Study of how rapidly reactions proceed – rate of reaction Details of process fro

Study of how rapidly reactions proceed – rate of reaction Details of process fro

Dorley, Rick, Minister has reference to this Academic Journal, PHwiki organized this Journal Study of how rapidly reactions proceed – rate of reaction Details of process from reactants to products – mechanism Thermodynamics determines the direction in which reactions proceed spontaneously in addition to equilibrium conditions, but not the rate at which equilibrium is reached. For a complete picture of a chemical reaction need in as long as mation on both the thermodynamics in addition to kinetics of a reaction. Chemical Kinetics N2(g) + 3H2(g) 2NH3(g) DGo = -33.0 kJ K298 = 6.0 x 105 Thermodynamically favored at 298 K However, rate is slow at 298K Commercial production of NH3 is carried out at temperatures of 800 to 900 K, because the rate is faster even though K is smaller.

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Thermodynamical functions are state functions (DG, DH, DE) Thermodynamics does not depend on the mechanism of the reaction. The rate of the reaction is very dependent on the path of the process or path between reactants in addition to products. Kinetics reveals in as long as mation on the mechanism of the reaction. Thermodynamics vs Kinetics A + B -> C + D K1 A + B -> E + F K2 If K1 > K2 =>products C & D are thermodynamically favored over E & F. What about the rates of the two reactions If products observed are C & D => reaction is thermodynamically controlled If products observed are E & F => reaction is kinetically controlled (1) 2NO(g) + O2(g) -> 2NO2(g) (2) 2CO(g) + O2(g) -> 2CO2(g) Both have large values of K Reaction (1) is fast; reaction (2) slow Reactions are kinetically controlled

Rates of Reactions Rate of a reaction: change in concentration per unit time A -> P average reaction rate = If concentration is in mol L-1, in addition to time in seconds, the rate has units of mol L-1 s-1. NO2(g) + CO(g) -> NO(g) + CO2 (g) NO2(g) + CO(g) -> NO(g) + CO2 (g) Time (s) [NO] mol L-1 0 0 50 0.0160 100 0.0240 150 0.0288 200 0.0320 Average rate 1st 50 seconds = 3.2 x 10-4 mol L-1 Average rate 2nd 50 seconds = 1.6 x 10-4 mol L-1 Average rate 3rd 50 seconds = 9.6 x 10-5 mol L-1

Instantaneous Rate – rate at a particular moment in time Kinetics deals with instantaneous rates, or simply rates NO2(g) + CO(g) -> NO(g) + CO2 (g) For a general reaction: aA + bB -> xC + yD

Factors affecting rates of reactions a) Nature of reactants 2NO(g) + O2(g) -> 2NO2(g) fast 2CO(g) + O2(g) -> 2CO2(g) slow b) Concentration of reactants: reactions proceed by collisions between reactants c) Temperature: In general, as T increases, rate increases d) Catalyst: increases rate of reaction e) Surface f) Nature of solvent Rate Laws in addition to Rate Constant rate = k [NO2] [CO] k is the specific rate constant NO2(g) + CO(g) -> NO(g) + CO2 (g) For a general reaction: aA + bB -> cC + dD rate = k [A]m [B]n For a reaction k has a specific value; k as long as the reaction changes with temperature Note: m need not equal a; n need not equal b Rate Law

Order of a Reaction rate = k [A]m [B]n Order of the reaction = m + n The reaction order is determined by the experimentally determined rate law N2O5(g) -> N2O4(g) + 1/2 O2(g) Rate = k [N2O5] reaction is a first order reaction For a 1st order reaction, units of k: time-1 C2H6(g) -> 2 CH3(g) rate = k [C2H6]2 second order reaction 2NO2(g) -> 2NO(g) + O2 (g) Rate = k [NO2]2 second order reaction For 2nd order reactions, units of k: concentration-1 time-1 Determination of order of a reaction 2HI(g) -> H2(g) + I2(g) At 443oC the rate of the reaction increases with HI concentration as follows: Data point 1 2 3 [HI] mol L-1 0.0050 0.010 0.020 Rate mol L-1 s-1 7.5 x 10-4 3.0 x 10-3 1.2 x 10-2 Determine the order of the reaction in addition to write the rate expression Calculate the rate constant, in addition to determine its units Calculate the reaction rate as long as a concentration of HI = 0.0020M

rate = k [HI]n a) rate1 = k ([HI]1)n rate2 = k ([HI]2)n rate2 / rate1 = ([HI]2)n / ([HI]1)n 3.0 x 10-3 / 7.5 x 10-4 = (0.010/0.0050)n 4 = 2n n = 2 rate = k [HI]2 b) 7.5 x 10-4 mol L-1 s-1 = k (0.0050 mol L-1)2 k = 30 L mol-1 s-1 c) rate = k [HI]2 = 1.2 x 10-4 mol L-1 s-1 2 NO(g) + O2(g) -> 2 NO2(g) Determine the rate expression in addition to 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) 1.0 x 10-4 1.0 x 10-4 2.8 x 10-6 1.0 x 10-4 3.0 x 10-4 8.4 x 10-6 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 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

First Order Reactions For the general reaction: A -> Products if the experimental rate law is: Rate = – d[A]/ dt = k [A] first order reaction Units of k as long as a 1st order reaction is time-1 d[A]/ dt = – k [A] [A]o where is the initial concentration of A at time t = 0 ln[A] = ln [A]o – kt [A] = [A]o e-kt N2O5(g) -> N2O4(g) + 1/2 O2(g) rate = k [N2O5] Radioactive decay is a first order process N = No e-lt where N is the number of radioactive nuclei at time t No is the initial number of radioactive nuclei l is the decay constant

Half life of a 1st order reaction Half life : time it takes as long as the concentration of the reactant A to fall to half its initial value t1/2 when [A] = [A]o/2 ln[A] = 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 What is the rate constant k as long as the first order decomposition of N2O5(g) at 25oC if the half life at this temperature is 4.03 x 104 s Under these conditions, what percent of the N2O5 molecules have not reacted after one day a) t1/2 = 0.6931 / k k = 1.72 x 10-5 s-1 b) [N2O5] = [N2O5]o e-kt [N2O5]/[N2O5]o = e-kt [N2O5]/[N2O5]o = 0.226 22.6% N2O5 molecules have not reacted after one day Second order reactions Rate = k[A]2 The half-life of a 2nd order reaction can be determined by setting [A] = [A]o/2 at t = t1/2 – d[A]/ dt = k [A]2 2nd order reaction as long as which the rate depends on one reactant rate = k [A] [B] or rate = k [A]2

ln [C2F4] slope = k 2C2F4 -> C4F8 ln[C2F4] vs time is not linear rate = k [C2F4]2 Zero order reactions A -> P If the rate law is – d[A]/ dt = k zero order reaction For a 0th order reaction: rate is independent of concentration [A] = [A]o – kt Zero order reaction: [A] = [A]o – kt t 1st order ln[A] = ln [A]o – kt [A] = [A]o e-kt

Catalytic Converters Incomplete combustion of gasoline produces CO, hydrocarbon fragments (CmHn) Also, high temperature in the engine causes oxidation of N2 to NO in addition to NO2 Introduce catalysts into the exhaust to convert these pollutants to less harmful compounds Without a catalyst conversion would be very slow Catalyst: pellets of Pt, Pd, Rh Autocatalysis Catalysis of a reaction by the products A-> P Rate = k [A][P] reaction rate increase as P is as long as med Concentration of reactants, products or intermediates vary periodically with time Autocatalysis plays the role of positive feedback A consequence of autocatalysis is an oscillating reaction BrO3- + HBrO2 + H3O+ -> 2BrO2 + 2 H2O 2BrO2 + 2 Ce3+ 2H3O+ -> 2HBrO2 + 2 Ce4+ + 2H2O http://www.chem.leeds.ac.uk/delights/texts/expt-11.html Briggs-Rauscher Reaction

Dorley, Rick Minister

Dorley, Rick is from United States and they belong to WTLS-AM and they are from  Tallassee, United States got related to this Particular Journal. and Dorley, Rick deal with the subjects like Religion

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