CHEMICAL KINETICS Goal of kinetics experiment is to measure concentration of a species at particular...
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Transcript of CHEMICAL KINETICS Goal of kinetics experiment is to measure concentration of a species at particular...
CHEMICAL KINETICS
Goal of kinetics experiment is to measure concentration of a species at particular time during a rxn so a rate law can be determined
Rate of Rxn: describes how fast reactants used up & pdts formed
rates are obtained from concen. vs fct of time
Chem Kinetics: 1)study of rates, 2) factors that affect rxn rates, & 3) mechanisms (steps) by which rxns occur
From a chem eqn, rate can be determined by following the concenof any subst that is quantitatively detected
4 factors that affect chem rxns1) nature of reactants2) concen of reactants3) temp4) catalyst present
Write rate law for rxn to describe how rate depends on concen.
Order of rxn cannot be deduced from chemical eqn. of rxn
Rate law expressions - calculate rate of rxn from rate constant & reactant concen - convert into eqn to determine concen of reactants @ any time
Rate Law is deduced experimentally from how its rate varies w/ concen
Order of rxn cannot be deduced from chemical eqn. of rxn
exponents x & y- usually integers- value of x is the order of rxn w/ respect to A- y??
Values for k, x, & y have no relation to coeff of balanced chem eqn.,remember, must be determined experimentally
For rxn: A + B -----> pdts
general form: rate = k[A]x[B]y
Exponent Define 0 1 2
rate not depend on [reacts] rate is directly proportional to [reacts]rate is directly proportional to square of concen; [reacts]2
overall order of rxn = x + ySum of orders of reacts
order in NO: overall order:
- order in rate law may not match coeff. in balanced eqn- no way to predict rxn orders overall from balanced eqn- orders must be determined experimentally
Examples of observed rate laws for following rxns
3NO(g) ------> N2O(g) + NO2 (g) rate = k[NO]2
2NO2(g) + F2(g) ------> 2NO2F (g) rate = k[NO2][F2] order in NO2: order in F2:
overall order:
2nd 2nd
1st 1st
2nd
quick summary
rate law can be determined by 2 methods:
1) Method of Initial Rates (if time)
2) using Integrated Rate Eqn
ZERO ORDER Has a rate which is independent of concentration of reactant(s),therefore, increasing concen. of rxning species not speed up rate
Rate is:k
k[A] rate O
k
t
A-r
d
dRate is a CONSTANT
A -----> pdts
integration gives eqn called integrated zero-order rate law
[A] = -kt + [A]o
Initial concentrconcentr of chemical@ particular time
[A] = -kt + [A]o
eqn line: y = mx + b
time, t
[A]
calculate k from plot of graph;straight line plot of [A] vs time,t; slope = -k
Determine units:s
M
T
[A]- k
d
d
half-life describes time needed for half of reactant to be depleted
k2
A t O
21
FIRST ORDERDepends on concentration of only 1 reactant, if other reactants present buteach will be zero-order
eqn for first-order reaction A -----> pdts
1st order rate constant,units of 1/time
t
[A] rate
d
d
integration gives eqn called integrated first-order rate law
ln[A] = -kt + ln[A]o
rate is: [A] t
[A]- k
d
dknow: [A] rate k
t [A]
[A]ln O k
d
eqn line: y = m x + b
calculate k from plot of graph;plot of ln[A] vs time,t; gives straight lineslope = -k
time, t
[A]
time, t
ln[A]
Determine units:s
1
sM
M
t
1
[A]
[A]- k
dd
d
half-life describes time needed for half of reactant to be depleted
k
0.693
k
2Ln t
21
SECOND ORDERA. A ------ > pdts depends on concentration of 2nd-order reactant
second-order rate law
22 [A] t
]A[ ]A[rate k
d
dk
integrated in the form:
[A]O @ t = 0 & [A] @ t: [A]
1
]A[
1 t
t ]A[
]A[
O
2
k
d kd
B. or, A + B = pdts two 1st-order reactants:
k[A][B] t
]B[
t
]A[
d
d
d
d
rate is:
ln r = ln k + 2 ln[A]
Another way to represents rate laws, take ln of both sides:
Plot 1/[A] vs time,t; slope = 2nd-order rate constant; +k
[A]
1 t
]A[
1
O
k
eqn line: y = m x + b
time, t
1/[A]
half-life for 2nd order dependent on one 2nd order reactant: O21
Ak
1 t
Determine units:sM
1
sM
M
t
1
[A]
[A]- k
22
d
d
FIRST ORDER REACTION
2 N2O5 (aq) --------> 4 NO2 (aq) + O2 (g)
ln[N2O5] 1/[N2O5], M-1
DATA Time, s [N2O5], M
0 0.0365600 0.02741200 0.02061800 0.01572400 0.01173000 0.008603600 0.00640
- 3.310- 3.597- 3.882- 4.154- 4.448- 4.756- 5.051
27.4 36.5 48.5 63.7 85.5116156
time,s
[N2O
5]
time,s
ln[N
2O5]
time,s
1/[N
2O5]
rate = k[N2O5]
time,s
ln[N
2O5]
slope = /s10*4.820- s 3000
(-3.310) - (-4.756)
tt
]Aln[]Aln[ 4-
03000
03000
SECOND ORDER REACTION
2 NO2 (g) --------> 2 NO (g) + O2 (g)
ln[N2O5] 1/[N2O5], M-1
DATA Time, s [N2O5], M
060120180240300360
- 4.605- 4.986- 5.263- 5.477- 5.655- 5.806- 5.937
100146193239286332379
0.01000.006830.005180.004180.03500.003010.00264
time,s
[NO
2]
time,s
ln[N
O2]
time,s
1/[N
O2]
rate = k[NO2]2
time,s
1/[N
O2]
slope = sL/mol 0.773 s 300
(100) - (332)
tt
[A]1 ]A[
1
0300
0300
ZERO ORDER REACTION
Can occur if:1) rate limited by [catalyst]2) photochemical rxn if rate determined by light intensity3) most often occur when subst as a metal surface or enzyme required for rxn to occur
2 N2O (g) --------> 2 N2 (g) + O2 (g)
N2O N2O N2O
N2O N2O N2O
N2O N2O N2O
N2O N2O N2O
N2O N2ON2O
N2O N2O N2O
N2ON2O
N2O N2O N2O
N2O N2O N2O
N2O N2O N2O
N2O N2O N2O
Pt metal surface
N2O N2O N2O
N2O N2O N2O
N2O N2O N2O
N2O N2O N2O
N2O N2ON2O
N2O N2O N2O
N2ON2O
N2O N2O N2O
N2O N2O N2O
N2O N2O N2O
N2O N2O N2O
Describe what is happening
Rxn occurs on a hot Pt surface, when surface completelycovered w/ N2O molecules, an increase of [N2O] has noeffect on rate, since only N2O molecules on the surface are reacting.
Therefore, the rate is constant because rsn is controlledby what happens on Pt surface rather than total [N2O].
time,s
[N2O
]
rate = k[N2O]0
k2
A t O
21
Determine: UNITS HALF-LIFE
0 ORDER
1ST ORDER
2ND ORDER
sL
mol
T
[A]- k
d
d
O21
Ak
1 t
smol
L
sM
M
t
1
[A]
[A]- k
22
d
d
s
1
sM
M
t
1
[A]
[A]- k
dd
d
k
693. t
21
Summary for reaction orders 0, 1, 2, & n
Zero-Order First-Order Second-Order nth-Order
Rate Law
IntegratedRate Law
Units ofRateConstant (k)
Linear Plotto determineky-intercept
Half-life
k
t
A-
d
d[A]
t
[A]- k
d
d 2[A]
t
[A]- k
d
d n[A]
t
[A]- k
d
d
[A] = [A]O - kt[A] = [A]Oe-kt
ln[A] = ln[A]O - ktt
[A]
1
]A[
1
O
k
order]1st [Except
t1)-(n [A]
1
]A[
11-n
O1n
k
sL
mol
s
1s
Lmol
11n
smol
L
[A] vs t -k
ln[A] vs t -k
k-
t vs]A[
1
order]1st [Except
t vs]A[
11n
k2
A t O
21
k
693. t
21 O2
1 Ak
1 t
[A]O ln[A]OO]A[
1
NOTES
Rate Rxn- describe rate rxn must determine concen of react/pdt at various times as rxn proceeds- devising methods is challenge for chemists-spectroscopic method: if 1 subst colored measure inc/dec in intensity of color
4 Factors: help control rates
Comparing the 2 experiments, [B] is ed by factor of:
1 1.0 * 10-2 M 1.0 * 10-2 M 1.5 * 10-6 M.s-1
2 1.0 * 10-2 M 2.0 * 10-2 M 3.0 * 10-6 M.s-1
3 2.0 * 10-2 M 1.0 * 10-2 M 6.0 * 10-6 M.s-1
METHOD OF INITIAL RATES
describing same rxn in each experiment, same rate law, form: rate = k[A]x[B]y
Notice, [A]O same in #1 & #2, what would affect the rxn rate?
deduce rate law from experimental rate data
Experiment [A]O [B]O initial rate
es in rxn rate due to diff initial concen of B
ratio [B] 2.0 10*0.1
10*0.22
2
rate es by factor of:
What order is rxn order in [B]?
rate ratio = ([B])yExponent y deduced from:
ratio rate 2.0 10*5.1
10*0.36
6
2.0 = (2.0)y solving, y = 1
rate = k[A]x[B]1
Experiments 1 & 3 show [B]O same but [A]O different
ratio [A] 2.0 10*0.1
10*0.22
2
[A] is ed by factor of:
rate es by factor of: ratio rate 4.0 10*5.1
10*0.66
6
What order is rxn order in [A]?
rate ratio = ([A])xExponent x deduced from:
4.0 = (2.0)xsolving, x = 2
rate = k[A]2[B]1
Rate constant, k, substitute data from any set of 3 sets into rate-law expression
or, rate = 1.5 M-2.s-1[A]2[B]rate1 = k[A]12[B]1
1