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 + ′