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Chemical Kinetics Chemical Kinetics P. J. Grandinetti Chem. 4300 P. J. Grandinetti (Chem. 4300) Chemical Kinetics 1 / 60 Chemical Kinetics How do reactants become products? Í What is the mechanism? What is the time scale of the reaction? Í How fast are reactants consumed and products formed? P. J. Grandinetti (Chem. 4300) Chemical Kinetics 2 / 60 Chemical Kinetics Consider generic reaction: ◦ 휉 nA = nA * a n = n◦ * b휉 → B B aA + bB cC + dD where ◦ 휉 nC = nC + c ◦ 휉 nD = nD + d for each substance we have where 휉 is extent (or advancement) of chemical reaction. ◦ 휈 휉 ni = ni + i ni is number of moles of ith species, or n◦ is initial number of moles of ith species, 1 ◦ i 휉 = .n * n / 휈 휈 i i i is stoichiometric coefficient for ith species in i reaction, Í 휈 > for products d휉 1 dni i 0 rate = = Í 휈 < 휈 i 0 for reactants dt i dt P. J. Grandinetti (Chem. 4300) Chemical Kinetics 3 / 60 Reaction Rates We define the reaction rate as d휉 1 dni rate = = 휈 dt i dt 1 dnNO dnO 1 dnN O e.g., 4 NO .g/ + O .g/ ,,,,,,,,,,,,,,,,,→ 2 N O .g/; rate = * 2 = * 2 = 2 5 2 2 2 5 4 dt dt 2 dt In definition so far reaction rate is extensive property. To make it intensive divide by system volume 0 1 rate 1 d휉 1 1 dni 1 d[i] R = = = 휈 = 휈 V V dt V i dt i dt R is an intensive reaction rate. It has dimensionality of concentration per time. [i] is molar concentration of ith species. More accurate to use activities instead of concentration, but concentration is good approximation in ideal systems. Reaction rates will depend on p, T, and concentration of reaction species. Rates also depend on phase or phases in which reaction occurs. Í Homogeneous reactions occur in a single phase Í Heterogeneous reactions involve multiple phases, e.g. reactions on surfaces. P. J. Grandinetti (Chem. 4300) Chemical Kinetics 4 / 60 Definition Rate Law relates rates to concentration of reactants and products Often we can define rate law in the form: R = k[A]훼[B]훽 5 k is the rate constant Overall reaction order is 훼 + 훽. 훼 and 훽 are partial reaction orders (usually integers but not necessarily). Reaction is 훼 order with respect to A Reaction is 훽 order with respect to B Determining the rate law for reaction or order of reaction must be done experimentally. Later, we will examine on a deeper level why reactions have a given order. P. J. Grandinetti (Chem. 4300) Chemical Kinetics 5 / 60 Zeroth order reaction d[A] Reaction: A ,,,,,,,,,,,,,,,,,→ P has rate law: R = * = k[A]훼 dt if 훼 = 0 then reaction is zeroth order in A and has rate law: d[A] = *k dt Rearranging and integrating [A]t t Ê d[A] = *k Ê dt [A]0 0 gives [A]t *[A]0 = *kt or [A]t = [A]0 * kt The unit of k would be mol/(L⋅s). P. J. Grandinetti (Chem. 4300) Chemical Kinetics 6 / 60 Zeroth order reaction 100 80 0th order 60 [A] 40 20 0 0 200 400 600 800 1000 time/seconds Zeroth order rate laws typically occurs when rate is limited by the concentration of a catalyst, e.g., ,,,,,,,,,,,,,,,,,→ 2 N2O(g) 2 N2(g) + O2(g) in presence of Pt surface catalyst. P. J. Grandinetti (Chem. 4300) Chemical Kinetics 7 / 60 First order reaction Consider this simple example A ,,,,,,,,,,,,,,,,,→ Products where d[A] rate = * = k[A]1 dt Integrated rate law is ¨ [A]t¨ d[A] t = *k dt Ê [A] Ê [A]0 0 [A]t *kt ln = *kt or [A]t = [A]0e [A]0 units of k would be 1/s. P. J. Grandinetti (Chem. 4300) Chemical Kinetics 8 / 60 First order reaction 100 80 0th order 60 [A] 40 20 1st order 0 0 200 400 600 800 1000 time/seconds P. J. Grandinetti (Chem. 4300) Chemical Kinetics 9 / 60 First order reaction 100 0th order 80 60 [A] 40 20 1st order 0 0 20 40 60 80 100 time/seconds P. J. Grandinetti (Chem. 4300) Chemical Kinetics 10 / 60 First order reaction – Log concentration plot 5 0th order 4 3 2 log([A]/M) 1st order 1 0 0 200 400 600 800 1000 time/seconds P. J. Grandinetti (Chem. 4300) Chemical Kinetics 11 / 60 First order reaction – Log concentration plot 5 0th order 4 3 1st order 2 log([A]/M) 1 2nd order 0 0 20 40 60 80 100 time/seconds P. J. Grandinetti (Chem. 4300) Chemical Kinetics 12 / 60 First order reaction - Half Life Sometimes it’s convenient to define a quantity called the half life, t1_2, particularly for 1st order reactions. t1_2 is the time needed for the reaction to be half complete, i.e., [A] 1 [A] 1 t1_2 0 1 when [A] = [A] then ln = ln 2 = ln = *kt t1_2 0 1_2 2 [A]0 [A0] 2 which gives t 1 = ln 2_k ù 0:693_k (only for 1st order reaction) 2 Half life of 1st-order reaction depends only on reaction rate constant and is independent of [A]0. This is not true for 2nd and 3rd order reactions. P. J. Grandinetti (Chem. 4300) Chemical Kinetics 13 / 60 First order reaction Example ◦ Benzene diazonium chloride undergoes 1st-order thermal decomposition in H2O at 50 C with : : a rate constant of k = 0 071/min. If the initial concentration is [A]0 = 0 01 M, how long must it be heated at 50 ◦C to reduce its concentration by half? : : t1_2 = ln 2_k = ln 2_.0 071_min/ = 9 9 min Note that result is independent of initial concentration, [A]0. P. J. Grandinetti (Chem. 4300) Chemical Kinetics 14 / 60 Second order reaction Here we have d[A] A ,,,,,,,,,,,,,,,,,→ Products with * = k[A]2 dt Integrating rate expression gives ¨ [A]t¨ d[A] t = *k dt Ê 2 Ê [A]0 [A] 0 1 1 = + kt [A]t [A]0 units of k would be L/(mol⋅s). Rearranging gives [A]0 [A]t = 1 + [A]0kt P. J. Grandinetti (Chem. 4300) Chemical Kinetics 15 / 60 Second order reaction 100 0th order 80 60 [A] 40 1st order 20 2nd order 0 0 20 40 60 80 100 time/seconds P. J. Grandinetti (Chem. 4300) Chemical Kinetics 16 / 60 Second order reaction – inverse concentration plot 2.0 2nd order 1.5 1/[A] 1.0 0.5 0th order 1st order 0.0 0 200 400 600 800 1000 time/seconds P. J. Grandinetti (Chem. 4300) Chemical Kinetics 17 / 60 Second order reaction – inverse concentration plot 2.0 2nd order 1.5 1/[A] 1.0 0.5 1st order 0.0 0 10 20 30 40 50 60 time/seconds P. J. Grandinetti (Chem. 4300) Chemical Kinetics 18 / 60 Another kind of second order reaction d[A] d[B] A + B ,,,,,,,,,,,,,,,,,→ Products where * = * = k[A][B] dt dt after some clever math we obtain 1 [B] [A] ln 0 t = kt [A]0 *[B]0 [A]0[B]t If [A]0 = [B]0 then solution is like previous 2nd-order example, i.e. 1 1 = + kt [A]t [A]0 P. J. Grandinetti (Chem. 4300) Chemical Kinetics 19 / 60 Another kind of second order reaction 30 25 20 15 10 5 0 0 5 10 15 20 25 30 time/seconds P. J. Grandinetti (Chem. 4300) Chemical Kinetics 20 / 60 Measuring Reaction Rate Order P. J. Grandinetti (Chem. 4300) Chemical Kinetics 21 / 60 Measuring Reaction Rate Order rate = k[A]훼[B]훽 5 [L]휆 need to determine orders, 훼, 훽,..., 휆 before measuring k. Half-life method Differential (initial rate) method Isolation method P. J. Grandinetti (Chem. 4300) Chemical Kinetics 22 / 60 Half-Life Method When rate law has the form rate = k[A]n then one can show that 2n*1 * 1 log t = log + .1 * n/ log [A] 10 1_2 10 .n * 1/k 10 0 and a plot of log10 t1_2 versus log10[A]0 gives a straight line with a slope of 1 * n. 1 Plot [A] versus t 2 Pick any [A] value and find the time interval, t1_2, for it to fall to half its value. 3 Pick another [A] value and repeat step 2. 4 Repeat step 3 several times and then plot log10 t1_2 versus log10[A]0 5 The slope will be 1 * n. P. J. Grandinetti (Chem. 4300) Chemical Kinetics 23 / 60 Half-Life Method 0.0 100 -0.2 90 t1/2= 0.100 min for [A] = 100 to 50 80 -0.4 70 t1/2= 0.125 min for [A] = 80 to 40 /min) 60 1/2 -0.6 [A] 50 t1/2= 0.167 min for [A] = 60 to 30 Log(t 40 -0.8 30 20 -1.0 10 0 -1.2 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 time/minutes Log([A]0/M) In this example, a 2nd-order rate (n = 2) gives slope of 1 * n = *1. P. J. Grandinetti (Chem. 4300) Chemical Kinetics 24 / 60 Differential (Initial Rate) Method n Since rate = k[A] then ln.rate/ = ln k + n ln[A]t, that is, slope is n. 100 3.00 d[A=100]/dt = 1000 mol/(L•s) 2.75 80 d[A=80]/dt = 640 mol/(L•s) 2.50 2.25 d[A=60]/dt = 360 mol/(L•s) 2.00 60 d[A=40]/dt = 160 mol/(L•s) 1.75 [A] 1.50 40 1.25 1.00 0.75 20 log(rate/ mol/(L•s)) 0.50 0.25 0.00 0 0.0 0.1 0.2 0.3 0.0 0.5 1.0 1.5 2.0 2.5 3.0 time/seconds log([A]0/M) For rate = k[A]n[B]m 5, vary [A] while holding other reactant concentrations constant.
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