Chapter x Kinetics of Complex reactions 8 10. 1 Typical complex reactions Outside classroom reading Levine: p559 17.9
Chapter X Kinetics of Complex Reactions Outside classroom reading: Levine: p.559 17.9 §10.1 Typical complex reactions
8 10.1 Typical complex reactions The kinds of typical complex reactions The most simple complex reaction is composed of two elementary steps aa+bB g+hh 1)Opposing reaction Reversible reaction K1B 2)Parallel reaction ompteting reactio C A-M>B-2>C 3)Consecutive Reaction
The kinds of typical complex reactions 2) Parallel Reaction Compteting reaction A B C 1 2 ⎯⎯→ ⎯⎯→ k k 3) Consecutive Reaction 1) Opposing Reaction Reversible reaction §10.1 Typical complex reactions The most simple complex reaction is composed of two elementary steps
8 10.1 Typical complex reactions 10.1.1 Opposing Reaction/reversible reaction The forward and the backward /reverse reaction take place simultaneously aa tbB hH for opposing reaction consisting of elementary k_AB=kgh reactions n=KAB K [G]HI KK k AlB k r=kgh As reaction proceeds, r, increases while r The connection between the equilibrium constant(K] and the rate coefficients of decreases. When r+ becomes equal to r simple reactions. This relation, named as equilibrium is reached kinetic equilibrium constant Can we extend this discussion to ammonia synthesis?
10.1.1 Opposing Reaction / reversible reaction The forward and the backward / reverse reaction take place simultaneously. for opposing reaction consisting of elementary reactions: [A] [B] a b r k + + = [G] [H] g h r k − − = As reaction proceeds, r+ increases while r− decreases. When r+ becomes equal to r− , equilibrium is reached. [A] [B] [G] [H] a b g h k k + − = [G] [H] [A] [B] g h a b c k K k + − = = − + = k k Kc The connection between the equilibrium constant (Kc ) and the rate coefficients of simple reactions. This relation, named as kinetic equilibrium constant. Can we extend this discussion to ammonia synthesis? §10.1 Typical complex reactions
810 1 Typical complex reactions 10.1.1 Opposing Reaction/reversible reaction (2)rate equation Under equilibrium conditions For first-first order opposing reaction k(a-x)=kx k(a-xe) B x X.-X t=0 k fe k t (x-x The total rate is a-x xe dx k(a-x)-k at
For first-first order opposing reaction: (2) rate equation t = 0 a 0 The total rate is k a x k x dt dx = + − − − ( ) e e k a x k x ( ) + − − = e e k a x ( ) k x + − − = Under equilibrium conditions e e dx ( ) x x k a dt x + − = t = t e a-xe xe t = t a-x x 10.1.1 Opposing Reaction / reversible reaction e e e ln ( ) x x k at x x + = − e e e ln ( ) a x x k at x x − − = − §10.1 Typical complex reactions
8 10.1 Typical complex reactions 10.1.1 Opposing reaction /reversible reaction a-x k Principle of relaxation method for n k In at (x-x) at x。-x studying fast reaction k+ and k can be determined by measuring The time required for the equilibrium to resume after disturbance x at t and at equilibrium concentration k tk==In er opposing reactions t (x-x 1-2 opposing reaction A kB+C =(k++k)=kt (x。-x) Why opposing reaction a+B C+D
k+ and k− can be determined by measuring x at t and at equilibrium concentration. Principle of relaxation method for studying fast reaction 10.1.1 Opposing Reaction / reversible reaction e e e ln ( ) x x k at x x + = − e e e ln ( ) a x x k at x x − − = − e e 1 ln ( ) x k k t x x + − + = − e e ln ( ) ( ) x k k t kt x x = + = + − − Why? §10.1 Typical complex reactions 1-2 opposing reaction 2-2 opposing reaction Other opposing reactions The time required for the equilibrium to resume after disturbance
8 10.1 Typical complex reactions 10.1.1 Opposing Reaction/reversible reaction ACTA CHEMICA SCANDINAVICA 12(1958)18341860 deco dt hcca, t h A Relaxation Method for the Determination of Moderately Rapid Reaction Rates near Chemical Equilibrium dcHco dt kCco, -k2cH, co, -k3CH, O, kac STIG LJUNGGREN and OLE LAMM dc The Royal Institute of Technology, Dirision of Physical Chemistry, Stockholm 70, Sweden dt A new relaxation method, based on the equilibrium displacement We will now introduce resoure rise, is described. Using the present ne twentieth of a second. The relaxation to equilibrium was followed +41 by a continuous registration of the conductivity of the sample solution This technique was employed to study the kinetics near equilibrium of the carbon dioxide-water system. It wag also found possible to cal CH, CO, co+42 culate the true dissociation constant of carbonic aci CO2+m2O、6 t HCO3 H++HcO k1=4.4×I0-38ec-mole11 ka= 2.4 sec-I KA=1.8×10 K4=1.6×104
§10.1 Typical complex reactions 10.1.1 Opposing Reaction / reversible reaction
8 10.1 Typical complex reactions 10.1.1 Opposing Reaction/reversible reaction 大孝化乎 Univ chem.2021,36(X.0001-0009 知识介绍 doi:10.3866 PKU DXHX202002076 www.dxhx.pku.edu.cn 平衡近似和稳态近似使用条件浅析 莹,张树永 山东大学化学与化工学院,济南250100 摘要:对化学动力学中经常使用的平衡近似和稳态近似的使用条件进行了讨论。从对峙反应达到和维持平衡的角度,给 出了平衡近似使用的条件,指出了原条件的不足之处。对于连续反应,采用定义无量纲量和作图的方法,进一步明确了 采用稳态近似的条件是中间体消耗速率常数(k2)比生成速率常数(k)大100倍,即k2>100k1。此时采用稳态近似处理的 误差小于1% 表1平衡近似使用条件比较 原条件 物理含义[.3 新条件-1 可能情况 物理含义 新条件-2 物理含义 k1+k2[B]>>k1[C]很小且不变 (k1+k-1)>>k2 k1>>k2 (2)是rd.s k1>>k2 (2)是rd k-1>>k2[B] (1)处于平衡 k-1>>k (1)处于平衡 平衡很快建立 k1>>k2 (2)是rds
§10.1 Typical complex reactions 10.1.1 Opposing Reaction / reversible reaction
810 1 Typical complex reactions 10.1.2 Competing reaction/Parallel reaction d[B K1 B KIAI C2H4+H2o . C C2HSOH dt-R2LA C2H OC2Hs+H2O k(a-x)+k2(a-x)=(k+k2)(a-x) CO,+H When k1>>k2 =k(a-x HCOOH CO+HO When k,<<k k2(a-x dt The rate of parallel reaction is determined mainly by the faster one
10.1.2 Competing reaction/Parallel reaction 1 [B] [A] d k dt = 2 [C] [A] d k dt = ( ) ( ) ( )( ) 1 2 1 2 k a x k a x k k a x dt dx = − + − = + − 1 2 When k k ( ) 1 k a x dt dx = − When 1 2 k k ( ) 2 k a x dt dx = − The rate of parallel reaction is determined mainly by the faster one. HCOOH CO2 + H2 CO + H2O §10.1 Typical complex reactions
810 1 Typical complex reactions 10.1.2 Competing reaction/Parallel reaction For production of B and C: dt(k,+k2)(a-x) K1 B =k1(a-x) Integration of the equation yields C B dz (k1+k2)t h,(a-x) dt 0 a-x (a-x)=aexp[-(,+k2)t] y+z = k,a[-(k+k2)门 k2 aexp[-(k,+ k2t
( )( ) 1 2 k k a x dt dx = + − Integration of the equation yields: k k t a x a ln ( ) = 1 + 2 − ( ) exp[ ( ) ] 1 2 a − x = a − k + k t A B C a 0 0 a-x y z x = y + z ( ) 1 k a x dt dy = − ( ) 2 k a x dt dz = − For production of B and C: exp[ ( ) ] 1 1 2 k a k k t dt dy = − + §10.1 Typical complex reactions 10.1.2 Competing reaction/Parallel reaction
810 1 Typical complex reactions 10.1.2 Competing reaction/Parallel reaction a-x)=aexp[-(k, +k,)t k, y {1-exp-(k1+k2) k1+k2 1-exp[-(K, +k2)t] k1+k2 B k selectivity of the reaction
( ) exp[ ( ) ] 1 2 a − x = a − k + k t {1 exp[ ( ) ]} 1 2 1 2 1 k k t k k k a y − − + + = 2 1 k k z y = selectivity of the reaction {1 exp[ ( 1 2 ) ]} 1 2 2 k k t k k k a z − − + + = A B C t c §10.1 Typical complex reactions 10.1.2 Competing reaction/Parallel reaction
