Integrated Rate Law: First-Order Given information - concentration and time ln[A] = -kt + ln[A]o Y = mx + b [A] = concentration of A at time t K = rate constant T = time [A]o = initial concentration of A First order: ln[A] varies linearly with time, rate = k[A]

Example 2N2O5(g) -> 4NO2(g) + O2(g) [N2O5] (mol/L) Time (s) 0.1000 0.0707 50 0.0500 100 0.0250 200 0.0125 300 0.00625 400 Example 2N2O5(g) -> 4NO2(g) + O2(g) Verify first order in [N2O5], and collect value of the rate constant, where rate = -[N2O5]/t 2. Calculate [N2O5] at 150s after the start of the reaction

Half-Life Definition: Amount of time required for a reactant to reach half of its original concentration (t1/2) For first order reactions, the general equation is t1/2 = 0.693/k Does not depend on the concentration

Example A certain first-order reaction has a half-life of 20.0 minutes. Calculate the rate constant for this reaction Answer: 3.47 X 10-2 min-1 2. How much time is required for this reaction to be 75% complete? Answer: 40. min

Integrated Rate Law: Second-Order 1/[A] = kt + 1/[A]0 1/[A] versus t will be linear with slope equal to k Rate = k[A]2 Half Life equation: t1/2 = 1/k[A]0 Each successive half-life is double the preceding one (when A0 is halved, t1/2 doubles)

Example 2C4H6(g) -> C8H12(g) [C4H6] (mol/L) Time (±1 s) 0.01000 0.00625 1000 0.00476 1800 0.00370 2800 0.00313 3600 0.00270 4400 0.00241 5200 0.00208 6200 2C4H6(g) -> C8H12(g) Is this reaction first or second order? Answer: second order 2. What is the value of the rate constant? Answer: 6.14 X 10-2 L/mol s 3. What is the half-life under these conditions? Answer: 1.63 X 103 s

Zero-Order [A] = -kt + [A]0 The rate is constant and does not change with concentration [A] versus t is a straight line Rate = k (rate is constant) Half-Life Equation: t1/2 = [A]0/2k

More Than One Reactant When there are large initial concentrations of all reactants but one, the rate law can be assumed to only be controlled by the one reactant Pseudo-first-order rate law

Summary – Memorize!! Pages 561-562 may help

2I(g) -> I2(g) Rate = k[I]2, where k = 7.0 X 109 Example 2I(g) -> I2(g) Rate = k[I]2, where k = 7.0 X 109 What are the proper units for k? If [I]0 = 0.40 M, calculate [I] at t=2.5 X 10-7 sec If [I]0 =0.40 M, calculate t1/2 If [I]0 =0.80 M, how much time would it take for 75% of the iodine atoms to react?