CHEMICAL KINETICS H 2 S (g) + Zn 2+ (aq) ⇆ ZnS (s) + 2H + (aq) Chemical reactions can be viewed...

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)


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)


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)


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)


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)


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)


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)


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)


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)



[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)



[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)



[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)


[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)


[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)


[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)


(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)


(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)


(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


y = ln km = -Ea/Rx = 1/Tb = ln A

4F-5 (of 12)

ln k = -Ea 1 + ln A _____ ___

R T


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)

CHEMICAL KINETICS H 2 S (g) + Zn 2+ (aq) ⇆ ZnS (s) + 2H + (aq) Chemical reactions can be viewed...

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Transcript of CHEMICAL KINETICS H 2 S (g) + Zn 2+ (aq) ⇆ ZnS (s) + 2H + (aq) Chemical reactions can be viewed...