The above is a bimolecular elementary reaction. A unimolecular elementary reaction might be HO 2 H + O 2 HO 2 just dissociates without any other influence. Rate Laws for Elementary Reactions: 1) A Fragments, depends only on A dc A dt = kc A [Like radioactive decay] 2) A+A Products, depends only on A, A collision = dc A dt = kc A 2
A + B Products. dc A dt = dc B dt = kc A C B. Elementary reaction is one place where stoichiometry and rate are related. However never know when you have an elementary reaction. Must guess and then verify with experiment. Elementary reactions are hypothetical constructs!
Unimolecular Decompositions A Fragments An example of Mechanisms, Steady State Approximation, and Elementary Reactions pyrazine Observed Experimentally to be first order in pyrazine.
da dt = k[a] or C - B C + B with d[c B] dt = k[c B] Lindemann [Lord Cherwell] ] suggested the following mechanism: Collisions in general produce molecules with higher than average energy. Consider a box full of A and let those molecules which have an energy E* greater than or equal to that necessary for decomposition be A*. Assume, however, that after A* is produced by a collision it hangs around for some time before decomposing. This time lag between activation and reaction may be thought of as the time necessary to transfer energy among the internal (vibrational( vibrational) ) coordinates.
If A * exists for a reasonable time it could suffer a collision and drop down to a lower energy where it cannot decompose. k Collision A(E 1 ) + A(E 2 ) 1 between A*(E*) + A(E 3 ) A(E 1 ) and A(E 2 ) creates activated Collision A*(E*) E 1, E 2, E 3 < E min ; E* E min = energy necessary for decomposition k 1 A * (E * ) + A(E 4 ) A(E 5 ) + A(E 6 ) k Collional cooling 2 step Reaction step Products Again E 4, E 5, E 6 < E min Competition between reaction of A*(E*) to form products and collisional cooling of A*(E*) to produce unreactive A(E 5 ) and A(E 6 ), and k 2 are kinetic rate constants) (k 1, k -1, and k
k 2 is decomposition step assumed irreversible. Mechanism Elementary Steps: A + A k 1 A + A A + A k 1 A + A Step 1 Step 2 A k 2 P dp dt = k 2 [A ] Step 3 Step 1 Step 2 Step 3 d[a ] dt = k 1 [A] 2 k 1 [A ][A] k 2 [A ] Don t t know what [A * ] is, however the number of [A * ] must be small or the reaction would go to completion very quickly. A* is a bottleneck for the reaction since product is only formed via A*.
Therefore we assume a steady state for [A * ] important trick or approximation: d[a * ]/dt = 0 = k 1 [A] 2 -k -1 [A * ][A]-k 2 [A * ] Solve for [A * ] [A * ] = k 1 [A] 2 / ( k 2 +k -1 [A] ) } Key Point: This is a most The Steady State Approximation (Assumes are making and destroying A * at the same rate.) Steady state approach allows us to solve for concentration of unknown species [A * ] in terms of known [A] concentration. Rate = dp dt = k 2[A ] = k k 2 1[A] 2 Fundamental result of Lindemann Unimolecular k 2 + k 1 [A] Reaction Mechanism Note, multistep mechanism leads to complex rate expression!
Two Limiting Cases I) k -1 [A] << k 2 Rate = dp dt = k 2 [A ] = k 2 k 1 [A]2 k 2 + k 1 [A] This says the rate of decomposition of A * is much faster than the rate of deactivation. Using k 2 >>k -1 [A] and the expression for dp dp/dt: Rate = dp/dt = k 2 [A*] = k 2 k 1 [A] 2 /(k 2 +k -1 [A]) Take k -1 [A] to be 0 compared to k 2 Thus, dp/dt dt k 1 [A] 2 }2nd order in [A] At this point, it looks like Mr. Lindemann will have to hand in his Theorists Club ID card since his scheme seems to predict a second order kinetic dependence on [A] 2!
Physical Interpretation of this particular limit: Mechanism Elementary Steps: Step 1 d[a*]/dt = k 1 [A] 2 A + A k 1 A + A A + A k 1 A + A Step 2 Slow enough to be ignored! A k 2 P Step 3 So fast that every A* formed Reacts immediately! Reaction rate here is just the rate at which [A*] is formed since every A* formed falls apart to product P immediately. The rate of o formation of A* is obtained from the first step: d[a*]/dt dt = k 1 [A] 2 So the reaction becomes a simple binary collison model, second order process in this limit.
(Case II) k -1 [A] >> k 2 Rate = dp dt = k 2 [A ] = This says there is an appreciable time lag between activation and reaction. Thus, a large amount of deactivation occurs. Take k Rate = dp/dt = k 2 [A*] = k 2 k 1 [A] 2 /(k 2 +k -1 [A]) 2 to be 0 compared to k -1 [A] Note that k -1 [A] is pressure dependent. Gets larger as pressure increases {P A =(n A /V)RT = [A]RT}. Thus, k -1 [A] >> k 2 is a good approxmation at high pressures. dp dt k k 2 1 [A] 1 k st order in A (apparent) 1 At high pressures expect to see this behavior. k 2 k 1 [A]2 k 2 + k 1 [A] [More careful investigation of unimolecular decomp.. showed 2 nd order kinetics at low pressure.]
Physical Interpretation of this particular limit: Mechanism Elementary Steps: A + A k 1 A + A A + A k 1 A + A Step 1 Step 2 k -1 [A*][A] = k 1 [A] 2 A k 2 P Step 3 So slow that steps 1 and 2 Can reach equilibrium. Reaction rate here is dp/dt dt = k 2 [A*] but the concentration of A* is given by the equilibrium condition, k -1 [A*][A] = k 1 [A] 2. So, solving for [A*] gives: [A*] = k 1 [A] 2 / k -1 [A] = k 1 [A]/ k -1 Result is that [A*] scales linearly with [A] and rate, dp/dt = k 2 [A*] also scales linearly with [A].
Catalysis Catalysis provides an additional mechanism by which reactants can be converted to products. The alternative mechanism has a lower activation energy than the reaction in the absence of a catalyst. A υ o B υ o no catalyst υ c -- catalyst present υ c (v 0 = -d[a]/ d[a]/dt dt with no catalyst) (v c = -d[a]/ d[a]/dt dt with a catalyst) Without a catalyst: Rate = v 0 With Catalyst: Rate=v 0 +v c but v c >>v 0
Not affected by catalyst E Not Potential Energy here is energy stored in the chemical bonds E a,f Potential Energy E a,f E a,f E Reactants Energy barrier without catalyst E a,r E a,r E a,f,f and E a,r are lowered by catalyst Products Energy barrier with catalyst Reaction coordinate Catalysts speed reactions by reducing the activation energy barrier.
Generally a catalyst is defined as a substance which increases the t rate of a reaction without itself being changed at the end of the reaction. This is strictly speaking not a good definition because some things catalyze themselves, but we will use this definition for now. Catalyst supplies a reaction path which has a lower activation energy than the reaction in the absence of a catalyst. Catalysis by Enzymes Enzymes may be loosely defined as catalysts for biological systems. They increase the rate of reactions involving biologically important systems.
Enzymes are remarkable as catalysts because they are usually (work only a particular kind of reaction.) amazingly specific (work only a particular kind of reaction.) They are also generally very efficient, achieving substantial They are also generally, achieving substantial Rate increases at concentrations as low as 10-8 M! Typical enzyme molecular weights are 10 4-10 6 gm/mole (protein molecules)
Summary of Enzyme Characteristics 1) Proteins of large to moderate weight 10 4-10 6. 2) Extremely efficient (work at 10-8 M) 3) Very specific (work only on special types of reactions). General Behavior of Enzyme Catalyzed Reactions E S (substrate) Products If the initial rate of the reaction is plotted versus the initial concentration of substrate S for a constant enzyme concentration, the following behavior is found:
Maximum initial rate Initial Rate (Half maximum initial rate) [S] concentration when V i = V S / 2 Substrate Concentration V i = - d[s]/dt dt measured at t=0. (Initial slope of [S] vs t plot)