Chemical Kinetics CHAPTER 14 Chemistry: The Molecular Nature of Matter, 6 th edition

Chemical Kinetics

CHAPTER 14

Chemistry: The Molecular Nature of Matter, 6th editionBy Jesperson, Brady, & Hyslop

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CHAPTER 14 Chemical Kinetics

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Learning Objectives: Factors Affecting Reaction Rate:

o Concentrationo Stateo Surface Areao Temperatureo Catalyst

Collision Theory of Reactions and Effective Collisions Determining Reaction Order and Rate Law from Data Integrated Rate Laws Rate Law Concentration vs Rate Integrated Rate Law Concentration vs Time Units of Rate Constant and Overall Reaction Order Half Life vs Rate Constant (1st Order) Arrhenius Equation Mechanisms and Rate Laws Catalysts

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Lecture Road Map:

① Factors that affect reaction rates

② Measuring rates of reactions

③ Rate Laws

④ Collision Theory

⑤ Transition State Theory & Activation Energies

⑥ Mechanisms

⑦ Catalysts

4Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Factors that Affect Reaction

Rates


Kinetics: Study of factors that govern oHow rapidly reactions occur and oHow reactants change into products

Rate of Reaction: o Speed with which reaction occursoHow quickly reactants disappear and

products form

Kinetics The Speed at Which Reactions Occur

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 5

Kinetics The Speed at Which Reactions Occur


AB

Reaction rate is measured by the amount of product produced or reactants consumed per unit time.

o [B] concentration of products will increase over time

o [A] concentration of reactants will decrease over time


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Kinetics Factors Affecting Reaction Rates

1. Chemical nature of reactantso What elements, compounds, salts

are involved?o What bonds must be formed,

broken?o What are fundamental differences in

chemical reactivity?


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2. Ability of reactants to come in contact(Reactants must meet in order to react)

The gas or solution phase facilitates thiso Reactants mix and collide with each other

easilyo Homogeneous reaction

oAll reactants in same phaseoOccurs rapidly

o Heterogeneous reactionoReactants in different phasesoReactants meet only at interface between

phasesoSurface area determines reaction rate o Increase area, increase rate; decrease area,

decrease rate


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3. Concentrations of reactantso Rates of both homogeneous and

heterogeneous reactions affected by [X ]

o Collision rate between A and B increase if we increase [A] or increase [B ].o Often (but not always) reaction rate

increases as [X ] increases


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4. Temperatureo Rates are often very sensitive to temperature

o Raising temperature usually makes reaction faster for two reasons:o Faster molecules collide more often and

collisions have more energyo Most reactions, even exothermic reactions,

require energy to occur

o Rule of thumb: Rate doubles if temperature increases by 10 °C (10 K)


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5. Presence of Catalystso Substances that increase rates of chemical

reactions without being used up

o Rate-accelerating agents

o Speed up rate dramaticallyoRate enhancements of 106 not uncommon

o Chemicals that participate in mechanism but are regenerated at the end


Measuring Reaction Rates



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Rates Measuring Rate of Reaction

o Rate = ratio with time unit in denominator

o Rate of Chemical Reactiono Change in concentration per unit time.

o Always with respect to a given reactant or producto [reactants] decrease with timeo [products] increase with time


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o Concentration in M unitso Time in s unitso Units on rate:

o [product] increases by 0.50 mol/L per second rate = 0.50 M/s

o [reactant] decreases by 0.20 mol/L per second rate = 0.20 M/s


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Rates Rate of Reaction

Always positive whether something is increasing or decreasing in [X ]

o Reactants o Reactant consumedo So [X ] is negativeo Need minus sign to make rate positive

o Products o Produced as reaction goes alongo So [X ] is positiveo Thus rate already positive


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Coefficients indicate the relative rates at which reactants are consumed and products are formed

o Related by coefficients in balanced chemical equationo Know rate with respect to one product or reactanto Can use equation to determine rates with respect to all

other products and reactants.

A + B C + D


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Rates Rate of Reaction: Example

• O2 reacts 5 times as fast as C3H8

• CO2 forms 3 times faster than C3H8 consumed

• H2O forms 4/5 as fast as O2 consumed

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)


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GroupProblem

Clorox bleach is sodium hypochlorite. It should never be mixed with acids, (like vinegar) because it forms chlorine gas:

NaClO + 2 HCl → Cl2 + H2O + NaCl

If Chlorine gas (Cl2) is formed at a rate of 5.0 x 10-4 mol/Ls what rate is HCl consumed?


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Rates Change of Reaction Rate with Time

Generally reaction rate changes during reaction, it isn’t constant

o Often initially fast when lots of reactant present

o Slower and slower as reactants are depletedWhy?

o Rate depends on the concentration of the reactants

o Reactants being used up, so the concentration of the reactants are decreasing and therefore the rate decreases

Measured in 3 ways: o instantaneous rate, average rate, initial rate


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Rates Instantaneous & Initial Reaction Rate

Instantaneous rateo Slope of tangent to curve at some specific time

Initial rateo Determined at time = 0

NO2 appearance

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0 200 400 600 800

Time (s)

[NO

2]


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Rates Average Rate of Reaction

Average Rate: Slope of line connecting starting and ending coordinates for specified time frame

rateΔtimeΔ[Product]

NO2 appearance

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0 200 400 600 800

Time (s)

[NO

2]

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Rates Example Reporting Different Types of Rates

Rate at any time t = negative slope (or tangent line) of curve at that point

Concentration vs. Time Curve for 0.005M phenolphthalein reacting with 0.61 M NaOH at room temperature

http://chemed.chem.purdue.edu


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[P] (mol/L) Time (s)

0.005 00.0045 10.50.004 22.3

0.0035 35.70.003 51.1

0.0025 69.30.002 91.6

0.0015 120.4

Initial rate = Average rate between first two data points

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Instantaneous Rate at 51.1 s

(90,0.0028)

(160,0.0018)

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Average Rate between 0 and 120.4 s


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GroupProblem

A reaction was of NO2 decomposition was studied. The concentration of NO2 was found to be 0.0258 M at 5 minutes and at 10 minutes the concentration was 0.0097 M. What is the average rate of the reaction between 5 min and 10 min?A. 310 M/minB. 3.2 × 10–3 M/minC. 2.7 × 10–3 M/minD. 7.1 × 10–3 M/min


Rate Laws



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Rate Laws Rates Based on All Reactants

A + B C + D

o Rate Law or Rate expressiono k is the rate constant

o Dependent on Temperature & Solvento m and n = exponents found experimentallyo No necessary connection between stoichiometric

coefficients (, ) and rate exponents (m, n)o Usually small integerso Sometimes simple fractions (½, ¾) or zero

= k[A]m[B]n


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Rate Laws Rates Based on All Reactants

Below is the rate law for the reaction 2A +B → 3C

rate= 0.045 M–1s–1 [A][B]If the concentration of A is 0.2 M and that of B is 0.3 M, and the reaction is 1st order (m & n = 1) what will be the reaction rate?

rate=0.0027 M/s 0.003 M/s

rate=0.045 M–1 s–1 [0.2][0.3]


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Rate Laws Order of Reactions

Rate = k[A]m[B]n

Exponents specify the order of reaction with respect to each reactant

Order of Reactiono m = 1 [A]1 1st order in [A]o m = 2 [A]2 2nd order in [A]o m = 3 [A]3 3rd order in [A]o m = 0 [A]0 0th order in [A]

[A]0 = 1 means A doesn't affect rate

Overall order of reaction = sum of orders (m and n) of each reactant in rate law


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Rate Laws Order of Reactions: Example

5Br– + BrO3– + 6H+ 3Br2 + 3H2O

x = 1 y = 1 z = 2o1st order in [BrO3

–]o1st order in [Br –] o2nd order in [H+]oOverall order = 1 + 1 + 2 = 4


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Rate Laws Order of Reaction & Units for k


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GroupProblem

The following rate law has been observed:Rate = k[H2SeO][I–]3[H+]2. The rate withrespect to I– and the overall reaction rate is:A. 6, 2B. 2, 3C. 1, 6D. 3, 6


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Rate Laws Calculating k

If we know rate and concentrations, can use rate law to calculate k

From Text Example of decomposition of HI at 508 °C

• Rate= 2.5 × 10–4 M/s• [HI] = 0.0558 M


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Rate Laws Determining Exponents in Rate Law

Experimental Determination of Exponentso Method of initial rateso If reaction is sufficiently slow

o or have very fast technique o Can measure [A] vs. time at very beginning of reaction

o Before it slows very much, then

o Set up series of experiments, where initial concentrations vary


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Rate Laws Determining Rate Law Exponents: Example

3A + 2B productsRate = k[A]m[B]n

Expt. #

[A]0, M [B]0, M Initial Rate, M/s

1 0.10 0.10 1.2 10–4 2 0.20 0.10 4.8 10–4 3 0.20 0.20 4.8 10–4

Convenient to set up experiments so The concentration of one species is doubled or tripled And the concentration of all other species are held constant

Tells us effect of [varied species] on initial rate


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Rate Laws Determining Rate Law Exponents

• If reaction is 1st order in [X],– Doubling [X]1 21

– Doubles the rate• If reaction is 2nd order in [X],

– Doubling [X]2 22

– Quadruples the rate• If reaction is 0th order in [X],

– Doubling [X]0 20

– Rate doesn't change• If reaction is nth order in [X]

– Doubling [X]n 2n times the initial rate


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Expt. #


1 0.10 0.10 1.2 104 2 0.20 0.10 4.8 104 3 0.20 0.20 4.8 104Comparing Expt. 1 and 2

o Doubling [A]o Quadruples rateo Reaction 2nd order in A = [A]2

4102.1108.4

1 Rate2 Rate

4

4

2m = 4 or m = 2


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Expt. # [A]0, M [B]0, M Initial Rate, M/s1 0.10 0.10 1.2 104 2 0.20 0.10 4.8 104 3 0.20 0.20 4.8 104

Comparing Expt. 2 and 3o Doubling [B]o Rate does not changeo Reaction 0th order in B = [B]0 = 1

1108.4108.4

2 Rate3 Rate

4

4

2n = 1 or n = 0



o Conclusion: rate = k[A]2

o Can use data from any experiment to determine k o Let’s choose first experiment

Expt. #


1 0.10 0.10 1.2 10–4 2 0.20 0.10 4.8 10–4 3 0.20 0.20 4.8 10–4


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Rate Laws Determining Rate Law Exponents: Ex 2

2 SO2 + O2 2 SO3 Rate = k[SO2]m[O2]n

Expt #

[SO2] M [O2] M Initial Rate of SO3 Formation, M s–1

1 0.25 0.30 2.5 103

2 0.50 0.30 1.0 102

3 0.75 0.60 4.5 102

4 0.50 0.90 3.0 102


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4 = 2m or m = 2


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3 = 3n or n = 1


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Rate = k[SO2]2[O2]1

• 1st order in [O2]• 2nd order in [SO2]• 3rd order overall• Can use any experiment to find k


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GroupProblem

Using the following experimental data, determine the order with respect to NO and O2 .

A. 2, 0B. 3,1C. 2, 1D. 1, 1

Expt #

[NO] M

[O2] M

Initial Rate M s–1

1 0.12 0.25 1.5 10–3

2 0.24 0.25 6.0 10–3

3 0.24 0.50 1.2 10–2

Chemical Kinetics CHAPTER 14 Chemistry: The Molecular Nature of Matter, 6 th edition

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Transcript of Chemical Kinetics CHAPTER 14 Chemistry: The Molecular Nature of Matter, 6 th edition