16
Hour Exam II Wednesday, March 15 7:00 – 9:00 pm 103 Mumford Hall AQG, AQI Allen AQJ Blair AQF 150 Animal Science Fisher AQB, AQC Koys AGD Pearson AQA conflict exam 4:30 – 6:30 162 Noyes

Hour Exam II

  • Upload
    oni

  • View
    31

  • Download
    0

Embed Size (px)

DESCRIPTION

Hour Exam II. Wednesday, March 15. 7:00 – 9:00 pm. 103 Mumford HallAQG, AQI AllenAQJ BlairAQF. 150 Animal Science FisherAQB, AQC KoysAGD PearsonAQA. conflict exam 4:30 – 6:30 162 Noyes. Kinetics. Thermodynamics :. spontaneity of reaction. G < 0 spontaneous. - PowerPoint PPT Presentation

Citation preview

Page 1: Hour Exam II

Hour Exam II

Wednesday, March 15

7:00 – 9:00 pm

103 MumfordHall AQG, AQIAllen AQJBlair AQF

150 Animal ScienceFisher AQB, AQCKoys AGDPearson AQA

conflict exam4:30 – 6:30162 Noyes

Page 2: Hour Exam II

Kinetics

Thermodynamics :

G < 0 spontaneous

2H2(g) +O2(g) 2H2O(l)

Ho So

Go = Ho - TSo

Thermodynamically spontaneous

spontaneity of reaction

Go = -474 kJ

a) low Tb) high T

very slow

favorableunfavorable

< 0< 0

Page 3: Hour Exam II

KineticsKinetics

2 H2(g) +

rates of reactionmechanism of reaction

Greaction < 0

kinetics

O2(g) 2 H2O(l)

thermodynamics thermodynamics

Greactants

Gproducts

Page 4: Hour Exam II

Kinetics

2H2(g) + O2(g) 2 H2O(l) Go = -474 kJspontaneous

P.E.P.E.n-en-e

qn qe

r

+ -

+

+

Page 5: Hour Exam II

Low Temperature

Page 6: Hour Exam II

High Temperature

Page 7: Hour Exam II

+ +

High Temperature

exothermicendothermic

Page 8: Hour Exam II

Kinetics

1. Temperature (K.E.)

rate of reaction

[reactant][product]

decreaseincrease

2. Concentration3. Orientation

Page 9: Hour Exam II

differential rate laws

A + B C

rate = d[A] =

d[B] = dt

d[C] dt

A + B 2 C

rate = - d[A] = dt

- d[B] = dt

d[C] dt

A + B C +

a) +b) - dt

- - +

12

Page 10: Hour Exam II

t (min)

0

ave. rate =

ave. rate =

81

= 16.2 min-1

-

16.2 min-1

x

x

x

xx

x

[ ]

0

50

100

time (min)1 2 3 4 5

100 100 01 72 72 282 52 52 483 37 37 634 27 27 73

5 19 19 81

[ ] = t

-[ ] = t

(19 )=5

5- 0- 0

- 100- 0

Page 11: Hour Exam II

ave. rate =

ave. rate =

1st minute

5th minute

= 28.0 min-1 1 - 0

- (72 )

t (min)

0 100 100 01 72 72 282 52 52 483 37 37 634 27 27 73

5 19 19 81

x

x

x

xx

x

time (min)1 2 3 4 5

[ ]

0

50

100

ave. rate = - = 16.2 min-1 overall(19- 100) 5 - 0

- 27) = 8.0 min-1 5 - 4

- (19

- 100

Page 12: Hour Exam II

slope of line tangent to curve =

at t = 0 initial rate

instantaneous rate

x

x

x

xx

x

time (min)1 2 3 4 5

[ ]

0

50

100

fastest rate

Page 13: Hour Exam II

0

k

0.330.330.330.330.330.33

rate [reactant] =

k = rate constant

k [reactant]

= rate [reactant]

t instantaneous (min) rate

1 72 23.82 52 17.23 37 12.24 27 8.95 19 6.3

100 33.0

Page 14: Hour Exam II

k = rate [reactant]

k

k

rate =

k, a, b and c

independent of [reactants] dependent on Temperature

k [A]a [B]b ...

determined experimentally

isolation method

[C]c-d[A]) = dt

Page 15: Hour Exam II

Exp. 1[A]i

Exp. 2[A]i [B]I initial rate(M) (M) (M s-1)

1.0 x 10-3

2.0 x 10-3

Exp. 3[A]i [B]I initial rate(M) (M) (M s-1)

1.0 x 10-3

(M)[B]i initial rate(M) (M s-1)

1.0 1.0

t (ms)

Con

cent

ratio

n (M

)[A

]

2.0 1.0

1.0 2.0

t (ms)

Con

cent

ratio

n (M

)[A

]

Page 16: Hour Exam II

Exp. 1[A]i [B]I initial rate(M) (M) (M s-1) 1.0 1.0

Exp. 2[A]i [B]I initial rate(M) (M) (M s-1) 2.0 1.0

1.0 x 10-3

2.0 x 10-3

Exp. 3[A]i [B]I initial rate(M) (M) (M s-1) 1.0 2.0 1.0 x 10-3

rate 2 =rate 1

2 x 10-3 =1 x 10-3

[2.0]a

[1.0]a

b = 0b = 1b = 2rate =

1st order reaction

1x10-3 (M s-1) =

k = 1 x 10-3 s-1

rate 3 =rate 1

1 x 10-3 =1 x 10-3

[2.0]b

[1.0]b

k [1.0 M]

a = 0a = 1a = 2

rate = k[A]a[B]b

k[A]