Chemical Kinetics Chapter 17 Chemical Kinetics Aka Reaction Rates.
CHE 112 - MODULE 3 CHAPTER 15 LECTURE NOTES. Chemical Kinetics Chemical kinetics - study of the...
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Transcript of CHE 112 - MODULE 3 CHAPTER 15 LECTURE NOTES. Chemical Kinetics Chemical kinetics - study of the...
CHE 112 - MODULE 3
CHAPTER 15
LECTURE NOTES
Chemical Kinetics Chemical kinetics - study of the rates of
chemical reactions and is dependent on the characteristics of the reactants
Reaction mechanisms - detailed pathway that atoms and molecules take as a chemical reaction proceeds
CD-ROM Screen 6.2 & 15.2
Effects on Chemical Kinetics
Concentration of reactants – usually the rate of a reaction increases with increased concentration of reactants
Concentration of a catalyst - catalyst speeds up the reaction
Temperature – rate will increase with an increase in temperature (increased KE)
Surface area – as the surface area increases, the reaction will proceed at a faster rate
Rates of Chemical Reactions
Rates - the change in concentration of a chemical per unit of time ex. (M/sec)– much like the speed of a car ex. (miles/hr)– much like interest rates ex. (5.4%/year)– much like sales taxes ex. (7cents/$)
Rates are just a ratio or fraction of one thing changing with respect to another
Decomposition of N2O5
2 N2O5 4 NO2 + O2
We look at the disappearance of N2O5
Rate of Reaction = change in [N2O5] change in time
Rate of Reaction = [N2O5] t
Stoichiometry
2 N2O5 4 NO2 + O2
2moles : 4moles : 1mole
Therefore, for every 2 moles of N2O5 decomposed, you have 1 mole of O2
formed. The rate of formation of O2 is equal to ½ the rate of decomposition of N2O5
[O2]/t = -½ [N2O5]/ t
Decomposition of N2O5
Therefore we can determine the rate of decomposition by the rate of formation of either NO2 or the O2 with consideration to stoichiometry.
2 N2O5 4 NO2 + O2
Rf = [NO2] = -2Rd Rf = [O2] = -½ Rd
t t
Calculations
2 N2O5 4 NO2 + O2
Where ti = 600s and tf = 1200s and the concentration of N2O5 decomposed from 1.24 x 10-2 M to 0.93 x 10-2 M
Rate of decomposition = [N2O5 ] /t= (0.93 x 10-2 M -1.24 x 10-2 M)
(1200s – 600s) = -5.2 x 10-5 M/s
Rate of formation O2 = -½ [-5.2 x 10-5 M/s]= 2.6 x 10-5 M/s
Plot of Concentration vs. Time
Figure 15.2 on pg 702 shows the graph of the disappearance of N2O5.
Average rate = change in concentration over an interval in time (c/t)
Instantaneous rate = concentration at an instant in time; tangent to the curve at a particular point in time (dc/dt)
Rate Law
Rate Law – equation that relates the rate of reaction to the concentration of reactants raised to various powers
Rate = k [N2O5]x
– where k is the rate constant and x is the order of the reaction
Determining Rate Law
We can perform an experiment to decompose N2O5. If we doubled the concentration of the reactant we can observe the change in rate as follows:
Initial Conc. Rate of decompositionExp.1 1x10-2 mol/l 4.8x10-6 mol/l•secExp.2 2x10-2 mol/l 9.6x10-6 mol/l•sec
Observation: The rate of reaction doubled.
Determining Order
When a reaction has this observable result, it is said to be first order.
As we double the concentration, the rate is 2x, where x is the classification of the order of that particular reaction.
Or R2/R1 = 2x, where x=1, then we can see that it is clearly first order.
Overall Rate Law for N2O5
We previously stated that:
Rate = k [N2O5]x
Where x = 1 determined experimentally, therefore this reaction is a first order reaction and follows the rules of first order kinetics.
Order of Reactions
Considering other reactions where we double the initial reactant concentration, we can observe the rate and order as follows:
Quadruple rate = 4 (2x=4, x=2) second order
Double rate = 2 (2x=2, x=1) first order
No change = 1 (2x=1, x=0) zero order
Half rate = 1/2 (2x=1/2, x=-1) -1 order
1.4 the rate = 1.4 (2x=1.4, x=1/2) half order
Review of Logarithms
Logs are the inverse function of an exponential function (y = 3x)
Algebraically it is written as x = 3y or y = log3x where y is the exponent on 3 that results in a value of x– ex. 52 = 25 or log5 25 = 2
Written as a function logs are expressed as f(x) = logbx
If b is not designated, b=10 is assumed. If b = e it is called the natural log (ln).
First Order Kinetics
2 N2O5 4 NO2 + O2
Rate of decomposition = - [N2O5]
t Rate = kt
Therefore - [N2O5] = -kt (by substitution)
t ln [N2O5]t = -kt + ln [N2O5]0 (derive by integration)
Integrated Rate Equations
For a zero order reaction:
[A]t = -kt + [A]0
For a first order reaction:
ln [A]t = -kt + ln [A]0
For a second order reaction:
1/[A]t = kt + 1/[A]0
Iodine Clock Reaction Prep
Each solution is prepared in a 100ml volumetric flask and qs with distilled water. Accurately weigh the following:
2.998g of NaI = 0.2M 1.169g of NaCl = 0.2M 0.2482g of sodium thiosulfate = 0.01M 3.485g potassium sulfate = 0.2M 5.404g potassium persulfate = 0.2M
Half-life of Reaction
Half-life (t½) = the time it takes for the initial concentration of a reaction to disintegrate to half its original concentration
First order reaction: t½ = 0.693/k
Collision Theory
Three conditions must be met for a reaction to take place:– The reacting molecules must collide with
one another– Molecules must collide with sufficient
energy– Molecules must collide with the proper
orientation
Activation Energy
Ea = the minimum amount of energy that must be absorbed by a system to cause it to react where k = Ae -Ea/RT
Activation energy can be determined from the Arrhenius Equation:
ln k = ln A - Ea/R (1/T)
ln (k 2/k1) = - Ea/R [1/T2 - 1/T1]
Conditions Increasing Rates
Temperature - increases the KE and enough energy to overcome Ea
Presence of a catalyst - provides different pathways with lower Ea
Surface area - increases the probability of a collision with proper orientation
Reaction Mechanisms
Reaction mechanisms = sequence of steps that show the intermediates formed ( bonds broken and bonds formed) between the reactant side and product side of any reaction
See CD-ROM Screens 15.12 – 15.13
Rate Equations for Elementary Steps
Elementary Step - a singular molecular event classified by the # of reactant molecules involved (molecularity)– unimolecular: AP where R = k[A]– bimolecular: A + BP where R = k[A][B]
or A + AP where R = k[A]2
– termolecular: 2A +B P where R = k[A]2[B]
See Table at the bottom of page 734
Rate-determining Step
Step 1: A + B X + M; k1 is slow
Step 2: M + A Y; k2 is fast
Overall Rxn. 2A + B X + Y
Because step 2 is fast, it does not contribute to the overall rate, therefore:
R = k1[A][B]