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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.

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Hour Exam II

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Hour exam ii

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

Kinetics

Thermodynamics :

spontaneity of reaction

G < 0spontaneous

2H2(g) +O2(g) 2H2O(l)

Go = -474 kJ

Go = Ho- TSo

a) low T

b) high T

Ho

< 0

favorable

So

< 0

unfavorable

Thermodynamically spontaneous

very slow


Kinetics1

thermodynamics

thermodynamics

Kinetics

rates of reaction

mechanism of reaction

O2(g)

 2 H2O(l)

Greaction < 0

2 H2(g) +

Greactants

Gproducts

kinetics


Kinetics2

qn qe

r

+

+

Kinetics

2H2(g) + O2(g)2 H2O(l) Go = -474 kJ

spontaneous

P.E.n-e

-

+


Hour exam ii

Low Temperature


Hour exam ii

High Temperature


High temperature

+

+

High Temperature

exothermic

endothermic


Kinetics3

Kinetics

rate of reaction

[reactant]

decrease

increase

[product]

1. Temperature (K.E.)

2. Concentration

3. Orientation


Differential rate laws

-

1

2

differential rate laws

A + B C

-

d[A] =

d[C]

dt

d[B] =

dt

a) +

b) -

+

rate =

dt

A + B 2 C

- d[A] =

dt

d[C]

dt

-d[B] =

dt

rate =

A + B C

+


Hour exam ii

100

[ ]

50

0

1

2

3

4

5

time (min)

[ ] =

t

-[ ] =

t

x

t (min)

0

100

100

0

x

72

1

72

28

x

2 52 52 48

x

x

3 37 37 63

x

4 27 27 73

5 19 19 81

81

- 0

= 16.2 min-1

ave. rate =

5

- 0

(19 )

- 100

ave. rate =

-

=

16.2 min-1

5

- 0


Hour exam ii

x

100

t (min)

0

100

100

0

x

[ ]

72

1

72

28

x

50

2 52 52 48

x

x

3 37 37 63

x

4 27 27 73

0

5 19 19 81

1

2

3

4

5

time (min)

overall

ave. rate =

-

(19

- 100)

= 16.2 min-1

5

- 0

1st minute

ave. rate =

= 28.0 min-1

- (72 )

- 100

1 - 0

5th minute

ave. rate =

- 27)

= 8.0 min-1

- (19

5

- 4


Hour exam ii

x

100

x

[ ]

x

50

x

x

x

0

1

2

3

4

5

time (min)

instantaneous rate

slope of line tangent to curve =

at t = 0

initial rate

fastest rate


Hour exam ii

t instantaneous

(min) rate

k

= rate

[reactant]

100

33.0

0

0.33

17223.8

25217.2

33712.2

4278.9

5196.3

0.33

0.33

0.33

0.33

0.33

rate [reactant] =

k [reactant]

k = rate constant


Hour exam ii

k = rate

[reactant]

k

independent of [reactants]

dependent on Temperature

k

[A]a

[B]b

[C]c

...

(-d[A])

k

rate =

=

dt

k, a, b and c

determined experimentally

isolation method


Hour exam ii

[A]

[A]

Concentration (M)

Concentration (M)

t (ms)

t (ms)

Exp. 1

[B]i

initial rate

[A]i

(M)

(M)

(M s-1)

1.0

1.0

1.0 x 10-3

Exp. 2

[A]i[B]I initial rate

(M)(M)(M s-1)

2.0

1.0

2.0 x 10-3

Exp. 3

[A]i[B]I initial rate

(M)(M)(M s-1)

1.0 x 10-3

1.0

2.0


Hour exam ii

Exp. 1

k

rate =

[A]a

[B]b

[A]i[B]I initial rate

(M)(M)(M s-1)

1.01.0

a = 0

a = 1

a = 2

2 x 10-3 =

1 x 10-3

[2.0]a

[1.0]a

rate 2 =

rate 1

1.0 x 10-3

Exp. 2

1 x 10-3 =

1 x 10-3

rate 3 =

rate 1

[2.0]b

[1.0]b

b = 0

b = 1

b = 2

[A]i[B]I initial rate

(M)(M)(M s-1)

2.01.0

rate =

k

[A]

2.0 x 10-3

1storder reaction

Exp. 3

[A]i[B]I initial rate

(M)(M)(M s-1)

1.02.0

1x10-3(M s-1) =

k

[1.0 M]

k = 1 x 10-3 s-1

1.0 x 10-3


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