The Integrated Rate Law Section 13.4

The Integrated Rate Law Integrated Rate Law-A relationship between the the concentrations the reactants and time To keep it simple, a single reactant decomposes into products A-->products Scientists use this law to estimate the time it will take to rid the atmosphere of Chloroflourocarbons (CFC’s)

First Order Reaction In a first order simple reaction, the rate is proportional to the concentration of A Rate=k[A] or -Δ[A]/Δt=k[A] since rate=-Δ[A]/Δt This is known as the differential rate law as well Using calculus, this equation can be integrated to ln[A] t / [A] 0 =-kt

First order reaction ln[A] t =-kt+ln[A] 0 This equation is the same as a straight line A first order simple reaction will always have a negative correspondence between time and ln[A]

Example Find the constant if the molarity of SO 2 Cl 2 is at 600 seconds and M initially? ln([A] t /[A] 0 )=-kt ln(.084/.100)=-k(600) k=2.91x10 -4 s -1

Second Order Reactions When the reaction is second order, the rate is proportional to the square of the the concentration of A Rate=k[A] 2 or =k[A] 2 Again, with calculus, this equation can be integrated to become: 1/[A] t =kt+1/[A] 0

Second Order Reactions The graph of a second order reaction is also a line 1/[A] t =kt+1/[A] 0 You have to plot the inverse of the concentration of the reactant as a function of time

Example If the rate constant of a reaction is M -1 s -1 and the initial molarity is 0.010M, what molarity would be completely decomposed at 300 seconds? 1/[A] t =kt+1/[A] 0 1/[A] t =(0.225)(300)+1/0.010 [A] t =5.97x10 -3

Zero Order Reaction In a zero order reaction, the rate is proportional to a constant Rate=k[A] 0 =k or =k Once more, this can be integrated into the zero-order integrated law [A] t =-kt+[A] 0 The graph of a zero order function is a line as well

Half-Life Half Life-Time it takes required for the concentration of a reactant to fall to its initial value There are three types of half lives First Order Reaction Half Life Second Order Reaction Half Life Zero Order Reaction Half Life LO 4.3 The student is able to connect the half-life of a reaction to the rate constant of a first-order reaction and justify the use of this relation in terms of the reaction being a first-order reaction. [See SP 2.1, 2.2]

First Order Reaction Half Life To get the First-Order Half Life equation, start with the original first-order equation Then substitute t 1/2 for t and ½[A] 0 for [A] 0 Ln(½[A] 0 )/[A] 0 =ln1/2=-kt 1/2 Solve for t 1/2 (use for ln1/2) t 1/2 =0.693/k

Second Order Reaction Half Life Once again, start with the original equation for a second order reaction Substitute t 1/2 for t and 1/2[A] 0 for [A] t 1/(½[A] 0 )=kt 1/2 +1/[A] 0 Solve for t 1/2 t 1/2 =[A] 0 /2k

Zero Order Reaction Half Life Start with the integrated zero order reaction equation Substitute the variables ½[A] 0 =-kt 1/2 +[A] 0 Solve for t 1/2 t 1/2 =[A] 0 /2k

Example Find the half life of a zero-order half life equation when the original molarity is 0.250M and the constant is 0.042M ⋅ sec -1. t 1/2 =[A] 0 /2k t 1/2 =0.250/2(0.042) t 1/2 =2.98sec