Basic Principle of Chemical Reaction and Air Pollution
: n n n n n (
Fe 3 O 4 (s) + 4CO(g) = 3Fe(s) +4CO 2 (g)? Al? TiO 2 (s) +2C= Ti (s) +CO 2 (g) 2 HOl ( ) 2 H( g) + O( g)? 2 2 2 N ( g) + 3 H ( g) 2 NH ( g) 2 2 3
1/ > Δh=h 2 -h 1 0 > ΔT=T 2 -T 1 0 > ΔV=V 2 -V 1 0 > ΔP=P 2 -P 1 0 : 1 2) 3 4
2 Δ C() s + O ( g) CO ( g) 2 2 Δ rh m = 218.66kJ mol 1867 Pt Δ H = 258.8kJ mol 2 H ( g) + O ( g 2 H O( l) 2 2 2 r m 1 1 Zn s Cu aq Zn aq Cu s 2+ 2+ () + ( ) ( ) + () HOs () HOl () > 273.15K 2 101.32kpa 2 NH NO () s NH ( aq) + NO ( aq) + 4 3 4 3 CaCO () s CaO() s + CO ( g) 1183K 3 101.32kPa 2 Δ H = 393.5kJ mol Δ H = 6.007kJ mol r m Δ H = 25.7kJ mol r 1 Δ H = 178.32kJ mol r r m m m 1 1 1 (/)
n n n HOs () HOl () > 273.15K 2 101.32kpa 2 NH NO () s NH ( aq) + NO ( aq) + 4 3 4 3 CaCO () s CaO() s + CO ( g) 1183K 3 101.32kPa 2 Δ H = 6.007kJ mol r m Δ H = 25.7kJ mol r 1 Δ H = 178.32kJ mol r m m 1 1 ( ~
3(S-entropy): n S = k ln Ω J/K 1.38 10-23 J/K n ΔS Δ S = q T r Ω - qr = SΔT
n n ΔS 0 n n ΔS = ΔS +ΔS < 0 o CH 2 O
, 100%
(): (0K) n S m S(0 K) = kln1 = 0 H 2 (g) O 2 (g) I 2 (s) I 2 (g) S m (J. mol -1. K -1 ) 130.684 205.138 116.135 260.65 H + (aq) Mg 2+ (aq) Ca 2+ (aq) Ba 2+ (aq) S (J. mol -1. K -1 ) 0-138.1-53.1 9.6 m 3 298.15K Br 2 (g) 245.468 CO 3 2- (aq) -56.9
n : : 1.S g > S l > S s 2.S 3.S 4. n>0, S>0 n=0s n<0, S<0 5.SS H O(g) H O(l) H O(s) 2 2 2-1 -1 S(Jmol K ) 188.7 69.96 39.33 CH 4 C2H 2 C3H 8 n-c4h 10 n-c5h12-1 -1 S(Jmol K ) 186.15 229.49 269.81 310.03 345.40 C(CH ) (CH ) CHCH CH CH (CH ) CH 3 4 3 2 2 3 3 2 3 3-1 -1 S(Jmol K ) 306.4 343 348.40
H f H m ΔH m kj/mol S S m ΔS m J/(mol. K) S S S >S
Δ r S m / ΔS n : aa(l) + bb(aq) = gg(s) + dd(g) Δ S = g S ( G, s) + d S ( D, g) a S ( A, l) b S ( B, aq) r m m m m m Δ S = v S Δ r m B m, B a. b.1 mol. c.. d. S ( T) Δ S (298.15 K) r m r m
1CaCO 3 ΔS (298.15 ) m K ΔH (298.15 ) m K S r 273.15K CaCO ( s) = CaO(s) + CO (g) Δ 3 2 1 fh m( kj mol ) -1206.9-635.5-393.5 1 1 m ( J mol K ) 92.9 39.7 213.6 Δ H = [( 393.5) + ( 635.5) ( 1206.9)] kj mol r m m = 177.9kJ mol Δ S = (213.6 + 39.7 92.9) J mol K 1 = 160.4 J mol K 1 1 1 1 Δ rhm > 0, Δ r S m > 0 1 Δ S (600 K) = 156.3J mol K 1 1 r m
ΔS ΔH + ΔS CaCO 3 (s)=cao(s)+co 2 (g) Δ r H m = 177.9 kj mol-1 >0 Δ r S m = 164.69 J mol-1 >0 298.15K 1074K 298.15KΔ r H m -298.15 Δ r S m =130.1kJ/mol >0 1074K Δ r Η m 1074 Δ r S m =0 kj/mol Δ r H m -TΔ r S m < 0 1200KΔ r Η m 1200 Δ r S m =20.63kJ/mol < 0
HCl(g) + NH 3 (g) =NH 4 Cl(s) Δ r H m = -176.89 kj mol-1 < 0 Δ r S m = -284.23 J mol-1 < 0 298.15K (),, 622.35K Δ r H m - 298.15 Δ r S m = -92.15kJ/mol < 0 Δ r H m - 622.35 Δ r S m = 0.00kJ/mol = 0 Δ r H m 900.00 Δ r S m = 78.92kJ/mol > 0 : ΔH-TΔS <0. ΔH-TΔS =0 ΔH-TΔS >0. :, ΔH-TΔS <0 ; :(H 2 -H 1 )-T(S 2 -S 1 )<0 (H 2 -TS 2 )-(H 1 -TS 1 )<0 (T,S,H)
1876 G /( Gibbs free energy) G = H TS G ~ Δ G =ΔH TΔS Δ G =Δ H TΔ S r m r m r m - Δ G < 0 Δ G = 0 Δ G > 0
2CaCO 3 ΔS (298.15 ) m K Δ H = 177.9kJ mol r r m m 1 Δ S = 160.4 J mol K 1 1 r m r m ΔH (298.15 ) m K Δ H > 0, Δ S > 0 Δ G =Δ TΔ S = 3 m rhm r m 177.9 (298.15)(160.4 10 ) =130.1kJ mol 1 25 0 C,100kPaCaCO 3
Δ G= ΔH-T ΔS H S G - + -, + - +, + + - - CaCO () s = CaO() s + CO ( g) 3 2 T< H/ S +, T > H/ S -, T< H/ S -, T> H/ S +, H ( g) + Cl ( g) = 2 HCl( g) 2 2 2 O ( g) = 3 O ( g) 3 2 T c =1074 K 2 CO( g) = 2 C( s) + CO ( g) N ( g) + 3 H ( g) = 2 NH ( g) 2 3 3 (T c ): ΔH ΔrHm(298.15 K) Δ G=ΔH Tc Δ S = 0 Tc = Δ S Δ S (298.15 K) r m Tc= 467K 2
/ ΔS > 0 ΔG < 0 ΔS = 0 ΔG = 0
n T,Pw -ΔG > -Δw - ΔG = - Δw - ΔG < - Δw (- Δw ) (-ΔG) Δ ' G = w max
Δ G=ΔH TΔ S =Δ U + pδv TΔS ΔU=q+w Δ G= q+ w+ pδv TΔS ' w= pδ V + w ' Δ G= q pδ V + w + pδv TΔS ' = q+ w TΔS qr = TΔS Δ ' G = w max Δ G= w '
2) ΔGΔG ΔG ΔGΔG x Q ~ x x p B ~ Bp ~ 100kPa c B ~ Bc ~ 1mol/dm -3 ()Q e.g. aa(l) + bb(aq) = gg(s) + dd(g) x d ( pd / p ) Δ rgm( T) =Δ rgm( T) + RTln ( / ) b c c B
CO ( g) + 2 NaOH ( c, aq) = Na CO ( c, aq) + H O( l) 2 1 2 3 2 2 ( c / c ) Na CO 2 3 Q = ( 2 / pco p ) ( cnaoh / c ) p= p i p i 1 1 2 px χ ι i χ i = ni 298K10L 200kPa2mol = i n
5 1298.15K Δ H + fg m Δ G ( 298.15K) = ν Δ G (298.15 K) r m B f m, B B kj / mol aa(l) + bb(aq) = gg(s) + dd(g) Δ G (298.15 K) = gδ G ( G, s,298.15 K) + dδ G ( D, g,298.15 K) r m f m f m aδ G ( A, l,298.15 K) bδ G ( B, aq,298.15 K) f m f m
1298.15K Δ r m 298.1 K = υb f m, B Δ B 2 ΔrG m( T) ΔrH m(298.15 K) T ΔrS m(298.15 K) 3 Δ G (298.15 K) =Δ H (298.15 K) 298.15 K Δ S (298.15 K) r m r m r m Δ H ( T) Δ H (298.15 K) Δ S ( T) Δ S (298.15 K) r m r m G ( 5 ) G (298.15 K) r m r m Δ G ( T ) =Δ G ( T) + RTlnQ r m r m
3. 723KΔ r G m, 2SO 2 (g) + O 2 (g) = 2SO 3 (g) /Pa 1.0 10 4 1.0 10 4 1.0 10 8-296.83 0-395.72 248.22 205.14 256.76 x x x x x x x x Δ r G m >0723K
4:SnO 2 Sn, (1) SnO 2 (s) Sn(s)+O 2 (g) 100kPa (2) ()SnO 2 (s)+c()=sn(s)+co 2 (g) (3) H 2 (g)sno 2 (s)+2h 2 (g)=sn(s)+2h 2 O(g)? (1) SnO 2 (s) Sn(s)+O 2 (g) Δ f H m -580.7 0 0 S m 52.3 51.55 205.03 Δ r H m =- Δ f H m (SnO 2 )=-(-580.7)=580.7kJ/mol Δ r S m = (205.03+51.55)-52.3=204.28J/k. mol,p(o 2 )=P Q=P(O 2 )/P =1; lnq=0 Δ r G m (T)=Δ r G m (T)+RTlnQ; Δ r G m (T)=Δ r G m (T)= ΔH -TΔS Δ r G m (T)<0, ΔH -TΔS <0 T c > ΔH / ΔS =2843K (Sn mp:505k, bp:2533k) (2) Sn T c = 904K (3) : T c = 840K
ΔG, ΔH, ΔS (1) (2), (3),ΔG, ΔH, ΔS ΔG, ΔH, ΔS ( (4),:, (),,,; H 2 O(l)H 2 O(g) (5)q,w,,, (6), ΔG, ΔH,ΔS ;, ΔG, ΔH, ΔS (7)
Δ r G m 1 2 3Δ r G m Δ r G m(41.8kj/mol) 4 1 HOl 2 () = H2( g) + O2( g), Δ rgm > 0 2 1 H2( g) + O2( g) = H2O( l), Δ rgm < 0 2
1 n (reversible reaction) 2KClO 3KCl + 3O Δ 3 2 x (CaSO 4, 1000 0 C 1840 0 C) (Ag + + Cl - =AgCl, 2NO 2 =N 2 O 4 )
n (chemical equilibrium) N 2 O 4 2NO 2 0 o C Δ r G m =0
2 12NO N O 2 2 4 K c [NO 2 ]/mol. L -1 [N 2 O 4 ]/mol. L -1 [N 2 O 4 ]/[NO 2 ] [N 2 O 4 ]/[NO 2 ] 2 (/mol. L -1 ) -1 0.052 0.595 11 220 0.024 0.127 5.3 220 0.068 1.02 15 220 0.101 2.24 22 220 (
n n K /K c K p = eq { p ( G)} { p ( D)} { p ( A)} { p ( B)} p eq a eq aa+ bb = gg + dd g eq d b K { c ( G)} { c ( D)} = { c ( A)} { c ( B)} eq g eq d c eq a eq b 2NO N O 2 2 4 NO 1 N O 2 2 2 4
n eq K { p / p } ν B p = p B B K /c 1. 1()K CaCO () s = CaO() s + CO ( g) 3 2 eq K = { p ( CO )/ p } 2 2 Fe ( aq) + Cl ( g) = 2 Fe ( aq) + 2 Cl ( aq) K = 2 + 3 + 2 eq { c ( Fe )/ c } { c ( Cl )/ c } 3+ 2 eq 2 = K =9.4*10 19 298.15K) eq 2+ 2 eq { c ( Fe )/ c } { p ( Cl2)/ p } eq ν Kc { cb / c } B B K =1.6*10-23 298.15K)
Ø Ø QK Q K K, K, K = f(t) Ø K α n α = 100% n 2K 1 3 N2( g) + H2( g) = NH3( g) N2( g) + 3 H2( g) = 2 NH3( g) 2 2 {p eq (NH K! {p eq (NH K! 2 = 3 ) / p! } 2 1 = 3 ) / p! } {p eq (N 2 ) / p! } 1/2 {p eq (H 2 ) / p! } 3/2 K 1! = K 2! {p eq (N 2 ) / p! }{p eq (H 2 ) / p! } 3
2 Q=K Δ ( ) 0 rgm T = Δ G ( T ) =Δ G ( T ) + RT lnq= 0 r m r m Δ ( ) ln rgm T = RT K Δ G ( T) = 2.303RTlg K r m rgm ln ( T K ) Δ = RT *
3 P45 (3)=(1)+(2) (3)=(1)-(2) (3)=2(1)-(2) (3) = -(2) K = K K 3 1 2 K = K / K 3 1 2 K = K / K 2 3 1 2 K = K 1
4 5298.15K600KCaCO 3 CaCO3() s = CaO() s + CO2( g),caco 3 1 298.15K Δ G (298.15 K) = 130.1kJ mol K Δ G (600 K) = 82.0kJ mol Δ G ( T ) = RT lnk ΔrGm(298.15 K) ln K1 (298.15 K) = = 52.52 RT 52.52 23 K (298.15 K) = e = 1.6 10 K r m 1 r r m m 16.44 2 (600 K) e 7. = = 2 1 0 8 1
+ AgCl() s Ag ( aq) + Cl ( aq) Δ f 109.7 Δ = ν Δ G = [( 131.2) + (77.1) ( 109.7)] r B f m B Δ G 55.6 10 J mol ln K = = = 22.48 K 3 1 r m 1 1 RT 8.314J mol K 298.15K = 1.7 10 10 K = 1.77 10 10
7:673K,2.00molCO(g)2.00molH 2 O(g); a.k, b.,co, c.h 2 O4.00mol,CO? :a. CO(g) + H 2 O(g) D CO 2 (g) +H 2 (g) 3 Δ f H m -110.52-241.82-393.50 0 (kj/mol) S m 197.67 188.72 213.64 130.57 (J/mol K) Δ r H m (298.15K) =-41.16kJ/mol Δ r S m (298.15K) =-42.07J/mol. K Δ r G m (673K) Δ r H m (298.15K)-673 Δ r S m (298.15K) =(-41.16)-673 (-42.07/1000)=-12.85kJ/mol lnk = - Δ r G /RT =-(12.85) 1000(J/mol) 8.31(J/mol. K) 673(K)=2.297 K =9.94
b): CO(g) + H 2 O(g) D CO 2 (g) + H 2 (g) mol: 2.00 2.00 0 0 mol : 2.00-x 2.00-x x x χ i : K 2.0 x 2.0 x x x 4.0 4.0 4.0 4.0 eq eq ( p / p )( p / p ) CO 2 2 = 2 2 eq eq ( p / p )( p / p ) CO H H O 2 2 p : (2.00-x)+(2.00-x)+ x + x =4.00mol = χ p eq CO CO total K χ CO H 2 2 = = = χ [( χco p ) / )][( ) / )] 2 total p χh p 2 total p = [( χ p ) / p )][( χ p ) / p )] CO χ χ H O 2 CO total H O total 2 x (2.00 x) 2 2 9.94 x =1.52mol CO 1.52 α = 100% = 76% 2.00
c. ): CO(g) + H 2 O(g) D CO 2 (g) + H 2 (g) mol: 2.00 4.00 0 0 mol : 2.00-x 4.00-x x x χ i : K χ CO χh 2.0 x 4.0 x x x 6.0 6.0 6.0 6.0 = 2 2 = = χ CO χ H O x 1 =1.84mol CΟ 2 2 x (2.00 x) 4.00 x) x 2 =4.82( α = 1.84 100% = 92% 2.00 9.94 : (2.00-x)+(4.00-x)+ x + x =6.00mol 1. Κ α 2. Κ α
4 n Δ rg ( ) lnq m =Δ rgm T + RT Δ G = r m RTln Q K Δ G = RT ln K Q < K, Δ G < 0, r m Q = K, Δ G = 0, r m Q > K, Δ G < 0, r m Q r m
n n n n
1) Q Δ rgm =Δ rgm + RTln Q= RTln K Q,Q. :3H 2 (g)+n 2 (g)=2nh 3 (g) ;Δ r G=0 Q eq [ p / p ] NH 3 = = eq eq [ p / p ][ p / p ] N 2 2 [N 2 ][H 2 ]();[NH 3 ](); Q c <K c, Δ r G<0., Q c > K c Δ r G>0. H 2 :,(); (),;( )(),. 3 K
2,; (1) : CO(g)+H 2 O(g)DCO 2 (g)+h 2 (g) eq eq Q 1 [ pco / p ][ p / ] 2 H p 2 Q1 = = K eq eq [ p / p ][ p / p ], Q 2 eq eq [2 pco / p ][2 p / ] 2 H p 2 Q2 = = K eq eq [2 p / p ][2 p / p ] :,, CO CO H O 2 H O 2
(2): N 2 (g)+3h 2 (g)d2nh 3 (g) :,, : : Q eq [ p / p ] 3 = = eq eq [ p / p ][ p / p ] Q 2 2 Q<K Δ r G<0,. N :,,,. NH H 2 3 K eq 2 [2 pnh / p ] 1 eq eq 3 [2 pn / p ][2 ph / p ] 4 3 = = 2 2 p/p 10 50 100 300 600 1000 NH 3 % 50.7 74.4 81.5 90.0 95.4 98.3 K
3) pv = nrt (1), p=(n/v)rt p i =(n i /V)RT V, n i /V,p i,;. (2),; V,n i /V,p i,.
4) K K Δ G = RT ln K r m Δ H Δ ln K = r m + r m Δ rgm =ΔrHm TΔrS RT R m Δ H Δ S r m r m A A = ; B = lnk = + B R R T, ΔH <0; A>0,T,K. ()., ΔH >0,, (). S 2SO 2 (g)+o 2 (g)=2so 3 (g) ; Δ r H m = -196.6kJ/mol / 0 C 400 600 800 1000 1100 K 397 9.53 0.195 0.187 0.098 Ν 2 (g)+o 2 (g)=2no(g) Δ r H m = 180kJ/mol 1538K K =0.86 10-4 2402K K =64 10 4
K 1, K 2 Δ r m r m ln K1 = + RT1 R Δ H H Δ Δ S S r m r m ln K2 = + RT2 R K Δ H 1 1 Δ H T T K R T T R TT ln 1 = r m ( ) = r m ( 1 2) 2 1 2 1 2 8. BeSO 4 (s)=beo(s)+so 3 (g)127 0 CK =3.87 10 16, 175kJ/mol327 0 C K Δ H 1 1 16 3 3.87 10 175 10 1 1 1 r m ln = ( ) ln = ( ) K2 R T1 T K 2 2 8.314 400 600 K 2 = 1.5 10 8
: 1. //, (). 2.,().. 3. 4.,;,,.. /(Le Chatelier H)
9. N 2 O 4 (g)=2no 2 (g), 101.3kPa325K N 2 O 4 50.2% 1K 25 101.3KPaN 2 O 4 1n(N 2 O 4 ) = b mol, α N 2 O 4 (g) = 2NO 2 (g) t 0 b 0 t b(1- α) 2bα b(1- α) +2bα=b(1+α) χ B (1-α)/(1+α) 2α/(1+α) p/kpa eq 2α p 2α p 2 p ( NO2 ) = 1+ α p eq 2 ( ) 2 [ p ( NO2 )] 1 p 4α p + α K = = = = eq 2 eq 1 α p p ( N 1 2O4) α p 1 α p p ( N2O4) = 1+ α p 1+ α p 2K 2 4α 5 101.3kPa K = = 1.37 2 1 α 100kPa α=25.1% 1.37
T 40 0 C Tc [(CH 3 CH 2 ) 2 NH 2 ] 2 CuCl 4 Vessel Ideation- One Winner of the 2009 World Kitchen Tea Off
n CO+NO=CO 2 +N 2 H 2 +O 2 =H 2 O n 1600 n (1/4) (7-10t) 2H 2 +O 2 =2H 2 O
n n n p p p p
1. : (/chemical reaction rate) mol. L -1. s -1 the average rate and instantaneous rate of chemical reaction H 2 O 2 2 H 2 O 2 (aq) 2 H 2 O(l) O 2 (g)
H 2 O 2 t/s 0 400 800 1 200 1 600 2 000 2 400 2 800 c(h 2 O 2 )/mol dm 3 2.32 1.72 1.30 0.98 0.73 0.54 0.39 0.28 t/s 400 400 400 400 400 400 400 c(h 2 O 2 )/mol dm 3 0.60 0.42 0.32 0.25 0.19 0.15 0.11 (c(h 2 O 2 )/t)/mol dm 3 s 1 15.0 10-4 10.5 10-4 8.0 10-4 6.3 10-4 4.8 10-4 4.8 10-4 2.8 10-4 = c(h 2 O 2 ) t c(h 2 O )/mol dm -3 2 2.50 1.50 0.50 400 1200 2000 2800 T/s
n υ aa + bb = gg + d D 1 Δc 1 Δc 1 Δc 1 Δc = = = = a Δt b Δt g Δt d Δt A B G D 1 dc 1 dc 1 dc 1 dc υ = = = = a dt b dt g dt d dt A B G D 2 HO( l) = 2 HOl ( ) + O( g) 2 2 2 2 1 dho do υ = = 2 dt dt 1 dho 2 2 υ = 2 dt 2 2
n [K 2 S 2 O 8 +KI Na 2 S 2 O 3 +] : 3
3. : Influential factors on chemical reaction rate p p p p n,
SO Cl = SO + Cl 2 2 2 2 υ = ksocl [ ] 2 2 υ = kc( SO Cl) 2 H Cl H Cl H Cl Cl H CO( g) + NO ( g) = CO ( g) + NO( g) 2 2 υ = kco [ ][ NO2 ]
aa + bb = gg + dd a υ = ka [ ] [ B] b k - a- Ab - Bn=a+b - ------ k: k k k
CHCl 2 5 CH 2 4+ HCl υ = kchcl [ ] 2 5 n=1 2NO NO + O 2 2 υ = kno [ ] 2 n=2 2 2 2NO + O 2NO 2 2 υ = KNO O 2 [ ] [ 2] n=2+1=3
H + Cl 2HCl 2 2 : aa + bb = gg + dd m υ = ka [ ] [ B] n 1 2 Cl2 + M Cl + M 2 3 H Cl HCl Cl 2NO + 2H = N + 2H O 2 2 2 1)2 NO + H = N + H O ( ) 2 2 2 2 2) HO + H = 2 HO( 2 2 2 2 2 Cl + H HCl + H 42Cl + M Cl + M + 2 + 2 H 2 +I 2 =2HI (1)I 2 =2I (2)2I+H 2 =2HI υ = kh [ ][ Cl] 2 2 [ ] [ 2] υ = kno H υ = kh [ 2][ I2] 2 1 2
2Na+2H 2 O=2NaOH+H 2 v=k 0 C 2 H 5 Cl=C 2 H 4 +HCl v=k[c 2 H 5 Cl] 1 NO 2 +CO=NO+CO 2 v=k[no 2 ][CO] 2 2NO+O 2 =2NO 2 v=k[no] 2 [O 2 ] 3 CH 3 CHO =CH 4 +CO v=k[ch 3 CHO] 3/2 3/2 mol. L -1. s -1. k k s -1 mol -1.L -1.s -1 n mol (1-n).L (n-1).s -1
1., 2., : C(s)+O 2 (g)=co 2 (g); ν=k[o 2 ] 2Na(s)+2H 2 O(l)=2NaOH+H 2; r= k 3., 2NO 2 (g)=2no(g)+o 2 (g r=k p [p(no 2 )] 2 ; 4.,,,, 5.
υ = ka [ ] N O N O + 2 5 2 4 1 O 2 Ra Rn+ 2 α CH 3 COOC 2H5 + H2O CH 3COOH + C2H5OH
dc υ = = kc dt dc = k dt c 0 ln c kt c = ln c= kt+ ln c0 c dc = k dt c c 0 0 t n lnc t-k n (t 1/2 )c 0 t1/2 = ln 2 / k = 0.693/ k n ks -1 ln c -k t
N+ n C+ H 14 1 14 1 7 0 6 1 C N+ e β 14 14 0 6 7 1 ( t1/2 = 5730a Shroud of Turin
: 8 (mg/100cm 3 ) t (h) 4 -lnc 8 12 16 c 0.480 0.326 0.222 0.151 a) t(h) a) t(h) 4 8 12 16 -lnc 0.734 1.121 1.202 1.890 -lnct [ A] b ln 0 [ A = kt ] k=0.964h -1 t 1/2 =0.693/k=7.2h b) kc)4h c) 0.37 0.37mg/100cm 3 1 [ A] 0 1 0.480 t ln ln 2.7 = h k [ A] = 0.964 0.370 = 242
n (1) (2), 3Fe(s) + 4H 2 O(g)Fe 3 O 4 (s) + 4 H 2 (g)
n 10K2-3 1: 2
n
n E a - E a k = Ae RT A- E a ln k = + ln A RT k1 Ea 1 1 Ea T1 T2 ln = ( ) = ( ) k R T T R TT 2 2 1 1 2 k Svante August Arrhenius Sweden 1859-1927 1930 Nobel prize chem Δ H Δ S RT R ln K = r m + r m K Δ H 1 1 Δ H T T K R T T R TT ln 1 = r m ( ) = r m 1 ( 2) 2 1 2 1 2
n v: E a, RT e kk k E a v:,ea RT e k, n : 1. k E 2. E a A. ln k a 3. kt,t,k. 4. lgk1/t,-e a /(2.303R), lga. 5. E a Ea Ea = + ln RT A
3-4 Ø (Collision Theory) Ø (Transition State Theory) Ø (Theory of Single Molecule Reactions )
3-4 : 1918Lewis 1. (//) Z r Z T=700 0 C [HI]= 1.0 10-3 mol L -1 2ΗΙ = Η 2 + Ι 2 5.8 10 4 mol L -1 s -1 1.2 10-8 mol L -1 s -1
3-4 : 2. f l (Ε c ): r Z f 1 µολ E c E a
3-4 : 3. P r = Z f P
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3-4 : 3. E b( E b( E b ref. 2010, 25, 35-42
3-4 : E b 42-420 kj/mol,63-250 kj/ mol E b E b /(kj mol -1 ) HCl (l) + NaOH (l) = NaCl (l) +H 2 O (l) 20 2H 2 O 2 (l) = 2H 2 O (l) + O 2 (g) 80 N 2 (g) + 3H 2 (g) = 2NH 3 (g) 335.5
3-4 : E b( E b( Δ r H Δ ρ Η = Ε β() Ε β()
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35 : MnO 2 V 2 O 2KClO 3 2 KCl + 3O 2 2 5 SO 2 +O 2 2SO 3
35 A+B AB: 1)I, E a ; 2)K,,II A+K AK E 1 AK+B AB+K E 2 E 1 E 2 Ea
35 :HI503K184kJ mol -1 Au104.6kJ mol -1 80kJ mol -1 1
35 1. 2. 3.
35 4. 5. 6. 7.
45 u I2 CH CHO CH + CO 3 4 E = 190 kj / mol E = 136 kj / mol 1 2 a a 2NO 2N + O 1 2 a a Au 2 2 2 E = 250 kj / mol E = 120 kj / mol u u TiO 2
45 ΧΟ+ΝΟ=ΧΟ 2 +Ν 2 La 2 O 3 Pd LaMnO
45 PtPdNi
45 37,,, ( )?
45
45