Chemistry 1AA3 2000/01

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1 Chemistry 1AA3 2000/01 Tutorial #5 Answers Week of February 12-16, 2001 Dr. M.A. Brook Dr. B. E. McCarry Dr. A. Perrott 1. The equation for the reaction of NO(g) with O2(g) at 660K is: 2 NO(g) + O2(g) 2 NO2(g) The reaction rate data below was measured by observing the disappearance of NO: Concentration (mol.l -1 ) Initial Rxn Rate Expt. No. [NO] [O2] (mol L -1 s -1 ) x x x 10-3 (a) Derive the rate law for the reaction in terms of the disappearance of NO(g). (b) What is the overall order of the reaction? (c) What is the numerical value of the specific rate constant k at 660K? Give the units of k. (d) If the concentrations of NO and O2 had been expressed in terms of partial pressure p(n0) and p(o2), both in units of atmospheres, what would the units of k be now? (a) From experiments 1 & 2, [O2] changes by 2x and rate of rxn changes by 2x: thus, rxn. rate is proportional to [O2] 1. From experiments 2 and 3, [NO] changes by 3x, while rate changes by (4.5 x 10-3 /5.0 x 10-4 =) 9x; i.e., rxn rate is proportional to [NO] 2. Rate Law: -1/2 d[no] = k [O2][NO] 2 (b) Overall order of reaction = = 3. Reaction is third order. (c) From experiment 1 and substituting into rate law: 2.5 x 10-4 mol.l -1.s -1 = k (0.020 mol.l -1 )(0.020 mol.l -1 ) x 10 mol.l.s k = (0.020 mol.l )(0.020mol. L ) k = 31.3 mol -2 L -1 s -2 = x 10 6 mol -2 L -1 s -2 (8.0x10 ) (d) If p(no) and p(o2) were used, initial rate would have units of atm.s -1, then k = rate = atm.s -1 = atm -2 s -1

2 (conc.)(conc.) 2 (atm)(atm) Sulfuryl chloride (SO2Cl2) decomposes to give SO2 and Cl2 at 320 C in a first order reaction with k = 2.20 x 10-5 s -1. SO2Cl2 SO2 + Cl2 (a) Calculate the half-life of this reaction. (b) How long will it take (in hours) for a 2.00 mole sample of SO2Cl2 in a 1.00 litre volume to decrease to mole at 320 C? (a) The half-life of a first order reaction is related to the rate constant by the equation: t1/2 = ln2 = = 3.15 x 10 4 s = 8.75 h rate 2.20 x 10-5 s -1 (b) ln (A0/At) = kt A0 = 2.00 moles At = mole ln(2.00 moles/0.125 mole) = (2.20 x 10-5 s -1 ) x t t = ln(2.00 moles/0.125 mole) = ln(16) = 1.26 x 10 5 s = 35.0 h 2.20 x 10-5 s x 10-5 s Cyclopropane undergoes a first order isomerization to propene at 300 C with k = 0.54 h -1. Cyclopropane propene (a) How many hours will have elapsed when half of the cyclopropane has reacted if the initial concentration of cyclopropane is 0.05M? (b) If the initial concentration of cyclopropane were quadrupled to 0.20M, how long would it take for half the cyclopropane to react? (c) How long would it take for the [cyclopropane] to change from an initial concentration of 0.20M to 0.020M? What is the [propene] at this time? (d) Draw a reaction progress diagram for this reaction, assuming the reaction proceeds in a single step with Hrxn = -45kJ/mol and Ea = 124 kj/mol. (a) Using the relationship: t1/2 = ln2 and k = 0.54 h -1 k t1/2 = (0.693)/0.54 h -1 = 1.28 h (b) For any first order reaction it does NOT matter how much material is present; the half-life (t1/2) is always the same (at the same temperature). That is, t1/2 = 1.28 h. (c) We need to use the formula which relates A0, At and time: ln (A0/At) = kt

3 -3- Rearranging this equation and substituting we get: k = ln(0.20/0.020) = 4.26 h 0.54 h Plutonium forms extremely toxic compounds. One of the isotopes of plutonium produced in "breeder" reactors decays to less toxic, although still harmful, uranium: 94Pu U á 90Th á further decay where the á particle is He 2+. If the "safe" level of a sample of buried 94Pu 239 is 0.01% of the original amount and the specific rate constant for decay of 94Pu 239 to 92U 235 is k1 = 2.84x10-5 yr -1, how long will it be before the plutonium-containing waste material can be safely unearthed? We use here the integrated form of the first order rate law: i.e. ln Ao / At = kt where Ao = initial concentration at t = 0 At = concentration at time, t We are given k = 2.84 x 10-5 yr -1 and At should be 0.01% of Ao That is, the ratio of Ao to At = 100:0.01 Thus, the time required for this to occur is: t = ln Ao / At. 1 k ln x10 = 5 1 yr 4 1 whence t = ln x10 yr t = 3.24 x 10 5 years 5. The combination of CF3 "free radicals", 2CF3 C2F6, occurs with rate constants of 5.9 x 10 9 M - 1.s -1 at 25 C and 7.1 x 10 9 M -1.s -1 at 60 C. (a) Use the Arrhenius form for the rate constant as a function of temperature to obtain the activation energy. (b) Predict the rate for -79 C where the product can be condensed to a liquid. (c) Would you describe CF3 as a reactive species? (d) CF3 radicals must be described as a very reactive species. An Ea of less than 5 kj.mol -1 means that it is reacting at about a diffusion-controlled rate. This is illustrated by the very small change that occurs in the rate constant over a very large change in temperature. From 194 K to 333 K and the rate constant changes only by 7.1/2.3 or approximately 3x. Slower reactions with larger Ea values change much more rapidly than this.

4 -4- (a) Using k = A exp(-ea) in ln form: ln k = -Ea + lna and substituting RT RT for the two temperatures, 298 K and 333 K (recall the Arrhenius relationship uses Kelvin NOT Celsius temperatures). ln (5.9 x 10 9 ) = -Ea. 1 + lna (1) R 298 ln (7.1 x 10 9 ) = -Ea. 1 + lna (2) R 333 Then (1) - (2) ln 5.9 x = -Ea. ( 1-1 ) 7.1 x R or ln (0.8309) = -Ea. ( ) x 333 or = -Ea x 333 whence Ea = x x 298 x = 4.36(4) kj.mol -1 (b) If the rate constant for -79 C is k-79, then we may use the data from the 25 C experiment together with the calculated Ea to obtain the value of k-79. Thus ln k-79 = kj. 1 + lna Eq n (1) R 194 and ln(5.9 x 10 9 ) = kj. 1 + lna Eq n (2) R 298 Therefore, subtracting Eq n (2) from Eq n (1) affords: ln k-79 - ln(5.9 x 10 9 ) = x 10 3 J.mol -1. ( 1-1 ) J mol -1 K K or ln k-79 = x x 298 = = whence k-79 = 2.30 x 10 9 M -1 s -1 at -79 C. (c) CF3 radicals must be described as a very reactive species. An Ea of less than 5 kj.mol -1 means that it is reacting at about a diffusion-controlled rate. This is illustrated by the very

5 -5- small change that occurs in the rate constant over a very large change in temperature. From 194 K to 333 K and the rate constant changes only by 7.1/2.3 or approximately 3x. Slower reactions with larger Ea values change much more rapidly than this. 6. For the reduction of ICl by H2 at 450K according to the following equation H2(g) + 2ICl(g) I2(g) + 2HCl(g) a proposed reaction mechanism is: (i) ICl + H2 HI + HCl (slow) (ii) ICl + HI I2 + HCl (fast) (a) What is the molecularity of each of the above elementary steps? (b) Derive the rate law in terms of the disappearance of H2 for the reaction. (a) Step (i) is BIMOLECULAR; step (ii) is also BIMOLECULAR. (b) Step (i) is the slow step and thus the Rate Determining Step (RDS) and therefore the observed reaction rate will be given by the rate law for this step: i.e., Reaction rate = - d[h2] = k [ICl] [H2] 7. For the reaction between H2(g) + NO(g) shown below, the rate of production of N2 was found to double when the concentration of H2 doubled but quadrupled when the concentration of NO doubled. What is the rate law for the reaction in terms of the formation of N2? 2H2(g) + 2NO(g) N2(g) + 2H2O(g) A suggested reaction mechanism is: (i) NO + NO N2O2 (fast) (ii) N2O2 + H2 N2O + H2O (slow) (iii) N2O + H2 N2 + H2O (fast) (a) Show that these elementary steps are consistent with the experimentally observed rate law. Rate Law: [H2] increases 2x, reaction rate increases by 2x Thus, reaction rate [H2] When [NO] increases 2x, reaction rate increases by 4x Thus, reaction rate [NO] 2

6 -6- i.e., Rate law is d[n2] / = k [H2] [NO] 2 (a) From the fast equilibrium step (i) we may write: Keq = [N2O2 ] [NO] 2 or [N2O2] = Keq [NO] (1) where [N2O2] is the steady state concentration of this intermediate. The slow, rate determining step is step (ii); thus, the observed reaction rate will be given by the Rate Law for step (ii), where k2 is the specific rate constant for this step; i.e., Reaction rate = d[n2] = k2 [H2][N2O2] and substituting the concentration of the intermediate N2O2 from equation (1) d[n2] = k2 [H2] Keq [NO] 2 = k2 Keq [H2][NO] 2 = k [H2][NO] 2 since Keq is a constant. Therefore, the rate law derived above is the same as the experimentally observed rate law, and therefore the proposed reaction mechanism is CONSISTENT with the observed law.

Unit - 4 CHEMICAL KINETICS VSA QUESTIONS (1 - MARK QUESTIONS) (aq) as product for the reaction : 5 Br (aq) + Br(aq) + 6H + (aq) 3 Br 2

Unit - 4 CHEMICAL KINETICS VSA QUESTIONS (1 - MARK QUESTIONS) 1. Define the term rate of reaction. 2. Mention the units of rate of reaction. 3. Express the rate of reaction in terms of Br (aq) as reactant