Download Chemical Kinetics - Neshaminy School District

Chemical Kinetics
• Rates of chemical reactions
• Mechanisms of chemical reactions
• Chemical reaction
• Particles involved can be ions, atoms and/or
molecules
Reaction Rate
• Measure of the # of moles of reactant, used up in a
volume of reaction mixture, per unit of time
• How much use in seconds
• Example
• Fast
– NaCl + AgNO3  white ppt. Almost instantly
• Slow
– Digestion of school cafeteria lunch
• Avg. rate = change in # moles of product/reactant
Dtime
•
= D (moles Product/reactant)
Dt
• Rates in terms of concentration
• A+BC+D
• Avg. rate = D[A]
Dt
• use [ ] to express conc. in M molarity
• = [A] final time – [A] initial time = M/s units
tf - ti
• Plot data on graph
• Graph curve used to determine instantaneous
rate at any point
• Instantaneous rate is the slope of the straight
line tangent that touches the curve at the point
of interest
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Rates and Stoichiometry
AgNO3 + NaCl  AgCl + NaNO3
Each reactant and product is 1:1 ratio
Therefore rate = D[AgNO3] = D[AgCl]
Dt
Dt
• Not all balanced equations are 1:1
•
2HI  H2 + I2
• In this case the disappearance of HI is twice
the rate of the appearance of both H2 and I2
• Rate = -1D[HI] = D[H2] = D[I2]
2 Dt
Dt
Dt
• General equation is
• aA + bB  cC + dD
• Rate= 1D[A] = 1D[B] = 1D[C] = 1D[D]
a Dt
b Dt
c Dt
d Dt
• Do following:
• 2N2O5  4NO2 + O2
• Rate of decomposition of N2O5 is
4.2 x 10-7M/s, what is the rate of appearance of
NO2 and O2
• 1D[NO2] = 1D[N2O5] = 4(4.2x10-7M/s) =
4 Dt
2Dt
2
• 8.4 x 10-7 M/s
• D[O2] = 1D[N2O5] = 1(4.2x10-7M/s) =
Dt
2Dt
2
• 2.1x10-7M/s
Dependence of rate on concentration
• Rates diminish as concentration of reactants is
decreased
• Rates generally increase when reactants
concentration increases
• Rates of reaction are dependent on the
concentrations of the reactants
• Called rate law
• Rate = k[reactant 1]m[reactant 2]n
• k is the rate constant
• Given a rate law for a reaction and its rate for a
set of reactant concentrations, the value of k at
a specific temp. and pressure can be calculated
• Magnitude of k provides valuable info about
reactions
• Exponents m and n are called reaction orders
• Sum of exponents is the overall reaction
order
• Reaction order exponents are determined
experimentally and do not necessarily correspond to
the coefficients of a balanced equation
• Most rate laws have reaction orders of 0, 1 or 2
• Some rate laws have fractional orders or even
negative
• Example: H2 + I2  2HI
• Rate = k[H2][I2]
• Since altering conc. of either reactant has same effect
on rate, they are both first order and have exponent
values of 1
• Overall reaction order is 1+1=2 ;second order
overall
• Rate law for a reaction must be determined
experimentally
• Observe the effect of changing initial concentration of
reactants on initial rate of a reaction
• Example: if changing the conc. of a reactant has no
effect on the reaction rate as long as some of the
reactant is present, it has a zero order
• If reaction is first order in a reactant, changes in conc.
of that reactant will produce proportional changes in
the rate
• Rate law second order, doubling conc. will increase
rate by a factor of 22=4
• Important to note; rate of reaction depends on conc.,
rate constant does not
• Determine rate law, magnitude of rate constant
and rate of reaction
• Given equation A + B  C
• Exp.# [A]M
[B]M Init.rate (M/s)
• 1
0.100
0.100
4.0x10-5
• 2
0.100
0.200
4.0x10-5
• 3
0.200
0.100
16.0x10-5
• Rate law: rate= k[A]m[B]n
• Need to solve for m and n
• Look at rate comparing [A] to [B]
• If [B] is changed(exp.2 doubled) it had no
effect([A] held constant) (remember only
change one variable at a time)
• Therefore rate law is zero order in B
• If [A] is doubled: rate increases fourfold
• Rate is proportional to [A]2
• Rate law of A is 2nd order
• Rate = k[A]2[B]0
• Can also find the values by taking the ratio of
the rates from two experiments
• Solving for n:
• Rate 2 = 4.0x10-5M/s =1
Rate 1 4.0x10-5M/s
• Using rate law:
• 1= rate 2 = k[0.100M]m[0.200M]n = [0.200]n =2n
rate 1 k[0.100M]m[0.100M]n [0.100]n
• 2n = 1 if n=0
• Solving for m:
• Rate 3 = 16.0 x 10-5M/s = 4
Rate 1
4.0 x 10-5M/s
• Using rate law
• 4 = rate3 = k[0.200M]m[0.100M]n = [0.200]m = 2m
rate1 k[0.100M]m[0.100M]n [0.100]m
• 2m = 4 so therefore m = 2
• Solving for magnitude of k
• k = rate = 4.0x10-5M/s = 4.0x10-3M-1s-1
[A]2 (0.100M)2
• Solving for rate of reaction if [A]=0.050M and
[B]=0.100M
• Rate = k[A]2 = (4.0x10-3M-1s-1)(.050M)2 =
• 1.0x10-5M/s
• Since B is not part of the rate law, it is immaterial to
the rate, provided there is at least some B to react
with A
Reaction rates
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Why necessary to understand?
Body uses chem. Reactions to stay healthy
Medicines
Industry
Curtail rust
Speed up production of products
5 Main Factors affecting rate
• 1. Nature of reactants
• In reactions bonds are rearranged
• Bond types affect rate due to electronegativities
• Slight e- arrangement, fast reaction at room
temp.
• Ex. Ionic double displacement
• Covalent bonds
• Large # of rearrangements
• Slow reactions at room temp.
• Ex. CH4 +2O2  2H2O + CO2
• 2. Concentration
• An increase conc. of one or more reactants
will usually but not always increase the
reaction rate if the mixture is homogenous
3. Temperature
• Increase temp. increase rate of most
reactions
• An increase in temp. of 100C will
approximately double reaction rate
• Due to collision theory
– More KE to change to PE
– More and faster collisions
– More chance for an Effective collision
–
Laser disc #9 Cpt. 30
4. Catalysts
• Substance which speeds up a reaction
without being permanently altered
• Lower the amount of energy (activation
energy) required for a reaction to occur
– Recoverable
– Ex. Enzymes, catalytic converter
• Inhibitors
– Chemicals that slow down reactions
5. Phase
• Reactants in same or more than one phase
• Homogenous
– Same phase
– Fast reaction rate
• Heterogeneous
– Reactants in different phases
• Slower reaction rate
– Due to reactants only being able to react at the interface site
(where meet)
• Pressure in gases
– Higher pressure forces closer-faster
– Low pressure further apart-slower
Reaction Mechanism
• Series of steps by which reacting particles
rearrange themselves to form the products
of a chemical reaction
• Hard to do
• Intermediate products often short lived
Effective Collision
• Based on collision theory
• Particles must collide for chem. reaction to
occur
• Ex.
–A+BC
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As A approaches B they begin to interact
e- shift positions
Old bonds are broken
New bonds are formed
Collision must be EFFECTIVE
• The reactants must approach each other at
Proper angle ( CORRECT
GEOMETRY) and with Sufficient energy
• Activation energy
• Minimum amount of energy required for a
reaction to occur
• Electron rearrangement needs energy
•
Laser disc #9 Cpt.20
Reaction mechanisms
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Series of steps
Reactants rearrange to form products
Most reactions follow a simple series of steps
Ex.
Chem.A + Chem.B  Chem.C
– Step 1
– Step 2
Reactants  Intermediate Prod
Intermediate 
Product
Net equation
• A + B  I1
• I1  C
• A + B  C
A +B C
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Step 1 A + B  I1
Step 2 I1 + A  I2
Step 3 I2  C
Net Eq 2A + B  C
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A collides with B to form intermediate product I1
I1 collides with another A particle to form I2
I2 rearranges into the final product C
When final equation is written it is in the net form
Done this way due to difficulty in actually
knowing what the intermediate is
• Intermediate products are highly unstable and
short lived
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Increase conc. usually increases rate
Not always true
Remember there are steps
Steps reactions occur at different rates
Slowest step is known as RATE
DETERMINING STEP
• To speed up reaction must increase conc. Of
reactants in rate determining step
•
Demo
• Example
2A + B + C  D
STEP 1
A + B  I1
STEP 2
I1 + A  I2 rate determining step
(slowest)
STEP 3
I 2 + C  I3
STEP 4
I3  D
Net Eq.
2A + B + C  D
• A + B have collided with an Effective collision
• The compound initially formed that has eundergoing rearrangement is called the Activated
Complex or Transition State Complex
– It is an intermediate
– Unstable and short lived
– 2 processes it can undergo
• Break down back into A + B
• Form product(s)
• Shown using a Potential Energy Diagram using
Activation energy
Thermodynamics
• Study of changes in energy in chemical
reactions and the influence of temperature
on those changes
• Every system or form of matter has stored
energy
• Energy in bonds
• P.E. due to phase, pressure and volume
• K.E. from random motion of particles
• Total of all these forms of energy referred to
as heat content
• Internal energy is symbolized as DU
• Heat of formation (DH, heat of reaction) is
known as ENTHALPY
• Change in energy when reactants form
products
• Enthalpy changes by amount of energy
gained or lost
• Energy given off: DH is neg.value
– Exothermic
• Energy absorbed: DH is pos.
– Endothermic
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Show in equation
• Expressed in kJ/mol
• Heat of formation depends on temp. and
press. at which reaction occurs and phase
• Standard Heat of Formation DHfo is at 250C
and 1 Atm.(101.325kPa)pressure
• Expressed as 298K and 101.3kPa plus phase
• Ex. Water liquid DHf0 = 286kJ/mol
H2(g) + 1/2O2(g) D H2O(l) + 286kJ
Since energy is released(part of product) DH is
neg. value ΔH= -286kJ
Elements in naturally found form have DHf0 of 0.
• Hess’s Law of Constant Heat Summation
– When a reaction can be expressed as the algebraic sum
of two or more other reactions, then the heat of the
reaction is the algebraic sum of the heats of these other
reaction
• If a reaction is carried out in a series of steps, DH
for the reaction will be equal to the sum of the
enthalpy changes for the individual steps
• Therefore we can calculate DH for a reaction by
“adding” up known DH of reactions
• Used to obtain energy changes that are difficult to
measure or predetermine
Ex.
CuO(s) + H2(g) g Cu(s) + H2O(g) DH=?
CuO(s) g
1/2O2 + Cu(s)
H2(g) + 1/2O2(g) g H2O(g)
DH = 155kJ
DH = -242kJ
• Calculate DH for :
• 2C(s) + H2(g)  C2H2(g) given the following:
• Correct mols and divide so function of coefficients mols
for wanted product is one
• C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(l) DH = -1299.6kJ
• C(s) + O2(g) 
CO2(g)
DH = -393.5kJ
• H2(g) + 1/2O2(g)  H2O(l)
DH = -285.9kJ
• Rearrange the equations so similar to algebra:
Entropy
• Measure of disorder, randomness or lack of
organization of a system
• Large entropy- great state of disorder
– Solid is organized pattern with low entropy and as
melts the particles become random and have higher
disorder of larger entropy; increases as change to gas
• Represented by DS
DS = Sf – Si
DS value pos. spontaneous change can occur
Gibbs Free Energy Equation
• Combines enthalpy and entropy changes to
determine spontaneity
• Neg.value spontaneous
• Pos. value non spontaneous
DG = DH - TDS