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4D-3 (of 21)
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CHEMICAL KINETICS
H2S (g) + Zn2+ (aq) ⇆ ZnS (s) + 2H+
(aq)
Chemical reactions can be viewed from different perspectives
4D-1 (of 21)
STOICHIOMETRYDescribes relationships based on conservation of atoms – predicts reaction quantities
THERMODYNAMICSDescribes energy and entropy changes – predicts if a reaction will occur
KINETICSDescribes how a reaction occurs – predicts the speed of a reaction
THERMODYNAMICS
KINETICS
CHEMICAL KINETICS 1 - Describes the speed of a chemical reaction and the factors that affect it2 - Determines the mechanism of a chemical reaction at the molecular level
COLLISION THEORY1 - Molecules must collide to react2 - Molecules must collide with sufficient energy to react (to break bonds)3 - Molecules must collide in the proper orientation
4D-2 (of 21)
4D-3 (of 21)
Measured by how fast a reactant reacts away, or how fast a product is produced
Rate = - Δ[reactant]_________________
Δt
= - d[reactant] _________________
dtRate = + Δ[product]
_________________
Δt
= + d[product] _________________
dt
REACTION RATES
4D-4 (of 21)
2N2O5 (g) → 4NO2 (g) + O2 (g)For the reaction
0.30 M/min- d[N2O5] =_____________
dtFind the reaction rate with respect to NO2 and O2
0.30 M N2O5_______________
min
x 4 M NO2 ____________
2 M N2O5
= 0.60 M NO2/min+ d[NO2] =____________
dt0.30 M N2O5_______________
min
x 1 M O2 ____________
2 M N2O5
= 0.15 M O2/min+ d[O2] =____________
dt
4D-5 (of 21)
Fastest Reaction Rates
Reaction rates are the slope of the tangent line (at any particular point) of a concentration vs. reaction time graph
0 Reaction Rate(equilibrium)
-0.30 M/min
+0.60 M/min+0.15 M/min
Concentration as a Function of Reaction Time
4D-6 (of 21)
Experimentally, rates of reactions are proportional to(1) Temperature(2) Concentration of reacting molecules
RATE LAW – An algebraic expression that relates the rate of a reaction (how fast a reactant disappears or how fast a product appears) to the concentrations of the reactants and the temperature
4D-7 (of 21)
2N2O5 (g) → 4NO2 (g) + O2 (g)
Rate T [N2O5 ]
Rate = k [N2O5 ]
SPECIFIC RATE CONSTANT (k) – The rate law constant that depends on temperature
FIRST-ORDER REACTION – One in which the rate is proportional to the concentration of reactants to the first power
4D-8 (of 21)
2NO2 (g) → 2NO (g) + O2 (g)
Rate = k [NO2 ]2
This is a SECOND-ORDER REACTION
H2 (g) + I2 (g) → 2HI (g)
Rate = k [H2 ] [I2 ]
1st order in H2
1st order in I2
2nd order overall
Rate laws can only be determined experimentally
4D-9 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/s)0.0400.0100.005
[NO]1.000.500.50
[Cl2]1.001.000.50
R = k [NO]x [Cl2]y
Choose 2 trials where [Cl2] is constant
0.040 = k [1.00]x [1.00]y
________________________________
0.010 = k [0.50]x [1.00]y
4 = 2x 2 = x
4D-10 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/s)0.0400.0100.005
[NO]1.000.500.50
[Cl2]1.001.000.50
R = k [NO]x [Cl2]y
Choose 2 trials where [NO] is constant
0.010 = k [0.50]2 [1.00]y
________________________________
0.005 = k [0.50]2 [0.50]y
2 = 2y 1 = y
4D-11 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/s)0.0400.0100.005
[NO]1.000.500.50
[Cl2]1.001.000.50
R = k [NO]x [Cl2]y
R = k [NO]2 [Cl2]1
The reaction is 2nd order in NO, 1st order in Cl2, and 3rd order overall
4D-12 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/s)0.0400.0800.640
[HCl]3.06.06.0
[NO2]1.01.02.0
R = k [HCl]x [NO2]y
Choose 2 trials where [NO2] is constant
0.080 = k [6.0]x [1.0]y
____________________________
0.040 = k [3.0]x [1.0]y
2 = 2x 1 = x
4D-13 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/s)0.0400.0800.640
[HCl]3.06.06.0
[NO2]1.01.02.0
R = k [HCl]x [NO2]y
Choose 2 trials where [HCl] is constant
0.640 = k [6.0]1 [2.0]y
____________________________
0.080 = k [6.0]1 [1.0]y
8 = 2y 3 = y
4D-14 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/s)0.0400.0800.640
[HCl]3.06.06.0
[NO2]1.01.02.0
R = k [HCl]x [NO2]y
R = k [HCl]1 [NO2]3
The reaction is 1st order in HCl, 3rd order in NO2, and 4th order overall
4D-15 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/s)0.0400.0800.640
[HCl]3.06.06.0
[NO2]1.01.02.0
R = k [HCl]x [NO2]y
R = k [HCl]1 [NO2]3
Find the value of k, with its units
R = k ________________
[HCl]1 [NO2]3
= 0.040 M/s ____________________
(3.0 M)1 (1.0 M)3
= 0.013 M-3s-1
4D-16 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/s)0.0400.0800.640
[HCl]3.06.06.0
[NO2]1.01.02.0
R = k [HCl]x [NO2]y
R = k [HCl]1 [NO2]3
Calculate the rate of the reaction when [HCl] = 1.0 M and [NO2] = 3.0 M
R = k [HCl]1 [NO2]3 = (0.0133 M-3s-1) (1.0 M)1 (3.0 M)3 = 0.36 Ms-1
4D-17 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/min)0.090.181.08
[F2]0.150.300.60
[Cl2]0.200.200.60
R = k [F2]x [Cl2]y
Choose 2 trials where [Cl2] is constant
0.18 = k [0.30]x [0.20]y
_____________________________
0.09 = k [0.15]x [0.20]y
2 = 2x 1 = x
4D-18 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/min)0.090.181.08
[F2]0.150.300.60
[Cl2]0.200.200.60
R = k [F2]x [Cl2]y
Choose 2 trials where [F2] is constant
1.08 = k [0.60]1 [0.60]y
_____________________________
0.18 = k [0.30]1 [0.20]y
6 = (2) 3y
1 = y
???????
3 = 3y
4D-19 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/min)0.090.181.08
[F2]0.150.300.60
[Cl2]0.200.200.60
R = k [F2]x [Cl2]y
R = k [F2]1 [Cl2]1
Find the value of k, with its units
R = k _____________
[F2]1 [Cl2]1
= 0.18 M/min ________________________
(0.30 M)1 (0.20 M)1
= 3.0 M-1min-1
4D-20 (of 21)
Find the rate law given the following experimental data
Initial Rate (M/min)0.090.181.08
[F2]0.150.300.60
[Cl2]0.200.200.60
R = k [F2]x [Cl2]y
R = k [F2]1 [Cl2]1
Calculate the rate of the reaction when [F2] = 0.20 M and [Cl2] = 0.40 M
R = k [F2]1 [Cl2]1 = (3.0 M-1min-1) (0.20 M)1 (0.40 M)1 = 0.24 Mmin-1
4D-21 (of 21)
2N2O (g) → 2N2 (g) + O2 (g)
Rate = k [N2O ]0
This is a ZERO-ORDER REACTION
Rate = k The rate law for this reaction is:
, or 0º
4E-1 (of 18)
-d[N2O] = k_________
dtd[N2O] = – k dt∫
0
t∫
0
t
[N2O]t – [N2O]0 = – kt – (– k0)
[N2O]t – [N2O]0 = – kt
[N2O]t = [N2O]0 – kt
[N2O]t = -kt + [N2O]0
To plot a linear graph: y = mx + b
y = [N2O]t
m = -kx = tb = [N2O]0
4E-2 (of 18)
For all reactions with zero-order kinetics:
[X]t = -kt + [X]o
a plot of [reactant]t vs. t will yield a line
N2O5 (g) → N2O (g) + 2O2 (g)
This is a FIRST-ORDER REACTION
Rate = k[N2O5 ]1The rate law for this reaction is:
, or 1º:
4E-3 (of 18)
-d[N2O5] = k[N2O5]_________
dt
d[N2O5] = – k dt ________
[N2O5]∫
0
t∫
0
t
ln[N2O5]t – ln[N2O5]0 = – kt – (– k0)
ln[N2O5]t – ln[N2O5]0 = – kt
ln[N2O5]t = ln[N2O5]0 – kt
To plot a linear graph:
y = ln[N2O5]t
m = -kx = tb = ln[N2O5]o
ln[N2O5]t = -kt + ln[N2O5]o
4E-4 (of 18)
For all reactions with first-order kinetics:
ln[X]t = -kt + ln[X]o
a plot of ln[reactant]t vs. t will yield a line
For the 1º decomposition of N2O5, k = 0.0124 s-1. Calculate the molarity of N2O5 remaining from 10.0 M N2O5 after 100. seconds
[N2O5]t = [N2O5]oe-kt = (10.0 M) e –(0.0124 s-1)(100. s) = 2.89 M
4E-5 (of 18)
This is the same as radioactive decay, which follows first order kinetics
First order kinetics can also be written as:
[X]t = [X]oe-kt
2HI (g) → H2 (g) + I2 (g)
This is a SECOND-ORDER REACTION
Rate = k[HI ]2The rate law for this reaction is:
, or 2º
4E-6 (of 18)
-d[HI] = k[HI]2
_______
dtd[HI] = – k dt______
[HI]2∫
0
t∫
0
t
–1 – –1 = – kt _____ _____
[HI]t [HI]0
–1 = –1 – kt _____ _____
[HI]t [HI]0
1 = kt + 1 ____ _____
[HI]t [HI]o
To plot a linear graph:
y = 1/[HI]t
m = kx = tb = 1/[HI]o
4E-7 (of 18)
For all reactions with second-order kinetics:
1 = kt + 1 ____ ____
[X]t [X]o
a plot of 1/[reactant]t vs. t will yield a line
Order
0
1
2
Rate Law
R = k
R = k [X]1
R = k [X]2
Equality
[X]t = -kt + [X]o
ln [X]t = -kt + ln [X]o
1 = kt + 1____ ____
[X]t [X]o
Linear Plot
[X] vs. t
ln [X] vs. t
1vs. t
____
[X]
4E-8 (of 18)
Find the rate law for the reaction A → B given the following:
[A] (M) :Time (s) :
0.25000.00
0.12505.00
0.062515.00
R = k [A]X
4E-9 (of 18)
Find the rate law for the reaction A → B given the following:
[A] (M) :Time (s) :
0.25000.00
0.12505.00
0.062515.00
Test for 0ºLinear plot would be [A] vs. tIf linear, the slope calculated with any 2 points will be constant
0.2500 M – 0.1250 M____________________________
0.00 s – 5.00 s
= -0.0250 Ms-1
0.1250 M – 0.0625 M____________________________
5.00 s – 15.00 s
= -0.00625 Ms-1
Slopes are not constant, not 0º
4E-10 (of 18)
Find the rate law for the reaction A → B given the following:
[A] (M) :Time (s) :
0.25000.00
0.12505.00
0.062515.00
Test for 1ºLinear plot would be ln[A] vs. t
ln (0.2500 M) – ln (0.1250 M)____________________________________
0.00 s – 5.00 s
= -0.139 s-1
ln (0.1250 M) – ln (0.0625 M)____________________________________
5.00 s – 15.00 s
= -0.0693 s-1
Slopes are not constant, not 1º
4E-11 (of 18)
Find the rate law for the reaction A → B given the following:
[A] (M) :Time (s) :
0.25000.00
0.12505.00
0.062515.00
Test for 2ºLinear plot would be 1/[A] vs. t
(1/0.2500 M) – (1/0.1250 M)____________________________________
0.00 s – 5.00 s
= 0.800 M-1s-1
(1/0.1250 M) – (1/0.0625 M)____________________________________
5.00 s – 15.00 s
= 0.800 M-1s-1
Slopes are constant, the reaction is 2º R = k [A]2
4E-12 (of 18)
REACTION MECHANISMS
Chemical reactions occur as a specific series of collisionsand each collision is considered a STEP
Each step has a MOLECULARITYa) If 1 molecule decomposes, the step is UNIMOLECULARb) If 2 molecules collide to react, the step is BIMOLECULARc) If 3 molecules collide to react, the step is TRIMOLECULAR (rare)
4E-13 (of 18)
ELEMENTARY REACTION – A reaction the occurs in only one step (or one collision)
REACTION MECHANISM – The series of steps that yield the balanced chemical reaction
RATE DETERMINING STEP (or RATE LIMITING STEP) – The slowest step in the reaction mechanism
The reactants in a reaction’s rate law are the reactants in the rate determining step
4E-14 (of 18)
2NO (g) + 2H2 (g) → N2 (g) + 2H2O (g)
This reaction occurs via a 3 step mechanism:
(1) 2NO ⇆ N2O2 (fast equilibrium)
(2) N2O2 + H2 → N2O + H2O (slow)
(3) N2O + H2 → N2 + H2O (fast)
R = k2 [N2O2] [H2]
This is not a reactant in the reaction, it is a REACTION INTERMEDIATE
K1
k2
k3
[N2O2] must be substituted out
4E-15 (of 18)
2NO (g) + 2H2 (g) → N2 (g) + 2H2O (g)
This reaction occurs via a 3 step mechanism:
(1) 2NO ⇆ N2O2 (fast equilibrium)
(2) N2O2 + H2 → N2O + H2O (slow)
(3) N2O + H2 → N2 + H2O (fast)
R = k2 [N2O2] [H2]
K1 = [N2O2] ________
[NO]2
K1
k2
k3
R = k2 K1 [NO]2 [H2]
K1 [NO]2 = [N2O2]
R = k [NO]2 [H2]
4E-16 (of 18)
CHCl3 (g) + Cl2 (g) → CCl4 (g) + HCl (g)
This reaction occurs via a 3 step mechanism:
(1) Cl2 ⇆ 2Cl (fast equilibrium)
(2) CHCl3 + Cl → CCl3 + HCl (slow)
(3) CCl3 + Cl → CCl4 (fast)
R = k2 [CHCl3] [Cl]
K1
k2
k3
K1 = [Cl]2
______
[Cl2]
R = k2 [CHCl3] K1½
[Cl2]½
K1 [Cl2] = [Cl]2
R = k [CHCl3] [Cl2]½
K1½
[Cl2]½ = [Cl]
4E-17 (of 18)
H2 (g) + l2 (g) → 2HI (g)
This reaction occurs via a 3 step mechanism:
(1) l2 ⇆ 2l (fast equilibrium)
(2) H2 + l ⇆ H2l (fast equilibrium)
(3) H2I + l → 2HI (slow)
R = k3 [H2I] [l]
K1
K2
k3
K2 = [H2I] _______
[H2][I]
K2 [H2] [I] = [H2I]
R = k3 K2 [H2] [l] [I]
K1 = [I]2
____
[I2]
K1 [I2] = [I]2
R = k3 K2 [H2] K1 [I2]
R = k [H2] [I2]
4E-18 (of 18)
ACTIVATION ENERGY (Ea) – The minimum energy needed by the reacting molecules for an effective collision
Reactants
Products
ΔH
Ea
4F-1 (of 12)
On the microscopic level, rates of reactions depend on: (1) The collision frequency of the reacting molecules (2) The fraction of collisions that have the proper orientation(3) The fraction of collisions that have the activation energy
On the macroscopic level, rates of reactions depend on: (1) Concentrations of reactants(2) Temperature
CALCULATING THE ACTIVATION ENERGY
4F-2 (of 12)
1889 SVANTE ARRHENIUSProposed that the specific rate constant, k, is the product of the collision frequency, the fraction of collisions with the proper orientation, and the fraction of collisions with the activation energy
k = zpe-Ea/RT
z = collision frequencyp = fraction of collisions with the proper orientation
e-Ea/RT= fraction of the collisions with the activation energy
A is the PRE-EXPONENTIAL FACTOREa is the ARRHENIUS ACTIVATION ENERGY
k = Ae-Ea/RT
4F-3 (of 12)
The activation energy can be determined graphically by knowing specific rate constants at different temperatures
k = Ae-Ea/RT
ln k = ln A - Ea ____
RT ln k = -Ea + ln A _____
RT
4F-4 (of 12)
ln k = -Ea 1 + ln A _____ ___
R T
The activation energy can be determined graphically by knowing specific rate constants at different temperatures
y = ln km = -Ea/Rx = 1/Tb = ln A
4F-5 (of 12)
ln k = -Ea 1 + ln A _____ ___
R T
The activation energy can be determined graphically by knowing specific rate constants at different temperatures
4F-6 (of 12)
-Ea = m _____
R Ea = -Rm
Ea = -(8.314 J/K)(-3413.5 K)
= 28,400 J
Reactants
Products
ΔH
Ea
ΔH
Ea
ExothermicNegative ΔH
EndothermicPositive ΔH
4F-7 (of 12)
Reactants
Products
REACTION ENERGY PROFILES
Reactants
Products
Enthalpy Change (ΔH)Forward Reaction Ea (Ea-for)Reverse Reaction Ea (Ea-rev)
Ea-for – Ea-rev = ΔH
4F-8 (of 12)
Reactants
Products
At the highest point of the graph, the reactants go through a high energy state called the TRANSITION STATEAt the transition state, the colliding molecules form a single unit called the ACTIVATED COMPLEX (ǂ)
ǂ
4F-9 (of 12)
The Activated ComplexA single unit in which old bonds are breaking and new bonds are forming
4F-10 (of 12)
CATALYSISIncreases the rate of a reaction by allowing the reaction to take place via a different pathway with a lower activation energy
EaEa
A catalyst brings a reaction to equilibrium faster, it does not change the equilibrium concentrations
4F-11 (of 12)
N2 (g) + 3H2 (g) → 2NH3 (g)
Metal surfaces can act as catalysts for gaseous reactions
4F-12 (of 12)