Measurement of Rate Constants Continuous-Flow: Stopped-Flow:

Measurement of Rate Constants Quench-Flow: Flash-Photolysis:

Measurement of Rate Constants Temperature or Pressure Jump:

Orders of Reaction Zero-order reactions: A  B rate = -d[A]/dt = k[A]n, and since n = 0, [A]0 = 1, therefore -d[A]/dt = k or –d[A] = k dt which upon integration yields: [A]t = A0 – k t -k [A] time [A0]

Orders of Reaction First-order reactions: A  B -d[A]/dt = k[A]n, where n = 1, [A]1 = [A], thus -d[A]/dt = k[A] or d[A]/[A] = -k dt which upon integration yields: In [A]/[A0] = – k t or [A] = [A0] e-kt time time [A] In [A] / [A0] - k

Orders of Reaction Second-order reactions: 2A  B -d[A]/dt = k[A]n, where n = 2, -d[A]/dt = k[A]2 or - d[A]/[A]2 = k dt which upon integration yields: 1 / [A] = kt + 1 / [A0] time 1 / [A] k 1/[A0]

Determination of Half-Lives Zero-order reactions: A  B t = t1/2 when [A] = ½[A0], therefore [A] = A0 – kt ½ A0 = A0 – k t1/2 k t1/2 = A0 - ½A0 t1/2 = A0 / 2k t1/2 [A0] 1/2k

Determination of Half-Lives First-order reactions: A  B t = t1/2 when [A] = ½[A0], therefore ln [A0]/2[A0] = - kt1/2 or ln ½ = - kt1/2 and since ln ½ = -ln2, t1/2 = 1/k ln2 = 0.693/k t1/2 [A0] 0.693/k

Determination of Half-Lives Second-order reactions: 2A  B t = t1/2 when [A] = ½[A0], therefore 1 / [A] = 1 / ½[A0] = 2 / [A0], so -1 / [A0] + 2 / [A0] = k t½ = 1 / [A0] t½ = 1/k · 1/[A0] [A0] t½ t1/2 1/[A0] 1/k

Determination of Reaction Order rate constant half-life -k [A] time [A0] zero order t1/2 [A0] 1/2k time [A] t1/2 [A0] 0.693/k first order [A0] t½ time 1 / [A] k 1/[A0] second order

LaPlace Transform Method for Solving Integrated Rate Equations Rate Law Transform (Diff. Eq.) (p = d/dt) Integrated Solve for Solution time variable Inverse Transform (from table)

LaPlace Transforms k1 define reaction A  B First-order reaction (irreversible) k1 define reaction A  B

LaPlace Transforms k1 A  B d[A]/dt = -k1[A] define rate law First-order reaction (irreversible) k1 A  B d[A]/dt = -k1[A] define rate law

LaPlace Transforms k1 A  B d[A]/dt = -k1[A] let p = d/dt First-order reaction (irreversible) k1 A  B d[A]/dt = -k1[A] let p = d/dt p[A] = -k1[A] LaPlace transforms assume [A] = 0 at t = 0, if not then subtract the value of p[A] at t = 0: p[A] – p[A0] = -k1[A]

LaPlace Transforms k1 A  B d[A]/dt = -k1[A] let p = d/dt First-order reaction (irreversible) k1 A  B d[A]/dt = -k1[A] let p = d/dt p[A] = -k1[A] LaPlace transforms assume [A] = 0 at t = 0, if not then subtract the value of p[A] at t = 0: p[A] – p[A0] = -k1[A] solve for [A] algebraically: p[A] + k1[A] = (p + k1)[A] = p[A0]

LaPlace Transforms k1 A  B d[A]/dt = -k1[A] let p = d/dt First-order reaction (irreversible) k1 A  B d[A]/dt = -k1[A] let p = d/dt p[A] = -k1[A] LaPlace transforms assume [A] = 0 at t = 0, if not then subtract the value of p[A] at t = 0: p[A] – p[A0] = -k1[A] solve for [A] algebraically: p[A] + k1[A] = (p + k1)[A] = p[A0] look up on inverse transform table: [A] = (p/(p + k1))[A0] = [A0]e-k1t

LaPlace Transforms k1 A  B [A] = [A0]e-k1t First-order reaction (irreversible) k1 A  B [A] = [A0]e-k1t time [A] where kobs = k1 and Amplitude = [A0]

First-order reaction (reversible) k+1 A  B k-1

First-order reaction (reversible) k+1 A  B k-1 1.) write the appropriate rate law: d[A]/dt = -k+1[A] + k-1[B] where [B] = [A0] – [A]

First-order reaction (reversible) k+1 A  B k-1 1.) write the appropriate rate law: d[A]/dt = -k+1[A] + k-1[B] where [B] = [A0] – [A] therefore d[A]/dt = -k+1[A] + k-1([A0] – [A]) = k-1([A0] - (k+1 + k-1)[A]

First-order reaction (reversible) k+1 A  B k-1 1.) write the appropriate rate law: d[A]/dt = -k+1[A] + k-1[B] where [B] = [A0] – [A] therefore d[A]/dt = -k+1[A] + k-1([A0] – [A]) = k-1([A0] - (k+1 + k-1)[A] 2.) rewrite as a LaPlace Transform where p = d/dt and subtract p[A0] if [A0]  0: p[A] – p[A0] = k-1[A0] - (k+1 + k-1)[A]

k+1 A  B k-1 3.) combine like terms: p[A] + (k+1 + k-1)[A] = p[A0] + k-1[A0] or (p + k+1 + k-1)[A] = (p + k-1)[A0]

k+1 A  B k-1 3.) combine like terms: p[A] + (k+1 + k-1)[A] = p[A0] + k-1[A0] or (p + k+1 + k-1)[A] = (p + k-1)[A0] 4.) solve for [A] algebraically: [A] = [A0]((p + k-1)/(p + k+1 + k-1))

k+1 A  B k-1 3.) combine like terms: p[A] + (k+1 + k-1)[A] = p[A0] + k-1[A0] or (p + k+1 + k-1)[A] = (p + k-1)[A0] 4.) solve for [A] algebraically: [A] = [A0]((p + k-1)/(p + k+1 + k-1)) 5.) look up the solution in the inverse transform table (i.e. for (p + a)/(p + b)): [A] = [A0]((a/b) – ((a – b)/b)e-bt) where a = k-1 and b = (k+1 + k-1)

k+1 A  B k-1 3.) combine like terms: p[A] + (k+1 + k-1)[A] = p[A0] + k-1[A0] or (p + k+1 + k-1)[A] = (p + k-1)[A0] 4.) solve for [A] algebraically: [A] = [A0]((p + k-1)/(p + k+1 + k-1)) 5.) look up the solution in the inverse transform table (i.e. for (p + a)/(p + b)): [A] = [A0]((a/b) – ((a – b)/b)e-bt) where a = k-1 and b = (k+1 + k-1) Thus by substitution: [A] = [A0]((k-1/(k+1 + k-1)) + (k+1/(k+1 + k-1))e-(k+1 + k-1)t

[A] = [A0]((k-1/(k+1 + k-1)) – (k+1/(k+1 + k-1))e-(k+1 + k-1)t A  B k-1 [A] = [A0]((k-1/(k+1 + k-1)) – (k+1/(k+1 + k-1))e-(k+1 + k-1)t Thus your observed rate constant (kobs) is a single exponential that is the sum of the forward and reverse rate constants, i.e. time [A] kobs = (k+1 + k-1) and the observed exponential transient will decay to a final value determined by the relative values of k+1 + k-1 (i.e the final equilibrium concentrations of A and B).

k+1 A  B k-1 kobs = (k+1 + k-1) To determine k+1 and k-1 explicitly additional information is required:

k+1 A  B k-1 kobs = (k+1 + k-1) To determine k+1 and k-1 explicitly additional information is required: 1.) Amplitude information from stopped-flow experiment can be used.

k+1 A  B k-1 kobs = (k+1 + k-1) To determine k+1 and k-1 explicitly additional information is required: 1.) Amplitude information from stopped-flow experiment can be used. 2.) The equilibrium constant for the reaction step determined independently.

k+1 A  B k-1 kobs = (k+1 + k-1) To determine k+1 and k-1 explicitly additional information is required: 1.) Amplitude information from stopped-flow experiment can be used. 2.) The equilibrium constant for the reaction step determined independently. 3.) If the reaction is actually second order (i.e. a binding step) but run under psuedo-first order conditions (i.e. [S] >> [E]), then kobs = (k+1[S] + k-1) and k+1 + k-1 can be determined from a plot of kobs vs. [S], where slope = k+1 and y-intercept = k-1. k+1 E + S  ES k-1

Sequential first-order reactions (irreversible) k+1 k+2 A  B  C 1.) write the appropriate rate law: d[A]/dt = -k+1[A] which is exactly the same rate law that describes a single irreversible step reaction.

Sequential first-order reactions (irreversible) k+1 k+2 A  B  C 1.) write the appropriate rate law: d[A]/dt = -k+1[A] which is exactly the same rate law that describes a single irreversible step reaction. 2.) Thus the integrated solution is: [A] = [A0]e-k+1t and we can determine k+1 directly from kobs by following the change in [A].

Sequential first-order reactions (irreversible) k+1 k+2 A  B  C 1.) write the appropriate rate law: d[A]/dt = -k+1[A] which is exactly the same rate law that describes a single irreversible step reaction. Thus the integrated solution is: [A] = [A0]e-k+1t and we can determine k+1 directly from kobs by following the change in [A]. But what about k+2? We need to be able to examine the change in [B] or [C]!

k+1 k+2 A  B  C 1.) write the appropriate rate law: d[B]/dt = k+1[A] – k+2[B]

k+1 k+2 A  B  C 1.) write the appropriate rate law: d[B]/dt = k+1[A] – k+2[B] 2.) rewrite as a LaPlace Transform where p = d/dt ( note that now [B0] = 0): p[B] = k+1[A] – k+2[B]

k+1 k+2 A  B  C 1.) write the appropriate rate law: d[B]/dt = k+1[A] – k+2[B] 2.) rewrite as a LaPlace Transform where p = d/dt ( note that now [B0] = 0): p[B] = k+1[A] – k+2[B] 3.) combine like terms: p[B] + (k+2)[B] = k+1[A]

k+1 k+2 A  B  C 1.) write the appropriate rate law: d[B]/dt = k+1[A] – k+2[B] 2.) rewrite as a LaPlace Transform where p = d/dt ( note that now [B0] = 0): p[B] = k+1[A] – k+2[B] 3.) combine like terms: p[B] + (k+2)[B] = k+1[A] and since as we already determined: [A] = [A0](p/(p + k+1)) we can write by substitution: (p + k+2)[B] = k+1[A0](p/(p + k+1))

k+1 k+2 A  B  C 4.) solve for [B] algebraically: [B] = k+1[A0](p/(p + k+1)(p + k+2))

k+1 k+2 A  B  C 4.) solve for [B] algebraically: [B] = k+1[A0](p/(p + k+1)(p + k+2)) 5.) look up the solution in the inverse transform table (i.e. for (p/((p + a)(p + b))): [B] = [A0]((1/(b – a)) e-at + (1/(a – b))e-bt) where a = k+1 and b = k+2

k+1 k+2 A  B  C 4.) solve for [B] algebraically: [B] = k+1[A0](p/(p + k+1)(p + k+2)) 5.) look up the solution in the inverse transform table (i.e. for (p/((p + a)(p + b))): [B] = [A0]((1/(b – a)) e-at + (1/(a – b))e-bt) where a = k+1 and b = k+2 Thus by substitution: [B] = [A0](((k+1/(k+2 - k+1)) e-k+1t + (k+1/(k+1 - k+2))e-k+2t) Therefore we observe a two exponential decay where the two observed rate constants correspond directly to k+1 and k+2!

Determinants of Matrices Consider a simple 2x2 matrix. The determinant of that matrix is given by: A B C D Det = = A•D – C •B

Cramer’s rule P = A •x + B •y Q = C •x + D •y P = A B x Q = C D y P B Two equations and two unknowns can be written in matrix form as: P = A •x + B •y Q = C •x + D •y P = A B x Q = C D y Where the solutions for x and y can be determined from the determinants of the following matrix equations: P B P•D – B •Q Q D x = = A•D – C •B A B C D A P A•Q – C •P C Q y = = A•D – C •B A B C D