#15 Chapter Chemical Kinetics
15.1) Reaction Rates 15.2) Rate Laws: Introduction 15.3) Determining the Form of the Rate Law 15.4) Integrated Rate Law 15.5) Rate Laws: Summary 15.6) Reaction Mechanisms 15.7) The Steady-State Approximation 15.8) A Model for Chemical Kinetics 15.9) Catalysis
What happens during a chemical reaction?
Example: 2NO2(g) 2NO(g) + O2(g)
Reactants are used up and products are formed The reaction takes time The final concentrations approach those on the righthand side of the reaction equation The rate of change in a molecules concentration (i.e. the slope) changes with time
300 oC Figure 15.1
The Rate is the Change in Concentration per Unit Time
Rate of consumption of NO2 = -[NO2]/t
Rate of production of NO = +[NO]/t Rate of production of O2 = +[O2]/t As t approaches zero, the instantaneous rate becomes the tangent: d[NO2]/dt
2NO2(g)
2NO(g) + O2(g)
Typically, were interested in the rate of the reaction itself. For the reaction: 2NO2(g) 2NO(g) + O2(g) Rate = -(1/2)d[NO2] = (1/2)d[NO] = d[O2] dt dt dt
The rate of change of concentration of each species is divided by its coefficient in the balanced chemical equation. Rates of change of reactants appear with negative signs. Product rates appear with positive signs.
The Differential Rate Law
Reaction: aA products Rate = -d[A] = k[A]n dt k = rate constant n = order of reaction in [A] k and n are determined experimentally!
Assumes reverse reaction is negligible (i.e. forward reaction is irreversible)
The Integrated Rate Law
Express rate law in the form of [A] vs. t
Example: First-Order Reaction (n = 1) [A]
Rate = -d[A] = k[A] dt
t
[A]0
1 d[A] = -k dt [A]
0
ln[A] - ln[A]0 = -kt
[A] ln = -kt [A]0
[A] = [A]0e-kt
How Do We Get the Time-Dependent Product Concentrations ?? Remember: The reaction must always be mass-balanced 2NO2 2NO + O2
[NO]t = [NO2]0 (starting mol of reactant) - [NO2]t (remaining) [O2]t = (1/2)[NO]t (also, let's assume constant volume)
[NO2]t = [NO]t [NO2]0e-kt = [NO2]0 - [NO2]0e-kt [NO]t = [NO2]0 {1- e-kt}
Let's check it: At t = 0: e-0 = 1, {1-1} =0, [NO]0 = 0, as presumed At t : e- = 0, {1-0} =1, [NO]0 = [NO]0: All NO2 reacted to NO
For a First-Order Reaction (n = 1): Exponential decay of concentration with time
[A] = [A]0e-kt ln[A] - ln[A]0 = -kt
To test if a reaction is first order, plot ln[A] vs. t It should be linear, with slope = -k and intercept = ln[A]0
Example of a 1st-order reaction
2N2O5 (soln) 4NO2 (soln) + O2 (g)
Rate = -d[N2O5]/dt = k[N2O5] (differential rate law) ln[N2O5] = -kt + ln[N2O5]0 (integrated rate law)
Data:
Figure 15.2
Plotting ln[N2O5] vs time gives a straight line
Figure 15.3
Linear! First order in N2O5 Slope = -k = -6.93 x 10-3 s-1
Half life of a 1st-order reaction
Half life of a reaction t1/2
time for a reactant to reach half of its initial concentration t = t1/2 when [A] = 1/2[A]0
[A] Since ln = -kt [A]0
t1/2 = (ln2)/k = 0.693/k
t1/2 is independent of starting concentration
Half life of N2O5 decomposition
Figure 15.4
Half life is independent of initial concentration
t1/2 = (ln2)/k = 0.693/6.93x10-3s-1 = 100s
Example: Radio-isotope Decay is First-Order R...