chemistry notes

4/1/2015

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MC assignments

• past due assignments:LT16 due 3/23 at 1 pm

LT17 due 3/25 at 1 pm


• available assignments:

LT19/20 due Fri 4/3 at midnight

HW‐Ch14 due Sat. 4/4 at midnight

• upcoming assignments:


Review

• method of initial rates to determine the order of reaction

• integrated rate laws

– first order

– second order

– zero order

• half life t½

Lecture 18: Monday of week 11

• Chapter 14.5 – 14.6

• Friday in Recitation: – Quiz 12,

• Laboratory:

• Next Exam: – Wednesday evening, Apr. 15, 6:45 – 7:45 pm

• Final Exam: http://www.registrar.iastate.edu/students/exams/springexams#alpha

– Wednesday evening, May 6, 7 – 9 pm

• put away all distractions

• set your clicker to Channel 17 or 27

I _____ the lecture notes Dr. Irmi has posted before lectures.

1. like

2. do not like

3. don’t know anything about

1. 2. 3.

79

49

23

This weather makes it ______ to come to class.

1. easier

2. harder

3. no difference

1. 2. 3.

66

40

56

N2O5(g)→ 4 NO2(g) + O2(g)The reaction is first order and k = 6.82 x 10‐3 s‐1 at 70 °C

what is t1/2 in min?

1. 101.6

2. 44.1

3. 0.74

4. 1.69

1. 2. 3. 4.

60

93

412

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N2O5(g) → 4 NO2(g) + O2(g)The reaction is first order and k = 6.82 x 10‐3 s‐1 at 70 °C

You start with 0.0250 mol N2O5(g) in 2LAfter 6.76 min, how much N2O5(g) is left?

1. 0.0125

2. 6.25 x 10‐3

3. 3.13 x 10‐3

4. 1.56 x 10‐3

1. 2. 3. 4.

2

119

2425

N2O5(g) → 4 NO2(g) + O2(g)The reaction is first order and k = 6.82 x 10‐3 s‐1 at 70 °C

You start with 0.0250 mol N2O5(g) in 2LHow many minutes will it take until only 0.10 mol is left?

1. 2.2

2. 1.1

3. 4.4

4. 6.6

1. 2. 3. 4.

135

3719

half life of a second order reaction

1½ ½

1

½ 1

depends on the initial concentration

How fast do chemical reactions proceed?

• Speeds depend on

– Physical state,

• Homogeneous or heterogeneousmixtures

– Reactant concentration,

– Reaction temperature

– Presence of a catalyst

• At molecular level:

– Frequency, energy, and orientation of collisions

Demo?

rate constants for isomerization of methyl isonitrile to acetonitrile

CH3−N≡C →CH3−C≡N

• the rate constant increases with increasing temperature

• the reaction proceeds faster at higher temperatures

the Arrhenius equation shows the temperature dependence of the rate constant k:

Ea Activation energyR gas constant 8.13 J/mol‐KT absolute temperatureA pre‐exponential factor

graphical determination of the activation energy

linear equation;

graph of ln k over 1/T

from the slope:

∗ ∗

Arrhenius Plot

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g

A first order reaction was studied at two different starting concentrations

and two different temperatures.

Which of the three graphs were taken at the same starting concentration?

1. 1 and 2

2. 2 and 3

3. 1 and 3

1. 2. 3.

19

1

151

p



Which of the three graphs were taken at the same

temperature?

1. 1 and 2

2. 2 and 3

3. 1 and 3

1. 2. 3.

156

48



Which graph(s) was taken at the lower temperature?

1. 1 and 2

2. 1 only

3. 3 only

1. 2. 3.

7

158

8

What is the Activation Energy?

Energy Profile of the Reaction initial state: CH3−N≡C molecule

add energy: bend the molecule

at the maximum of the energy curve: activated complex, or transition state

energy is released: new bond is formed

final state: CH3−C≡N molecule

Where does the energy come from?• for a reaction to proceed, the activation energy has to be added to the molecule

• Thermal energy of the molecules can supply this energy

• kinetic molecular theory shows this nicely for gasses

• at a given temperature a fraction of molecules have a kinetic energy that exceeds the activation energy.

• at a higher temperature moremolecules have enough kinetic energy

• the fraction of molecules that have a kinetic energy that exceeds the activation energy is

•

• this is almost the Arrhenius equation

• pre‐exponential factor A:

• not every collision is “successful”

• has the proper units

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uses of Arrhenius equation

• determine the activation energy of a reaction

– measurement of k at different temperatures

– use slope of Arrhenius plot

• determine the rate constant at a different temperature

– use rate constant k at T1 and Ea

Reaction Mechanisms

• What goes on at the molecular level?– some bonds are broken

– other bonds are formed

– there may be intermediates

• Elementary Reactions– step by step reactions

– one, two, or threemolecules involved

– molecularity of the elementary reaction

– each elementary reaction proceeds at its own rate

molecularity of elementary reactions

• a molecule “falls apart” AB → A + B – unimolecular

– rate = k [AB]

• two molecules collide: C + D → CD– bimolecular

– rate = k [C][D]

• three molecules collide: D + E + F → DEF– termolecular (very rare)

– rate = k [E][D][F]

Team ScoresPoints Team Points Team

5.33 easier

5.27 no difference

5.15 harder

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