Chemical Kinetics Chapter 14 AP Chemistry. Chemical Kinetics Kinetics – the area of chemistry...
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Transcript of Chemical Kinetics Chapter 14 AP Chemistry. Chemical Kinetics Kinetics – the area of chemistry...
Chemical Kinetics
Chapter 14AP Chemistry
Chemical Kinetics
• Kinetics – the area of chemistry concerned with the rate (or speed) of a reaction.
• Kinetics vs. Thermodynamics
• Applications: Medicine, Chemical Engineering
Reaction Rate Factors
• Physical state of reactants– Surface area
• Concentration– Rate increases with concentration increase
• Temperature– Rate doubles every 10oC increase
• Catalyst– Increase the reaction rate w/o being used up
Reaction Rates
• Speed is the change in a particular quantity with respect to a change in time.
• In chemistry, we define the reaction rate– The change in concentration of the reactants or
products over time– Units are usually M/sec
– Rate =
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
• A graph of concentration vs. time is often plotted.
• Slope of the tangent line at any point along the curve is the instantaneous rate.
• Rate decreases with time.– Reactants decrease with
time
Average Rate
• Because the rate of reaction changes with time, it is useful to consider an average rate.
• Rate =
• The average rate for a reaction is usually take as the early stages of a reaction.
Measuring Rates
• To determine the progress of a reaction, we can measure two quantities:1. Disappearance of the reactant2. Formation of products
• Reaction rate is a positive value.• Reaction rate is the same, no matter the method of
measurement.
aA + bB cC + dD
• The rate of reaction is given by the following equalities:– A: Rate =
– B: Rate =
– C: Rate =
– D: Rate =
– Rate =
H2O2(g) H2(g) + O2 (g)
• Write the rate in terms of each species.
SO2(g) + O2(g) SO3(g)
• Write the rate in terms of each species.
H2(g) + O2(g) H2O(g)
• Hydrogen is burning at the rate of 0.85 M/sec. Rate of oxygen consumption? Rate of water vapor formation?
How?
• How is it possible to measure the concentration of reactants or products?
• There are a variety of methods.• One of the more common methods is
spectroscopy. • Measures the ability to absorb/transmit light
and converts the data to a concentration.
Spectrophotometer (Spec-20)
Beer-Lambert Law• There is a linear relationship between the
concentration of a sample and its absorbance.– A = -logT– Beer’s Law: A = εbC– Standards to find slope– Convert T to A to C
NH3(g) N2(g) + H2(g)
• Nitrogen is forming at the rate of 0.264 M/sec. Rate of ammonia consumption? Rate of hydrogen formation?
Rate Law
• General Equation: aA + bB cC + dD
• Rate =
• m is the order of A• n is the order of B• (m+n) is the overall reaction order• k is the rate constant – specific for a rxn, Temperature dependent
Rate = k[A]m[B]n
• The rate law must be experimentally determined.
• m and n are NOT the stoichiometric coefficient• Unit of rate constant k– M-p s-1 or 1/(Mp s )– p = (m+n) – 1
• Rate depends on reactant conc…k does not depend on reactant conc.
Rate = k[A][B]• What happens to the rate if we…
a) Double conc of A (everything else the same)?
b) Double conc of B (everything else the same)?
c) Triple conc of A and double conc of B?
d) Order of A? B? Overall?
Rate = k[A]2[B]• What happens to the rate if we…
a) Double conc of A (everything else the same)?
b) Double conc of B (everything else the same)?
c) Triple conc of A and double conc of B?
d) Order of A? B? Overall?
Rate = k[A]0[B]3
• What happens to the rate if we…
a) Double conc of A (everything else the same)?
b) Double conc of B (everything else the same)?
c) Triple conc of A and double conc of B?
d) Order of A? B? Overall?
Rate = k[A]m[B]n
• Two ways to determine the rate law1. Initial rate method– Can have many reactants
2. Graphical Method– Can only have one reactant
• Solving a rate Law:– Need to determine the orders of the reactants– Need to determine the rate constant k
• Determinethe rate law:
2 NO(g) + 2 H2(g) N3(g) + 2H2O(g) Experiment [NO] [H2] Initial Rate (M/s)
1 0.10 0.10 1.23 x 10-3
2 0.10 0.20 2.46 x 10-3
1 0.20 0.10 4.92 x 10-3
• Determinethe rate law:
a A(g) + b Bg) c C(g) + d D(g) Experiment [A] [B] Initial Rate (M/s)
1 0.40 0.30 1.00 x 10-4
2 0.80 0.30 4.00 x 10-4
1 0.80 0.60 1.60 x 10-3
• Determinethe rate law:
S2O82-
(aq) + 3 I1-(aq) 2 SO4
2-(aq) + I3
1-(aq)
Experiment [NO] [H2] Initial Rate (M/s) 1 0.080 0.034 2.2 x 10-4
2 0.080 0.017 1.1 x 10-4
1 0.160 0.017 2.2 x 10-4
Change in Conc. with time
• So far, we have considered rate based on the change in concentration and rate constants.
• Using calculus, we can convert these same equations to more useful forms.
• This is the graphing method to determining the order of a reaction.
• Specific for only one reactant: [A]
Rate Laws
• Differential Rate Law:
• Expressed how rate depends on concentration.
• Integrated Rate Law
• Integrated form of the differential. Has specific variables.
𝑅𝑎𝑡𝑒= −∆ሾ𝐴ሿ∆𝑡 = 𝑘[𝐴]
ln[𝐴]𝑡 = −𝑘𝑡+ ln[𝐴]0
Zero Order Reaction
• Rate only depends on the rate constant…not on the concentration of A
• Differential:
• Integrated:
First Order Reaction
• Rate only depends on the rate constant and on the concentration of A
• Differential:
• Integrated:
First Order
Plot of [A] vs. t Plot of ln[A] vs. t
Second Order Reaction
• Rate only depends on the rate constant and on the square concentration of A
• Differential:
• Integrated:
Second Order
Plot of ln[A] vs. t Plot of 1/[A] vs. t
Usefulness of the Integrated Rate Laws
[A]t = -kt + [A]0
• We can know the concentration at any time point for a given reaction.
• We can determine the order of a reaction.
• We can determine the half life of a reaction.
Determining the Order
[A]t = -kt + [A]0
• This tells us that a plot of concentration of A vs time will yield a straight line.
• Because this is the zero order rate equation, a plot of [A] vs. t will yield a straight line.
y = mx + b
Determining the Order
• First Order: ln[A]t = -kt + ln[A]0
– A plot of ln[A] vs. t will give a straight line.
• Second Order:
– A plot of 1/[A] vs. t will give a striaght line.
Half life, t1/2
• The time required for the concentration of a reactant to reach one-half its value: – [A]t1/2 = ½[A]0
• This is a convenient way to describe the rate of a reaction.• A fast reaction will have a short half life.
Derivation: t1/2 of First Order ln[A]t = -kt + ln[A]0
SummaryOrder Rate Law Integrated Rate Law Half Life
Zero Rate = k [A]t = -kt + [A]0 t1/2 = [𝐴]02𝑘
First Rate = k[A] ln[A]t = -kt + ln[A]0 t1/2 = 0.693𝑘
Second Rate = k[A]2 1[𝐴]𝑡 = 𝑘𝑡+ 1[𝐴]0 t1/2 = 1𝑘[𝐴]0
** Note: The half life of a first order says that it does NOT depend the concentration of the reactant A. So, the concentration decreases by ½ each regular time interval, t1/2.
A first order reaction has k = 6.7 x 10-4 s-1. • How long will it take for the conc to go from
0.25M to 0.15M?
• If the initial conc is 0.25M, what is the conc after 8.8 min?
A first order reaction has [A] = 2.00M initially. After 126 min, [A] = 0.0250M.
• What is the rate constant k?
• What is the half life?
A second order reaction has k = 7.0 x 10-9 M-1s-1. • If the initial conc is 0.086M, what is the conc
after 2.0 min?
A first order reaction has t1/2 = 35.0 sec.• What is the rate constant k?
• How long would it take for 95% decomposition of the reactant?
A first order reaction has a half life of 19.8 min. What is the reaction rate when [A] = 0.750M?
Temperature
Collision Model
• Based on Kinetic Molecular Theory
• Molecules must collide to react
• Greater the collisions, greater the rate
• As concentration increases, rate increases
• As temperature increases, rate increases
Orientation
• Most collisions do not lead to reactions• Molecular orientation of collision is important
Still not enough
• Usually, a collision in the correct orientation is still not enough to cause a reaction.
• Kinetic energy of a collision must cause bonds to break.
• For a reaction to occur, there must be enough kinetic energy to be greater than some energy.
• Activation Energy, Ea, is the minimum energy required to initiate a reaction.
Activation Energy
Transition State
• Transition state is also called the activated complex
• High energy intermediate state• The activation energy represents the higher
energy state of the transition state
A A + B BA A
B BA B + A B
‡
Arrhenius Equation• This equation combines all factors contributing to
the reaction rate.
k = A e−Ea/RT
• k = rate constant• Ea = activation energy• R = Gas constant• T = Temperature in Kelvin• A = frequency factor “constant”– probability of correctly oriented collisions
Other versions
• Original Equation: k = A e−Ea/RT
• Linear Equation:
• If we know the Ea and k2 at T2, we can calculate k1 at T1:
y = m * x + b
Catalysis
• Catalyst – speed up a reaction without being consumed
• Beneficial or harmful, ex. Body• Homogenous catalyst – same phase as reactant– Phase transfer catalyst
• Heterogeneous catalyst – different phase than reactant– Saturation of alkenes and alkynes
Catalyst
• Lowers the activation energy• Potential energy diagram is 3-D
Enzymes
• Biological Catalyst• Reactants called “substrate”• S + E SE P + E• Active site – binding location• SE = enzyme-substrate complex• Lock and Key Model
vs. Induced Fit Model
Reaction Mechanism
• Balanced rxn tells us the species before a rxn starts and after the rxn ends.– Does NOT show how the reaction occurs.
• Reaction Mechanism – the steps a reaction progresses.
• The steps can be of varying speeds
Elementary Reactions
• Reactions occur because of collisions– Must be correctly oriented– Must have enough energy to reach TS
A + B C + D This is a single collision reaction Elementary reactions – rxn occurring in a single
step
Molecularity
• Number of molecules that participate in a single reaction.
• Unimolecular, bimolecular, termolecular• Probability decreases with molecularity
A P2A PA + B PA + 2B P
Multiple Steps
• Some reactions occur in multiple steps• Multistep Mechanism – a reaction consisting
of a series of elementary reactions– Must add to give the overall reaction.– Items that are produced within a mechanism and
consumed within a mechanism are intermediates.– Intermediates are not R or P.
NO2(g) + CO(g) NO(g) + CO2(g)
Rate Laws
• Earlier we stated that rate laws can not be determined from a chemical equation.
• Why?– There is a possibility for multistep mechanism– We must consider the speed of each step
• However, we can derive the rate law from a the mechanism of a reaction.
It’s Elementary, my dear chemists!
• If a rxn is an elementary rxn, the rate law is based off the molecularity (coefficients)
• A P Rate = • 2 A P Rate = • A + B P Rate = • 3 A P Rate = • A + 2 B P Rate =• A + B + C P Rate =
Rate-Determining Step (RDS)
• Let’s go shopping.• Mechanisms can have a slow step.• RDS = slowest step – governs rate law• If first step is slow, intermediates short lived• If later step is slow, there is a buildup if
intermediates.• Each step has its own rate constant
Slow Initial Step
NO2(g) + CO(g) NO(g) + CO2(g)
Experimentally Determined Rate Law: Rate = k[NO2]2[CO]0
Fast Initial Step
• The slow step rate law still governs the rate law as before.
• However, because the slow step is the second step, there are intermediates in the rate law.
• Intermediates are short lived and can not be measured easily – not good in rate law
Fast Initial Step
• Make assumptions: there is a dynamic equilibrium established in the fast step– The intermediate is more likely to decompose
(reverse of fast step is also fast = k-1) than be consumed in step 2 (k2)
– The forward rxn rate in step one (k1) equals the rate of the reverse reaction: Rate forward = Rate Reverse
2 NO(g) + Br2(g0 2 NOBr(g)
• Experimental Rate Law: Rate = k[NO]2[Br2]