium 1. Use the diagram at right to answer these questions: X a. Is this reaction endothermic or exothermic? Z b. Which is higher in energy; the reactants or products? c. What is point "X" called? d. What does line "Z" represent? e. What does line "D" represent? energy D reaction progress 2. For the following, indicate the most likely result in reaction rate: speed up or slow down increasing temperature: adding water to reactants: using higher concentration of reactants: crushing the reactants into a powder: 3. A + B ---> C Rate = k[a] 2 [B] 1 a. Which will affect the reaction rate more; doubling [A] or doubling [B]? b. What is the overall order of this reaction? 4. C + D ---> E initial [C] (mol/l) initial [D] (mol/l) initial rate (mol/l s) experiment x 10-6 experiment x 10-6 experiment x 10-6 a. the reaction order with respect to [C]? with respect to [D]? overall? b. What is the value of the rate law constant? 5. F (s) + G (aq) 3 H (g) + J (g) KJ a. Write the expression for K eq for the above : b. Name 5 ways to [J]:

11 6. NH4 + (aq) + NO2 - (aq) N2(g) + 2 H2O (l) At 400 K, the 1.0 L reaction vessel is found to contain 1.55 mol NH4 +, mol NO2 -, and 3.20 mol H2O. Given the equilibrium constant = 39.5, calculate the concentration of the N2. 7. Calculate the Ksp for barium hydroxide at O C. See reference sheet for soubility data. 8. Calculate the solubility of cobalt sulfide in 25 C, in units of g/l. See reference sheet for Ksp value. 9. CO2(g) + H2(g) CO (g) + H2O (g) At 900 K, Keq for this reaction is You start with 0.10 M CO2 & 0.10 M H2. What are the concentrations of all species at equilibrium?

12 the clock reaction kinetics lab Name: In this lab you will investigate ways to change how fast a reaction occurs. After establishing the time of the control (full concentration & room temperature), you will change the concentrations of the reactants. Then, you will change the temperature of the reactants. Solution A : is an acidified M NaHSO3 solution with starch indicator. Solution B is M KIO3 solution. Part 1: Establish the time on control (undiluted, room temperature) Use of A. Use of B. V A [A] V B [B] time (s) M M Part 2: Effect of concentration of solution B Use of A. Use indicated ml of B diluted to with water. M f = M V V t V A [A] V B [B] time (s) M M M M 20 ml 15 ml 10 ml 5 ml Part 3: Effect of concentration of solution A Use of B. Use indicated ml of A diluted to with water. M f = M V V t V A [A] V B [B] time (s) 20 ml 15 ml 10 ml 5 ml M M M M Part 4: Effect of temperature Use of cold A and of cold B. Next, use of hot A and of hot B. Bonus! Got the time? temp cold time (s) Show Mr. A that you can get the reaction to occur in seconds hot

13 Graph: Plot your data for part 2 & part 3, and create a molarity vs. time graph. Be sure to include the data point from the control (part 1). Scale your axises to best fit your data. Once your data is plotted, try to connect your points with a smooth curve or a straight line (whichever seems more appropriate). Do NOT do a dot-to-dot graph. time (sec) [A] (mol/l) [B] (mol/l) time (sec) Questions: 1. Which has a greater effect on reaction time: the [NaHSO3] or the [KIO3]? 2. Explain why changing the molarity of the reactants effected the reaction rate, in terms of collision theory. 3. Explain why changing the temperature of the reactants had an effect on the reaction rate, in terms of collision theory. 4. Based on your graph (and on common sense), what would happen to time (sec) if the molarity was zero?