Kinetics Worksheet 1. Distinguish between average rate and instantaneous rate. Which of the two rates gives us an unambiguous measurement of reaction rate? Why? Average rate is exactly what it sounds like. Instantaneous rate is the speed at which something is consumed or produced at any point in the reaction. 2. Suggest experimental means by which the rates of the following reactions could be followed: a) CaCO3(s) → CaO(s) + CO2(g) Check the pressure created by CO2 over a period of time. b) Cl2(g) + 2 Br –(aq) → Br2(aq) + 2 Cl –(aq) Chlorine gas is yellowish-green, bromine is a brownish gas. Use a colorimeter to check for variations in absorbance. 3. On which of the following properties does the rate constant of a reaction depend? a) b) c) 4. reactant concentration nature of the reactants temperature. Based on the equation shown below, 2 ClO(g) → Cl2(g) + O2(g) and the following data, use graphical analysis to determine the order of the reactant: Time (s) 1.2 9.6 22.4 32.0 40.0 [ClO] 0.849 0.710 0.579 0.520 0.477 5. For a hypothetical chemical reaction that has the stoichiometry 2 X + Y → Z, the following initial rate data were obtained. All measurements were made at the same temperature. Initial Rate of Formation of Z, (mol/L-sec) 7.0 x10-4 1.4 x10-3 2.8 x10-3 4.2 x10-3 a) Initial [X]o, (mol/L) 0.20 0.40 0.40 0.60 Initial [Y]o, (mol/L) 0.10 0.20 0.40 0.60 Identify the rate law for this reaction from the data above. r = k [X]0 [Y]1 b) Calculate the specific rate constant for this reaction and the units. k = 7.0 x10-4 mol/L-s/[0.10 mol/L] = 7.0 x10-3/s c) How long must the reaction proceed to produce a concentration of Z equal to 0.20 M, if the initial reaction concentrations are [X]o = 0.80 M, [Y]o = 0.60 M and [Z]o = 0 M? 2X+Y→Z Z : Y [Y] = 0.60 - 0.20 = 0.40 ln [A]0/[A]t = kt ln [0.60]o/[0.40]t = 7.0 x10-3 t t = 0.41/7.0 x10-3 = 58 s 6. The rate constant for the second order reaction 2 NOBr(g) → 2 NO(g) + Br(g) is 0.80 L/mol-s at 10° C. a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. 1/[B]t - 1/[B]0 = kt 1/[B]t - 1/[0.086] = 0.80 M-s (22 s) [B]t = 0.034 b) Calculate the half-lives [NOBr]0 = 0.054 M when [NOBr]0 = 0.072 M and t1/2 = 1/k [B] t1/2 = 1/0.80 M-s [0.072] = 17. s t1/2 = 1/0.80 M-s [0.054] = 23. s 7. The rate constant for the second order reaction 2 NO2(g) → 2 NO(g) + O2 (g) is 0.54 L/mol-s at 300° C. How many seconds would it take for the concentration of NO2 to decrease from 0.62 M to 0.28 M? 1/[0.28] = 1/[0.62]0 + 0.54 t t = 3.6 s 8. The rate law for the reaction 2 NO(g) + Cl2(g) → 2 NOCl(g) is given by rate = k [NO][Cl2]. a) What is the order of the reaction? 2. The order of the reaction is the sum of the exponents in the rate law. b) A mechanism involving the following steps has been proposed for the reaction → NOCl2(g) NO(g) → 2 NOCl(g) NO(g) + Cl2(g) NOCl2(g) + If this mechanism is correct, what does it imply about the relative rates of these two steps? The rate law identifies those substances participating in the slow step of a reaction mechanism. 9. The potential energy profile for H2 + 2 ICl below a) How many elementary steps are there? 3 A B C b) → → → B C D How many intermediates are formed? 2 B c) → → C Which step is the rate determining step? C I2 + 2 HCl is shown d) Is the overall reaction exothermic or endothermic? Net loss of heat from A to D 10. For the reaction X2 + Y + Z → XY + XZ it is found that doubling the concentration of X2 doubles the reaction rate, tripling the concentration of Y triples the rate, and doubling the concentration of Z has no effect. a) What is the rate law for this reaction? r = k [X2] [Y] b) Why is it that the change in the concentration of Z has no effect on the rate? Occurs after the slow step. c) Suggest a mechanism for the reaction that is consistent with the rate law. X2 + Y → XY + X slow → fast X + Z XZ 11. The decomposition of N2O to N2 and O2 is a first order reaction. At 730° C the half-life of the reaction is 3.58 x103 min. If the initial pressure of N2O is 2.10 atm at 730° C, calculate the total gas pressure after one half-life. Assume that the volume remains constant. 2 N2O → 2 N2 + O2 moltotal = 1 + 1 + 0.5 = 2.5 mol 2.10 atm/2 mol gas (2.5 mol gas) = 2.63 atm after one half-life