Upload
suzanna-daniels
View
213
Download
1
Tags:
Embed Size (px)
Citation preview
Kinetics
3 OUT OF 75 M/C QUESTIONSFREE RESPONSE—ALMOST
EVERY YEAR
Chapter 12
Chemical Kinetics
Study ofReaction ratesReaction mechanism—series of steps by which a reaction takes place
Rate of Reaction the change in concentration of a
reactant or product per unit time
Rate = [A]t2 – [A]t1 / (t2 – t1)
OR Rate = D[A] / Dt Usually considered positive even if
concentration is decreasing.
The square brackets indicate concentration of a substance.
Example Problem
Nitrogen dioxide decomposes to nitrogen monoxide and oxygen gas. At the beginning of the reaction, a pressurized 1-L flask contains 2.0 mol of nitrogen dioxide. At the end of 20 minutes, the flask contains 1.5 mol of nitrogen dioxide and 0.5 mol of nitrogen monoxide. What is the rate of the reaction in terms of M/s of nitrogen dioxide?
Example Problem Solution• Important information:
2.0 mol/ L nitrogen dioxide initially After 20 minutes, 1.5 mol/L of nitrogen dioxide• Change in concentration: 2.0 mol – 1.5 mol = 0.5 mol• Change in time: 20 min• Rate: 0.5 mol / 20 min = 0.025 mol/min
Average vs. Instantaneous Rate
Rate is usually not constant, but decreases over time
Average rate—total change over total time (like the previous example)
Instantaneous rate—rate at a particular instant in time
Instantaneous Rate
Calculated by line’s slope tangent to the point in time on the rate curve Page
558
Think about a slope: change in "y” over change in “x”—It’s the same thing as the average rate if y is the concentration and x is the time.
Rate Law
A mathematical expression which describes the dependence of reaction rate on concentration of reactants
Rate = k[A]n
k = proportionality constant
This is a very important concept!
The [A] means concentration of substance A. The exponent “n” is known as the order of the reaction. The order must be determined experimentally from data.
The lower case k is the rate constant. It must be lower case because upper case K is an equilibrium constant.
Rate Law[A] = concentration of
reactant(s)n = order of reactantBoth k and n must be
experimentally determinedCannot determine from
balanced equation!!!
Write the rate law
2NO2 2NO + O2 if NO2 is second order
Rate = k [NO2]2
2H2 + 2NO N2 + 2H2O if NO is second order and H2 is first order
Rate law always begins with “rate = k….”
Next, put the formula for each reactant in brackets and raise it to the power of the order.
You don’t need to include zero order reactants since anything raised to the zero power is equal to one.
Combining Rate Laws
Rate = D[A] / DtRate = k[A]n
D[A] / Dt = k[A]n
More product terms can be added
Since both of these are equal to the rate, they are equal to each other.
Use this equation to solve for either rate, the rate constant or a concentration.
Example
Rate = k[A]n[B]m[C]p
(Greater value of exponent means greater effect on rate.)
To find exponents, determine what happens if concentration of a reactant doubles.
Determining orders of reactants
Find two experiments in which one concentration remains the same while the other changes.
Rate 2Rate 1
= (D concentration)n
Example Calculation
TrialInitial Concentration of
Reactants (M)[A] [B] [C]
Initial Rate
(M/min)
1 0.10 0.10 0.10 0.01
2 0.10 0.10 0.20 0.01
3 0.10 0.20 0.10 0.02
4 0.20 0.20 0.10 0.08
To determine orders of reactions, by method of initial rates: 1. Look at the data and find two trials
that start with the same concentrations in all but one reactant.
2. See trials 1 and 2. The concentration of reactant C was doubled while A and B stayed the same. There was no effect on the rate, so reactant C is zero order. Leave it out of the rate law.
Continued on next slide
Example Calculation
TrialInitial Concentration of
Reactants (M)[A] [B] [C]
Initial Rate
(M/min)
1 0.10 0.10 0.10 0.01
2 0.10 0.10 0.20 0.01
3 0.10 0.20 0.10 0.02
4 0.20 0.20 0.10 0.08
3. Compare trials 1 and 3. While the concentrations of A and C stayed the same, the concentration of B was doubled and the rate doubled. Since the effect was the same, B is a first order reactant. (2 x B = 2 x rate; 21 = 2)—The exponent is the order.
4. Compare trials 3 and 4. The concentrations of B and C are the same while the concentration of A was doubled. The rate increased by a factor of 4 so A is a second order reactant. (2 x A = 4 x rate; 22 = 4)—The exponent is the order.
5.Rate = k [A]2 [B]1
Example
Overall Rate:
Rate = k[A]2[B]1[C]0
Overall order of reaction:2 + 1 + 0 = 3
Applying Rate Laws
Once the rate law is established, use data to solve for k and to find rates at different conditions.
Example
Calculate the rate constant k for the previous reaction.
Rate = k[A]2[B]Trial
Initial Concentration of Reactants (M)
[A] [B] [C]
Initial Rate
(M/min)
1 0.10 0.10 0.10 0.01
ExampleSince [C] has no effect on rate, we
can leave it out when solving for k.
Rate = k[A]2[B]k = Rate / [A]2[B]k = .01M/sec / (.10M)2 (.10M)k = 10 M-2sec-1
To find the rate constant, k, choose any of the trials and substitute the concentrations and initial rate into the equation. Solve for k.
Pay attention to units and include them for k.
Write the Rate Law Equation
Look at data sets from different trials:
NH4+ + NO2
- N2 + 2 H2O
Experiment #
Initial Conc. Of
NH4+
Initial Conc. Of
NO2-
Initial Rate (mol/L *s)
1 0.100 M 0.0050 M 1.35 x 10-7
2 0.100 M 0.010 M 2.70 x 10-7
3 0.200 M 0.010 M 5.40 x 10-7
Answer
Both reactants are first order, so Rate = k [NH4
+]1[NO2-]1
Solve for k.
Answer
Rate = k [NH4+]1[NO2
-]1
Using information from trial #1:1.35 x 10-7 M/s = k (0.100 M)(0.0050
M)k = 2.7 x 10-4 M-1s-1
2 ClO2 + 2 OH- ClO3- +
H2O
[ClO2] in mol/L [OH-] in mol/L Initial Rate in M/s
0.0500 0.100 5.75 x 10-2
0.100 0.100 2.30 x 10-1
0.100 0.0500 1.15 x 10-1
Write the rate law equation and find the rate constant.
Answer
Rate = k [ClO2]2 [OH-]k = 230 M-2 s-1
Types of Rate Laws
Differential Rate Law—expression of how rate depends on concentration (previous problems)
Integrated Rate Law—expression of how concentration depends on time
When one is determined, the other can be calculated.
Integrated Rate LawShows how concentration of A depends on time
Usually given [A]0 and kEquation (whichever form) is in the form y = mx + b so it will be a straight line when graphed.
Calculating Integrated Rate Law—First Order
ln[A] = -kt + ln[A]0
ln = natural logarithmt = time[A] = conc. at time t[A]0 = conc. at time 0
Integrated Rate Law Equation
Equation can also be expressed as a ratio:
ln ([A]0 / [A]) = kt
Half LifeThe time required for a reactant to reach half its original concentration
t1/2
Calculating Half LifeWhen t = t1/2, then [A] = [A]0/2Use equation ln ([A]0 / [A]) = kt
ln ([A]0 / [A]0/2) = kt1/2
So…ln(2) = kt1/2
Notice, half life does not depend on concentration.
Calculating Integrated Rate Law—Second Order
Rate = k[A]2
Integrated Rate Law
1/[A] = kt + 1/[A]0
Integrated Second Order
Plot of 1/[A] versus time is a straight line with slope = k
Half life for second order:
t1/2 = 1/k[A]0
For 2nd order, half life does depend on initial concentration.
Zero Order Rate LawsOften encountered when a
reaction rate is limited because a surface (such as a platinum catalytic converter) is required.
When surface is covered, increasing concentration of a reactant has no effect.
If the reaction takes place on a surface, increasing concentration will not affect reaction rate.
Zero Order Rate LawsRate = k[A]0 = k(2) = k
Integrated Rate Law:
[A] = -kt + [A]0Half Life:
t1/2 = [A]0 / 2k
More Complex Rate LawsMany reactions have several
reactants—each affects rateTo study them, high concentrations of all but one reactant are used.
Highly concentrated reactants stay practically constant, so the remaining reactant can be studied.
Pseudo-First-OrderIf conc. of B & C are large,Rate = k[A]n[B]m[C]p =
k’[A]Calculate k’ and solve for k
Good Summary—Table 12.6
ImportanceHelps determine reaction mechanism
Find rate determining step in a series of reactions so total process can be sped up
Reaction MechanismsSum of elementary steps must give overall balanced equation
Must agree with experimentally determined rate law
Reaction MechanismsMost reactions are much
more complicated that they appear from their equations.
Intermediate—a species that is neither a reactant nor a product but that appears and then is consumed in the course of a reaction
Reaction MechanismsEach reaction is called an
elementary step.Rate of each elementary step can
be written from its molecularity.Molecularity = the number of
species that must collide to cause the reaction
MolecularityUnimolecular—depends on one
molecule; Rate = k[A]Bimolecular—depends on two
molecules; Rate = k[A]2 or Rate = k[A][B]
Termolecular—depends on 3 molecules; Rate = k[A]2[B] (etc)
Rate Determining StepThe slowest step in a
reaction mechanismDetermines overall rate much like pouring water through a funnel—limiting factor
Rate Law
Comes from the slow step and every step prior to it.
Only consider species that are in the overall reaction.
Only consider reactants.
Species Not in Overall Reaction
Intermediate—a species that is produced and then consumed in a reaction—appears in mechanism but not in overall reaction—appears as a product and then a reactant
Catalyst—a species that is used during one step of a mechanism but is reproduced later—appears as a reactant and then a product
Determining Mechanism See if rates of elementary
steps agree with observed rate law
If yes, it is an acceptable mechanism
Never proven—only possibly correct
Example Problem
2NO2 + F2 2NO2F
Rate = k[NO2][F2]Possible Mechanism??
NO2 + F2 NO2F + F (slow)
F + NO2 NO2F (fast)
ExampleOverall Reaction: I2 + H2 2HI
Step 1: I2 2I
Step 2: I + H2 H2I
Step 3: H2I + I 2HIDoes this give the overall equation?If step one is rate determining, what is
the rate law? Step 2?Identify any catalysts or intermediates.
Example2 NO2 (g) + F2 (g) 2NO2F (g)
Rate Law: Rate = k [NO2] [F2]Possible Mechanism:
NO2 + F2 NO2F + F (slow)
F + NO2 NO2F (fast)
Is this a possible mechanism?
Identify any catalysts or intermediates.
ExampleNO2(g) + CO(g) NO (g) + CO2(g)
Step 1: NO2 + NO2 NO3 + NO (slow)
Step 2: CO + NO3 NO2 + CO2 (fast)If the mechanism is correct, what is
the rate law?Identify any catalysts or intermediates.
ExampleWrite the rate law and overall reaction:
A + A + B C + D (fast)C + E D + B (slow)Identify any catalysts or intermediates.
Collision ModelMolecules must collide in order to react
Anything that increases frequency or energy or effectiveness of collisions increases reaction rate.
Effective CollisionsOnly a small fraction of collisions produce reactions.
Many ineffective collisions occur because energy is too low or orientation is wrong.
Some orientations for a collision between BrNO molecules. Orientations (a) and (b) can lead to a reaction, but (c) cannot.
Activation EnergyThe minimum amount (threshold) of energy that a system must have to produce a reaction
Using k to Calculate Activation Energy
k = zpe-Ea/RT
k = rate constantz = collision frequencyp = steric factor (fraction of
collisions with proper orientation)
e-Ea/RT = fraction of collisions with
enough energy to produce reaction
Arrhenius Equationk = Ae-Ea/RT
zp replaced by AA is the frequency factor for the
reactionEa is activation energyR is universal gas constant (8.3145)Taking ln of both sides gives a
y = mx +b form line equation
Arrhenius Equation
ln(k) = -Ea/RT+ lnA
In slope intercept form:Slope = Ea
Intercept = A
To Find Activation Energy
Measure rate constant (k) at several temps
Plot ln(k) versus 1/T
To Find Activation Energy
Use equation:
ln(k2/k1) = Ea/R (1/T1 – 1/T2)
Example Problem—p. 588
To Speed up a Reaction:
1. Increase temperature2. Increase pressure3. Increase concentration4. Increase surface area5. Add a catalyst
CatalystA substance that speeds up
a reaction without being consumed in the reaction
Enzymes—biological catalysts
Works by providing the reaction with a new pathway with lower activation energy
Effect of a CatalystNotice that energy difference between products and reactants is unchanged.
Effect of a Catalyst-
Heterogeneous Catalysis
Usually gaseous reactants adsorbed on a metal surface
Example—hydrogenation of ethylene
Breaking H-H bond requires lots of energy, but metal-H interactions weaken bond and allow bonds to break at lower energy
Homogeneous Catalysis
Catalyst and reactants exist in the same phases
Example: Catalysis of ozone by nitric oxide and freons (CCl2F2)