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Determining Rate Laws

Determining Rate Laws. 2 NO(g) + O 2 (g)  2 NO 2 (g) Determine the rate expression and the value of the rate constant from the data below. [NO] (mol L -1 ) [O 2 ](mol L -1 ) initial rate (mol L -1 s -1 ) (1) 1.0 x 10 -4 1.0 x 10 -4 2.8 x 10 -6

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Determining Rate Laws

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  1. Determining Rate Laws 2 NO(g) + O2(g)  2 NO2(g) Determine the rate expression and the value of the rate constant from the data below. [NO] (mol L-1) [O2](mol L-1) initial rate (mol L-1 s-1) (1) 1.0 x 10-4 1.0 x 10-4 2.8 x 10-6 (2) 1.0 x 10-4 3.0 x 10-4 8.4 x 10-6 (3) 2.0 x 10-4 3.0 x 10-4 3.4 x 10-5 Rate = k [O2]m [NO]n To determine the rate law from the data, first determine the dependence of the rate on each reactant separately. rate2/rate1 = k [O2]2m [NO]2n / k [O2]1m [NO]1n 8.4 x 10-6 / 2.8 x 10-6 = (3.0 x 10-4)m/ (1.0 x 10-4)m 3= 3m => m = 1; 1st order in O2

  2. rate3/rate2 = k [O2]3m [NO]3n / k [O2]2m [NO]2n 3.4 x 10-5 / 8.4 x 10-6 = (2.0 x 10-4)n/ (1.0 x 10-4)n 4= 2n => n = 2; 2nd order in NO Rate = k [O2][NO]2 Order of reaction = 3 2.8 x 10-6 mol L-1s-1 = k [1.0 x 10-4 mol L-1] [1.0 x 10-4 mol L-1]2 k = 2.8 x 106 L2 mol-2s-1

  3. Concentration and Time - Integrated Rate Laws Integrated rate laws: variation of concentration of reactants or products at any time Derived from the experimental rate laws Zero order reactions A  P - d[A]/ dt = k d[A] = - k dt [A]t = [A]o - k t integrated rate law [A]o : concentration of A at t = 0 Slope = -k

  4. First Order Reactions Rate = - d[A]/ dt = k [A] first order reaction Units of k for a 1st order reaction is time-1 d[A]/ dt = - k [A] d[A]/[A] = - k dt Solution is: [A]t ln [A]t = [A]o e-kt = - kt [A]o [A]owhere is the initial concentration of A at time t = 0

  5. N2O5(g)  N2O4(g) + 1/2 O2(g) [A]t = [A]o e-kt ln[A]t = ln [A]o - k t

  6. 0.693 t1/2 = k Half life of a 1st order reaction Half life : time it takes for the concentration of the reactant A to fall to half its initial value t1/2 when [A]t = [A]o/2 ln[A]t = ln [A]o - kt ln [A]o/2 = ln [A]o - k t1/2 ln(1/2) = - k t1/2 ln(2) = k t1/2 t1/2 = ln(2) / k

  7. Mercury (II) is eliminated from the body by a first-order process that has a half life of 6 days. If a person accidentally ingests mercury(II) by eating contaminated grain, what percentage of mercury (II) would remain in the body after 30 days if therapeutic measures were not taken? k = ln 2 / t1/2 = (ln 2) / (6 days) Fraction remaining after 30 days [A]t / [A]0 = e-kt = 0.03 Answer: 3% remaining in the body after 30 days.

  8. Radioactive decay is a first order process Nt = No e-kt where Nt is the number of radioactive nuclei at time t No is the initial number of radioactive nuclei k is called the decay constant t1/2 = ln 2/ k

  9. Carbon-14 dating uses the decay of 14C 14C is produced in the atmosphere at an almost constant rate As a result proportion of 14C to 12C is ~ constant 14C enters living systems as 14CO2; all living systems have a fixed ratio of 14C to 12C; about 1 14C to 101212C When the organism dies, there is no longer exchange of C with surroundings, but 14C already in the organism continues to decay with a constant half life, so ratio of 14C to 12C decreases.

  10. rate = k [A] [B] or rate = k [A]2 Second order reactions 2nd order reaction for which the rate depends on one reactant Rate = k[A]2 - d[A]/ dt = k [A]2 d[A]/[A]2 = - k dt

  11. The half-life of a 2nd order reaction when [A] = [A]o/2

  12. ln[A]t = ln [A]o - kt 1st order 2nd order

  13. 2C2F4 C4F8 ln[C2F4] vs time - nonlinear slope = k rate = k [C2F4]2

  14. Reaction Mechanisms Determines rate laws; use experimental rate law to determine mechanism Reactions often proceed in a series of steps - elementary reactions For example: O2 + light  O2* O2*  O. + O. 2(O. + O2 + M  O3 + M) Overall reaction: 3O2 + light  2O3 O. is an intermediate species; involved in a step in the mechanism, but does not appear in the overall reaction

  15. Overall reaction 6 Fe2+(aq) + Cr2O72-(aq) + 14H+(aq)  6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l) The rate of an elementary reaction is directly proportional to the product of the concentrations of the reactants, each raised to a power equal to its coefficient in the balanced equation for that step The number of reacting species in an elementary reaction - molecularity

  16. Types of elementary reactions Unimolecular reaction; molecularity = 1 O2*  O + O Rate = k[O2*] Bimolecular reaction; molecularity = 2 NO(g) + O3(g)  NO2(g) + O2(g) Rate = k [NO] [O3] Termolecular reaction; molecularity = 3 O + O2 + M  O3 + M rate = k [O] [O2] [M] Termolecular reactions are low probability reactions; require three species to come together simultaneously

  17. From rate law to reaction mechanism Products of a reaction can never be produced faster than the rate of the slowest elementary reaction - rate determining step Experimental data for the reaction between NO2 and F2 indicate a second-order rate Overall reaction: 2 NO2(g) + F2(g)  2FNO2(g) Rate = k [NO2] [F2] How can a mechanism be deduced from the rate law? Rate law indicates that the reaction cannot take place in one step

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