Download + NO 2

KINETICS CHEMISTRY—
INTRODUCTION
KINETICS
• STUDY OF REACTION RATES
• HOW FAST DOES IT HAPPEN? WHAT VARIABLES INFLUENCE THE RATE? WHAT IS THE PATH
THE REACTION TAKES TO CONVERT REACTANTS TO PRODUCTS?
• RATE
• CHANGE IN AN AMOUNT OVER A PERIOD OF TIME
• EX. DISTANCE TRAVELED, SPACE TRAVEL
REACTION RATES
• CHANGE IN THE CONCENTRATION OF A CHEMICAL COMPOUND IN THE REACTION OVER A
PERIOD OF TIME
• FOCUS ON ONE REACTANT OR ONE PRODUCT IN THE REACTION
• WANT TO KNOW RATE OF DISAPPEARANCE FOR REACTANT/RATE OF APPEARANCE FOR A
PRODUCT
• RATE = ∆X
∆T
UNITS = M/TIME
EXAMPLE 1: A + B → D + E
• A) RATE OF DISAPPEARANCE FOR A
• RATEA = -∆[A]
∆T
• B) RATE OF APPEARANCE FOR D
• RATED = ∆[D]
∆T
EXAMPLE 2: 2AB → A2 + B2
Time = 0
seconds
15.0 M
0M
0M
Time= 60
seconds
5.0 M
5.00M
5.00M
• DETERMINE THE RATE OF DISAPPEARANCE OF AB
• DETERMINE THE RATE OF APPEARANCE OF A2 AND B2
REACTION RATE AT ANY MOMENT IN TIME
• SLOPE OF LINE TANGENT TO THE REACTION CURVE
COLLISION THEORY
• REACTANT MOLECULES MUST COLLIDE TO PRODUCE A CHEMICAL
REACTION
• THE CONCENTRATIONS OF REACTANTS AFFECT THE # OF COLLISIONS
AMONG REACTANTS
• FOR REACTIONS OCCURRING IN ONE STEP—RATE OF REACTION IS
PROPORTIONAL TO PRODUCT OF REACTANT CONCENTRATIONS
• RATE = K[A] [B]
• RATE OF ANY REACTION STEP DEPENDENT ON COLLISION FREQUENCY
VARIABLES AFFECTING REACTION RATE
1) COLLISION RATES BETWEEN REACTANTS
2) % OF COLLISIONS WITH REACTANTS ARRANGED IN PROPER
ORIENTATION TO PRODUCE REACTION.
3) % OF COLLISIONS WITH ENERGY ENERGY (ACTIVATION ENERGY)
TO PRODUCE REACTION.
WHEN DO COLLISION RATES INCREASE?
• INCREASE IN CONCENTRATIONS OF REACTANTS
• TEMPERATURE INCREASES
•WHY?
MOST COLLISIONS DO NOT RESULT IN A
CHEMICAL REACTION!
• SMALL PERCENTAGE OF COLLISIONS ACTUALLY CONVERT REACTANTS TO
PRODUCTS. WHY?
1)
MOLECULAR ORIENTATION
• RANDOM ORIENTATION
• NOT ALL COLLISIONS HAVE CORRECT ORIENTATION
2)
MOLECULAR ENERGY AT COLLISION
• MOLECULES HAVE DIFFERENT KINETIC ENERGIES
• COLLISION ENERGY IS ENERGY SOURCE TO GET A REACTION STARTED
ACTIVATION ENERGY (EA)
• THE AMOUNT OF COLLISION ENERGY NEEDED TO OVERCOME EA SO
THE REACTION CAN OCCUR
• AMOUNT OF ENERGY NEEDED FOR A CHEMICAL REACTION TO
HAPPEN, ENERGY NEEDED TO CONVERT REACTANTS TO PRODUCTS.
ACTIVATION ENERGY--ENDOTHERMIC
WHEN WILL REACTIONS OCCUR?
1) MUST HAVE A COLLISION
2) COLLISION MUST HAPPEN WITH THE CORRECT MOLECULAR
ORIENTATION TO GENERATE A REACTION
3) COLLISION ENERGY ≥ EA
ARRHENIUS EQUATION
• RATE CONSTANT AND REACTION RATE ARE TEMPERATURE DEPENDENT.
• ENABLES THE ACTIVATION ENERGY FOR A REACTION TO BE
DETERMINED BASED ON THE RELATIONSHIP BETWEEN REACTION RATE
AND TEMPERATURE.
ARRHENIUS EQUATION
•LNK = -EA ( 1/T ) + LNA
R
•
•
•
•
•
K = RATE CONSTANT
EA = ACTIVATION ENERGY (J)
R = 8.314 J/MOLK
T = KELVIN
Z = PROPORTIONALITY CONSTANT, CHANGES BASED ON REACTION
ARRHENIUS EQUATION
• DIFFERENT FORM OF EQUATION CAN BE USED TO OBSERVE HOW
TEMPERATURE CHANGES AFFECT THE RATE CONSTANT (K)
•LN (K1/K2) = EA (1/T2 – 1/T1)
R
EXAMPLE 1
• CALCULATE ACTIVATION ENERGY (EA) FOR HI DECOMPOSITION WITH
THE FOLLOWING DATA.
Temperature (K)
Rate Constant (M/s)
573
2.91 x 10-6
673
8.38 x 10-4
773
7.65 x 10-2
COLLISION THEORY
• IN ORDER FOR TWO PARTICLES TO REACT CHEMICALLY, THEY MUST
COLLIDE. NOT ONLY MUST THEY COLLIDE, BUT IT MUST BE AN “EFFECTIVE
COLLISION.” THAT IS, THEY MUST HAVE THE CORRECT AMOUNT OF
ENERGY AND COLLIDE WITH THE PROPER ORIENTATION IN SPACE.
• ANY FACTOR WHICH INCREASES THE LIKELIHOOD THAT THEY WILL
COLLIDE WILL INCREASE THE RATE OF THE CHEMICAL REACTION.
Presented by Mark Langella, PWISTA.com
FACTORS WHICH AFFECT THE RATE OF A
CHEMICAL REACTION (BONDS MUST BREAK)
• SURFACE AREA/ CONTACT AREA (OPPORTUNITY FOR COLLISIONS)
• CONCENTRATION ( INCREASE FREQUENCY)
• TEMPERATURE ( INCREASE FREQUENCY)
• CATALYST ( EFFECTIVE COLLISIONS)
• NATURE OF REACTANTS ( EFFECTIVE COLLISIONS)
Presented by Mark Langella, PWISTA.com
WHAT CAN INFLUENCE REACTION RATES?
1) TEMPERATURE
2) CONCENTRATION
3) CATALYST
4) SURFACE AREA
5) VOLUME/PRESSURE
6) REACTANT PROPERTIES
12_300
T1
T2 > T1
T2
0
0
Ea
Energy
Plot showing the number of collisions with a particular
energy at T1& T2, where T2 > T1 -- Boltzman Distribution.
Presented by Mark Langella, PWISTA.com
RATE LAW (CONT.)
• A+B→ C+D
• RATE = K [A]M[B]N
• RATE = RATE OF DISAPPEARANCE OF REACTANTS
• K = RATE CONSTANT, SPECIFIC TO REACTIONS AND TEMPERATURE
• M = REACTION ORDER IN TERMS OF A
• N = REACTION ORDER IN TERMS OF B
• M + N = OVERALL REACTION ORDER
REACTION ORDER
• INDICATES HOW CONCENTRATION CHANGES AFFECT CHANGES IN THE REACTION RATE
• ORDERS: 0, 1, 2
• OVERALL ORDER OF REACTION = Σ INDIVIDUAL ORDERS OF EACH REACTANT
• ORDER OF A REACTION IN TERMS OF A REACTANT ≠ REACTANT’S COEFFICIENT IN CHEMICAL
EQUATION
REACTION ORDERS (CONT.)
• ZERO-ORDER REACTION
• RATE NOT DEPENDENT ON REACTANT’S CONCENTRATION
• CONSTANT REACTION RATE
• FIRST-ORDER REACTION
• CONCENTRATE AFFECTS REACTION RATE
• EXAMPLE: DOUBLE CONCENTRATION, DOUBLE THE RATE.
• SECOND-ORDER REACTION
• CONCENTRATION AFFECTS REACTION RATE
• EXAMPLE: DOUBLE CONCENTRATION, QUADRUPLE THE RATE
RATE CONSTANT (K) UNITS
Reaction Order
Basic Formula
Units
0
Rate = k
Ms-1
1
Rate = k [A]
s-1
2
Rate = k [A]2
M-1s-1
3
Rate = k [A]3
M-2s-1
EXAMPLE 3:
2NO(G) + O2(G)  2NO2(G)
BASED ON THE REACTION’S RATE LAW OF
RATE = K(NO)2 (O2)
CLASSIFY THIS REACTION’S ORDER.
EXAMPLE 4:
• DETERMINE THE RATE LAW, REACTION ORDER, AND RATE CONSTANT (K) FOR THE
FOLLOWING REACTION AT A SPECIFIC TEMPERATURE---• 2NO(G) + 2H2(G) 
N2(G) + 2H2O(G)
Experiment
[NO]initial
[H2]initial
Rate initial
1
0.20M
0.30M
0.0900 M/s
2
0.10M
0.30M
0.0225 M/s
3
0.10M
0.20M
0.0150 M/s
EXAMPLE 5:
• DETERMINE THE RATE LAW FOR THE FOLLOWING REACTION---• NH4+(AQ) + NO2-(AQ) 
N2(G) + 2H2O(L)
Experiment
[NH4+]initial
[NO2-]initial
Rate initial
1
5 x 10-2 M
2 x 10-2 M
2.70 x 10-7
M/s
2
5 x 10-2 M
4 x 10-2 M
5.40 x 10-7
M/s
3
1 x 10-1 M
2 x 10-2 M
5.40 x 10-7
M/s
INTEGRATED RATE LAW
• ENABLES THE DETERMINATION A REACTANT’S CONCENTRATION AT
ANY MOMENT IN TIME
• ENABLES THE DETERMINATION OF THE TIME IT TAKES TO REACH A
CERTAIN REACTANT CONCENTRATION
• ENABLES THE DETERMINATION OF THE RATE CONSTANT OR REACTION
ORDER
1ST ORDER INTEGRATED RATE LAW
• ONLY USED WITH 1ST ORDER REACTIONS
• FOCUS ON INITIAL CONCENTRATION AND ΔC FOR ONE REACTANT
• INITIAL CONCENTRATION OF REACTANT KNOWN---- CAN DETERMINE
REACTANT CONCENTRATION AT ANY TIME
• INITIAL AND FINAL REACTANT CONCENTRATIONS KNOWN---CAN
DETERMINE RATE CONSTANT
1ST ORDER INTEGRATED RATE LAW
• RATE = -Δ[A]
= K [A]
ΔT
-TAKE EQUATION AND INTEGRATE WITH CALCULUS TO GET….
• LN[A]T – LN[A]0 = - KT
• [A]0 = INITIAL CONCENTRATION (T = 0)
• [A]T = CONCENTRATION AFTER A PERIOD OF TIME
EXAMPLE 1: A  B + 2D
• USING THE DATA PROVIDED FOR A 1ST ORDER REACTION, DETERMINE
THE RATE CONSTANT AND [A] AT TIME = 5.0 X 102S.
Time (s)
[A] (M)
0
0.020
5.0 x 10
0.017
1.0 x 102
0.014
1.5 x 102
0.012
2.0 x 102
0.010
HALF-LIFE
• RADIOACTIVE DECAY IS A 1ST ORDER PROCESS
• HALF-LIFE (T1/2)—
• TIME IT TAKES FOR HALF OF A CHEMICAL COMPOUND TO DECAY OR
TURN INTO PRODUCTS
• FOCUS ON REACTANT
• CONSTANT, NOT DEPENDENT ON [ ]
• RATE CHANGES WITH TEMPERATURE SO HALF-LIFE VARIES BASED ON
TEMPERATURE
EXAMPLE 2:
• FIND THE HALF-LIFE FOR THE FOLLOWING REACTION WITH A REATE
CONSTANT (K) OF 1.70 X 10-3 S-1
2ND ORDER INTEGRATED RATE LAW
• USED ONLY FOR SECOND ORDER REACTIONS
• FOCUS ON INITIAL CONCENTRATION AND ΔC FOR ONE REACTANT
WITH REACTION 2ND ORDER WITH RESPECT TO IT.
• INITIAL CONCENTRATION OF REACTANT KNOWN---- CAN DETERMINE
REACTANT CONCENTRATION AT ANY TIME
• INITIAL AND FINAL REACTANT CONCENTRATIONS KNOWN---CAN
DETERMINE RATE CONSTANT
2ND ORDER INTEGRATED RATE LAW
• RATE = -Δ[A]
= K [A]2
ΔT
-TAKE EQUATION AND INTEGRATE WITH CALCULUS TO GET….
•
1
[A]T
__ -
1__ = KT
[A]0
• [A]0 = INITIAL CONCENTRATION (T = 0)
• [A]T = CONCENTRATION AFTER A PERIOD OF TIME
EXAMPLE 3:
2NO2(G)  2NO(G) + O2(G)
• USING THE DATA PROVIDED, FIND THE RATE CONSTANT IF THE RATE
LAW = K[NO2]2.
Time (s)
[NO2]
0.0
0.070
1.0 x 102
0.0150
2.0 x 102
0.0082
3.0 x 102
0.0057
EXAMPLE 4:
• NO2 REACTS TO FORM NO AND O2 BY SECOND-ORDER KINETICS
WITH A RATE CONSTANT = 32.6 L/MOLMIN. WHAT IS THE [NO2]
AFTER 1 MINUTE IF THE INITIAL [NO2] = 0.15M?
REACTION MECHANISM
• PATHWAY OR SERIES OF STEPS THROUGH WHICH REACTANTS
CONVERTED TO PRODUCTS
• NOT ALL STEPS PROCEED AT THE SAME RATE
• SUM OF THE STEPS = OVERALL REACTION
• RATE OF ANY STEP IS DIRECTLY PROPORTIONAL TO THE REACTANT
CONCENTRATIONS IN THE STEP.
EX. 1: 2NO(G) + O2(G)2NO2(G)
• STEP 1: 2NO  N2O2
• STEP 2: N2O2 + O2  2NO2
__________________________
• 2NO + O2  2NO2
RATE-LIMITING STEP
• STEP IN A REACTION MECHANISM THAT “LIMITS” HOW FAST
PRODUCTS ARE FORMED.
• “LIMITS” THE RATE OF REACTANTS CONVERTING TO PRODUCTS
INTERMEDIATES
• CHEMICAL COMPOUNDS FORMED AND CONSUMED IN A REACTION
MECHANISM
• APPEAR ON BOTH SIDES OF CHEMICAL EQUATION
• TRANSITION COMPOUNDS BETWEEN REACTANTS AND PRODUCTS
• UNSTABLE, ONLY EXIST A SHORT TIME
DETERMINING THE RATE LAW FOR A
REACTION USING A REACTION
MECHANISM
1)OVERALL REACTION RATE LAW
• ONLY DETERMINE EXPERIMENTALLY
2)OVERALL REACTION RATE LAW CAN BE APPLIED TO FIND THE STEPS IN
A REACTION MECHANISM
• RATE LAW INDICATES SLOWEST STEP IN THE MECHANISM
• DETERMINE THE FAST STEPS REMAINING IN THE MECHANISM
RATE LAWS FOR ELEMENTARY STEPS IN MECHANISM
• PREDICTABLE
• CAN USE COEFFICIENTS ONLY FOR THESE STEPS
• AA + BB  DD + EE
RATE = K[A]A [B]B
• OVERALL REACTION RATE = RATE OF SLOWEST STEP
EX. 2: NO2 + CO NO + CO2
• STEP 1: 2NO2  NO + NO3 (SLOW)
• STEP 2: NO3 + CO  NO2 + CO2 (FAST)
DETERMINE THE RATE LAW FOR EACH STEP.
EX.3 BASED ON THE FOLLOWING REACTION
MECHANISM….
• F2 + NO2  NO2F + F (SLOW)
• NO2 + F  NO2F
(FAST)
a)WRITE THE OVERALL REACTION.
b)DETERMINE THE RATE LAW FOR EACH STEP
c) WHAT IS THE RATE LAW FOR THE OVERALL REACTION?
CATALYSTS
• CHEMICAL COMPOUNDS (ATOMS, MOLECULES, IONS) THAT INCREASE
ONLY THE REACTION RATE.
• INCREASES BOTH FORWARD AND REVERSE RATES FOR A CHEMICAL
REACTION
• NOT CONSUMED IN THE REACTION, NOT ALTERED
• PRESENT AT THE START AND END OF REACTION
• HAVE NO EFFECT ON EQUILIBRIUM CONSTANTS (K), ΔH, OR ΔS
CATALYSTS (CONT.)
• CHANGE THE PATH A CHEMICAL REACTION TAKES TO GET REACTANTS
TO PRODUCTS
• LOWERS ACTIVATION ENERGY (EA) OR ENERGY NEEDED FOR THE
REACTION TO START
• ONLY SMALL AMOUNT OF THE COMPOUND NEEDED TO INCREASE
THE RATE FOR A REACTION WITH A LOT OF REACTANT
• RECYCLE AND REUSE
CATALYST REACTION PATHWAY
CATALYSTS (CONT.)
• EXAMPLE: H2O2 DECOMPOSITION
Catalyst
Ea (kJ/mol)
Reaction Rate
None
75.3
1
Catalase
8
6.3 x 1011