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

• 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

COLLISION THEORY

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.

VARIABLES AFFECTING REACTION RATE

• INCREASE IN CONCENTRATIONS OF REACTANTS

• TEMPERATURE INCREASES

•WHY?

WHEN DO COLLISION RATES INCREASE?

• 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

MOST COLLISIONS DO NOT RESULT IN A CHEMICAL REACTION!

• 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 (EA)

ACTIVATION ENERGY--ENDOTHERMIC

1) MUST HAVE A COLLISION

2) COLLISION MUST HAPPEN WITH THE CORRECT MOLECULAR ORIENTATION TO GENERATE A REACTION

3) COLLISION ENERGY ≥ EA

WHEN WILL REACTIONS OCCUR?

• 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/MOLK

• 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

ARRHENIUS EQUATION

• CALCULATE ACTIVATION ENERGY (EA) FOR HI

DECOMPOSITION WITH THE FOLLOWING DATA.

EXAMPLE 1

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

00

EaEnergy

T2 > T1

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 __ - 1__ = KT

[A]T [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/MOLMIN.

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