PaRT 11-Chemical Kinetics Reference: Chapter 18 in book
PART 11 PART 11 – Chemical Kinetics Chemical Kinetics Reference: Chapter 18 in book Reference: Chapter 18 in book
Rate of Chemical Equations 1. Representation of Reaction Rate △n n2-n △t (unit: mol:S-1) (unit: mol- L-1S or mol L-1. min-1. mol L-1.h- (unit: atm: s-1 or kpa S-1, mmHg S-1)
Rate of Chemical Equations Rate of Chemical Equations 1. Representation of Reaction Rate: Δ n n n v = = (unit: mol·s-1 ) t n Δ Δ n n t t 2 1 2 1 − − vC = (unit: mol·L-1·s-1 C C t t 2 1 2 1 − − or mol·L-1·min-1, mol·L-1·h-1) vP = (unit: atm·s-1 o r k pa·s-1 , mmH g·s-1 ) P P t t 2 1 − 2 P ( p g ) t t 2 1 −
2N2O5(9)4NO2(g)+O2(9) v=-AN2O5/tv=^N○2]/△t △[O2]/△t P △PNo/△t vp=△PNo2/△t △Po,/^t Most of reactions do not have constant rate. above are all tor average reaction rates
2 N O ( ) 4 NO () + O ( ) 2O5 (g) → 4 NO2 (g) + O2 (g) v = Δ[N O ] / Δt v ’ = Δ[NO ] / Δt v ’’ v = Δ[O ] / Δt c = – Δ[N2O5] / Δt vc = Δ[NO2] / Δt vc = Δ[O2] / Δt vP = – ΔPN2O5 / Δt vP’ = ΔPNO2 / Δt vP’’ = ΔPO2 / Δt P P P Most of reactions do not have constant rate. Above 3 are all for average reaction rates
Example: H202 >H20+202 [,O2l changes with time: Co=0.80 molL- After 20 min: C=0.40 mol L-I v1=-(04-0.8)20=002 molL-I.min1 After another 20 min: C=0.20 mol L - v2=-(02-0.4)20=001 molL-l. min1 After another 20 min: 0.10 molL 3=-(0.1-0.2)20=0.005molL1min1
Example: Example: H2O2 → H2O + ½O2 [H O ] h ith ti C 0 80 l L 1 [H2O2] changes with time: C0 = 0.80 mol·L-1 After 20 min: After C = 0 40 mol·L-1 20 min: C1 = 0.40 mol·L v1 = – (0.4 – 0.8)/20 = 0.02 mol·L-1·min-1 v1 (0.4 0.8)/20 0.02 mol L min After another 20 min: C2= 0.20 mol·L-1 v2 = – (0.2 – 0.4)/20 = 0.01 mol·L-1·min-1 After another 20 min: C3= 0.10 mol·L-1 4 v3 = – (0.1 – 0.2)/20 = 0.005 mol·L-1·min-1
乙dC dt △C △t dc 2N2O5(9)→>4NO2(g)+O2(9) N205 d[n2O5/dt No2 d[No2]/dt Vo2=d[o2/dt
C v Δ = __ Δt dC v = dt 2 N2O5 (g) → 4 NO2 (g) + O2 (g) 2 5 (g) 2 (g) 2 (g) vN2O5 = – d[N2O5] / dt vNO2 = d[NO2] / dt 5 vO2 = d[O2] / dt
2 N2O5(9)>4 NO2( 9)+O2(g) d[n2O5]/2 dt = d[NO2]/4-at =d[o2]/dt Therefore, for a general chemical reaction aA+bB→>gG+hH V=-d[a]/adt=-d[B]/b dt=d[G]g dt=d[H]/h dt
2 N2O5 (g) → 4 NO2 (g) + O2 2 N (g) 2O5 (g) → 4 NO2 (g) + O2 (g) v = – d[N2O5]/2·dt = d[NO2]/4·dt = d[O2 v d[N ]/dt 2O5] / 2 dt d[NO2] / 4 dt d[O2] / dt Ì Therefore, for a general chemical reaction: aA + bB aA + bB → gG + hH gG + hH v = –d[A] / a d[A] / a dt · = –d[B] / b d[B] / b dt · = d[G] / g d[G] / g dt · = d[H] / h d[H] / h dt · 6
Rate Laws aa t bb gG t hh Net rate: V=V forward-V, reverse At initial moments: Reactant - Product [A]/adt=k·[A·[B]→ Differential Rate Laws k is rate constant (a function of temperature m, n are orders of reaction (m is the order of reaction for A, n is the order of reaction for B)
Rate Laws Rate Laws aA + bB ' gG + hH Net rate: v = vforward – vreverse At initial moments: Reactant → Product −d[A] / a·dt = k · [A]m · [B ]n Æ Differential Rate Laws k is rate constant k is rate constant (a function of temperature) (a function of temperature) m, n are orders of reaction (m is the order of reaction for A, 7 n is the order of reaction for B )
Question a reaction at room temperature is as follows: S2O82+3→2SO42+l3 The rates under different initial concentrations are [2O32] v(d[S2082-]/dt) 0.076 0.060 2. 8x 10-5 mol/L min 2.0038 0.060 1.4x105mo∥Lmin 3.0.076 0.030 1.4x10-5 mol/L min Write the rate laws for this reaction d[s208]/dt=k IsoM[n
Question: A reaction at room temperature is as follows: S O 2 SO 2 2O82- + 3 I- → 2 SO42- + I3- The rates under different initial concentrations are: The rates under different initial concentrations are: [S2O82-] [I-] v (–d[S2O82-] / dt ) 1. 0.076 0.060 2.8 x 10-5 mol/L min 2 0 038 0 060 1 4 10 5 . 0.038 0.060 1.4 x 10 l/L i -5 mol/L min 3. 0.076 0.030 1.4 x 10-5 mol/L min Write the rate laws for this reaction: 8 –d[S2O82-] / dt = k [S2O82-]m [I- ]n
Solution By simple observation, we can see m=1. n=1 Total order of this reaction m+n= 2 d [s2o82]/dt= k[S2O82][[] k=28×105/(0.076)(0.060)=614X103mo1L1 dS2Og2]dt=6.14×10s22]] d[/dt=3(-d[S2O82]/d)=3X6.14X103S2O32][
Solution: By simple observation, we can see: m = 1, n = 1 Total order of this reaction: m + n = 2 – d [S2O82-] / dt = k [S2O82- ]1 [I- ]1 k = 2 8 x 10-5 / (0 076)(0 060 ) = 6 14 x 10-3 mol-1·L·s-1 k = 2.8 x 10 / (0.076)(0.060 ) = 6.14 x 10 mol L s – d[S2O82-]/dt = 6.14 x 10-3[S2O82- ]·[I- ] d[I- ]/dt = 3 (– d[S O 2-] / dt) = 3 x 6 14 x 10-3[S O 2- ]·[I- d[I ]/dt = 3 (– d[S ] 2O8 ] / dt) = 3 x 6.14 x 10 [S2O8 ] [I ] 9
About the rate constant k Physical meaning the reaction rate when concentration of each reactant is 1 mol/L Unit: s-1(1st order reaction mol-1.L S-1(2nd order reaction) mol-zL2 S-1 3rd order reaction) Some other reactions have more complicated rate laws, for instance, for H2(g)+ br2( g))2 HBr(g) the rate law is: kH,Br,1/ 1+k THBr [Br2]
About the rate constant k: Ì Physical meaning: the reaction rate when concentration of each reactant is 1 mol/L concentration of each reactant is 1 mol/L. Ì Unit: s-1 (1st order reaction) mol-1·L·s-1 (2nd order reaction) mol-2·L 2·s-1 (3rd order reaction) Some other reactions have more complicated rate laws, for instance, for H 2 (g) + Br 2 (g) Æ 2 HBr (g) , the rate law is: [HBr] k[H ][Br ] 1 2 2 2 v = 10 [Br ] [HBr] 1 k 2 + ′