Kinetics

KineticsReaction rate and rate laws

Reaction rate:rate = concentration of speciest

aA + bB cC + dD

rate = [C] = [D] = -[A] = -[B] ct dt at bt

Dependence on concentration of reactants:single reactant, A:rate = k[A]m; k = rate constant, m = ordertwo reactants, A & B:rate = k[A]m[B]nm + n = overall order

Generally, m and n are positive integers but can be zero or fractions.

Two important things about this form of the rate law:1. The concentrations of the products do not appear in the rate law b/c the reaction rate is being studied under conditions where the reverse reaction does not contribute to reaction rate.2. The values of m & n must be determined by experiment.

Determination of m and k from rate-concentration data:CH3CHO(g) CH4(g) + CO(g)

rate [CH3CHO]2.0 M/s1.0 M0.5 M/s0.50 M0.080 M/s0.20 M

rate1 = k[A]1m= [A]1 mrate2 k[A]2m [A]2

2.0 / 0.50 = (1.0/0.50)m

4.0 = 2.0mm = 2 so rate = k[CH3CHO]2

plug any data set back in to solve for k.

Integrated rate laws:zero order reactions:[A] = [A]o - ktplot of [A] vs. t is linearslope = -k, b = [A]o

first order reactions:ln[A] = ln[A]o - ktplot of ln[A] vs t is linearslope = -k, b = ln[A]o

second order reactions:1/[A] = kt + 1/[A]oplot of 1/[A] vs. t is linearslope = k, b = 1/[A]o

Zero order reactions most often involve a catalyst. When the catalyst is completely in use the concentration of reactants does not matter. When the same reaction occurs with no catalyst, the reaction is not zero order.

For first order reactions:t1/2 is independent of the original concentrationt1/2 is inversely related to k. If t1/2 is small, k is large and vice versa

Here is the notes quiz

https://docs.google.com/spreadsheet/embeddedform?formkey=dHhnUlQyTmFDREUzWmg4Qncwc2JRcUE6MQ