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