Print Chapter 16 Chemical Kinetics.docx

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1. The Rate of a Reaction: Factors That Affect Reaction Rates 2. Nature of the Reactants 3. Concentrations of the Reactants: The Rate-Law Expression 4. Concentration Versus Time: The Integrated Rate Equation 5. Collision Theory of Reaction Rates 6. Transition State Theory 7. Reaction Mechanisms and the Rate-Law Expression 8. Temperature: The Arrhenius Equation 9. Catalysts

Concentrations of Reactants & Products

Chapter 16 : Chemical Kinetics

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Cdiamond  O 2g   CO 2g  G o298  396 kJ VERY SLOW

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Rate = •

-  A  a t



-   B b t

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+ C c t



+  D d t

The rate of a simple one-step reaction is directly proportional to the concentration of the reacting substance.

A (g)  B(g) + C(g)

Rate  A or Rate = kA • • • • • •

INSTANTANE OUS Consider the hypothetical reaction, aA(g) + bB(g) → cC(g) + dD(g) equimolar amounts of reactants, A and B, will be consumed while products, C and D, will be formed as indicated in this graph:

[C] & [D]

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[A] is the symbol for the concentration of A in M ( mol/L). Note that the reaction does not go entirely to completion. The [A] and [B] > 0 plus the [C] and [D] < 1. Reaction rates are the rates at which reactants disappear or products appear. Mathematically, the rate of a reaction can be written as:

H +aq  + OH-aq   H 2 O l  G o298 = -79 kJ

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[A] & [B]

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Time

1. The Rate of a Reaction Kinetics is the study of rates of chemical reactions and the mechanisms by which they occur. The reaction rate is the increase in concentration of a product per unit time or decrease in concentration of a reactant per unit time. A reaction mechanism is the series of molecular steps by which a reaction occurs. Thermodynamics (Chapter 15) determines if a reaction can occur. Kinetics (Chapter 16) determines how quickly a reaction occurs. Some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible.

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0

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1

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[A] is the concentration of A in molarity or moles/L. k is the specific rate constant. k is an important quantity in this chapter. For a simple expression like Rate = k[A] If the initial concentration of A is doubled, the initial rate of reaction is doubled. If the reaction is proceeding twice as fast, the amount of time required for the reaction to reach equilibrium would be: A. The same as the initial time. B. Twice the initial time. C. Half the initial time. If the initial concentration of A is halved the initial rate of reaction is cut in half.

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If more than one reactant molecule appears in the equation for a one-step reaction, we can experimentally determine that the reaction rate is proportional to the molar concentration of the reactant raised to the power of the number of molecules involved in the reaction.

2 Xg   Yg  + Zg  Rate  X or Rate = kX 2

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Rate Law Expressions must be determined experimentally. – The rate law cannot be determined from the balanced chemical equation. – This is a trap for new students of kinetics. • The balanced reactions will not work because most chemical reactions are not one-step reactions. Other names for rate law expressions are: – rate laws – rate equations or rate expressions Important terminology for kinetics. The order of a reaction can be expressed in terms of either: each reactant in the reaction or the overall reaction. Order for the overall reaction is the sum of the orders for each reactant in the reaction. For example:

2 N 2 O5g   4 NO2g  + O 2g  Rate = kN 2 O5 

This reaction is first order in N 2 O5 and first order overall. A second example is

CH3 3 CBraq   OH-aq   CH3 3 COHaq   Br-aq  Rate = k[CH3 3 CBr ] This reaction is first order in CH3 3 CBr, zero order in OH- , and first order overall.

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A final example of the order of a reaction is:

2 NOg  + O 2g   2 NO2g  Rate = k[NO]2 O 2 

This reaction is second order in NO, first order in O 2 , and third order overall REMEMBER, ALL RATE EXPRESSION S ARE DETERMINED EXPERIMENT ALLY

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Given the following one step reaction and its rate-law expression. – Remember, the rate expression would have to be experimentally determined.

2 A g   Bg   Cg  Rate = kA

2

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A. B. C. D.

Because it is a second order rate-law expression: – If the [A] is doubled the rate of the reaction will increase by a factor of 4. 22 = 4 – If the [A] is halved the rate of the reaction will decrease by a factor of 4. (1/2)2 = 1/4

Factors That Affect Reaction Rates The nature of the reactants. The concentration of the reactants. The temperature of the reaction. The presence of a catalyst.

A. Nature of Reactants • This is a very broad category that encompasses the different reacting properties of substances. • For example sodium reacts with water explosively at room temperature to liberate hydrogen and form sodium hydroxide. •

2 Na s   2 H 2 O   2 NaOHaq   H 2g  This is a violent and rapid reaction. The H 2 ignites and burns.

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Calcium reacts with water only slowly at room temperature to liberate hydrogen and form calcium hydroxide.

Example 16-2: The following data were obtained for the following reaction at 25oC. What are the rate-law expression and the specific rate constant for the reaction? 2 A(g) + B(g) + 2 C(g) → 3 D(g) + 2 E(g)

Ca s   2 H 2 O    Ca OH2 aq   H 2g  This is a rather slow reaction. •

The reaction of magnesium with water at room temperature is so slow that that the evolution of hydrogen is not perceptible to the human eye.

Mg s   H 2O   No reaction •

Example 16-3: consider a chemical reaction between compounds A and B that is first order with respect to A, first order with respect to B, and second order overall. From the information given below, fill in the blanks. You do it!

However, Mg reacts with steam rapidly to liberate H2 and form magnesium oxide. C Mg s   H 2O(g ) 100   MgO s   H 2g  o

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The differences in the rate of these three reactions can be attributed to the changing “nature of the reactants”. B. Concentrations of Reactants: The Rate-Law Expression This is a simplified representation of the effect of different numbers of molecules in the same volume. – The increase in the molecule numbers is indicative of an increase in concentration. A(g) + B (g) → Products

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Example 16-1: The following rate data were obtained at 25oC for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction? 2 A(g) + B(g) → 3 C(g)

Experiment Number

Initial [A] (M)

Initial [B] (M)

Initial rate of formation of C (M/s)

1

0.10

0.10

2.0 x 10-4

2

0.20

0.30

4.0 x 10-4

3

0.10

0.20

2.0 x 10-4

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Concentration vs. Time: The Integrated Rate Equation The integrated rate equation relates time and concentration for chemical and nuclear reactions. – From the integrated rate equation we can predict the amount of product that is produced in a given amount of time. Initially we will look at the integrated rate equation for first order reactions. These reactions are 1st order in the reactant and 1st order overall. An example of a reaction that is 1st order in the reactant and 1st order overall is: a A → products This is a common reaction type for many chemical reactions and all simple radioactive decays. Two examples of this type are: 2 N2O5(g) → 2 N2O4(g) + O2(g) 238 U → 234Th + 4He

–

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In this example we will use grams rather than mol/L. Example 16-6: The half-life for the following first order reaction is 688 hours at 10000C. Calculate the specific rate constant, k, at 10000C and the amount of a 3.0 g sample of CS2 that remains after 48 hours. CS2(g) ® CS(g) + S(g) You do it!

The integrated rate equation for first order reactions is:

ln

A0  a k t A

where: [A]0= mol/L of A at time t=0. [A] = mol/L of A at time t. k = specific rate constant. t = time elapsed since beginning of reaction. a = stoichiometric coefficient of A in balanced overall equation. •

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1  A 0  ln   a k  A 

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1  A 0   ln  a k  1/2A 0 

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Where: [A]0= mol/L of A at time t=0. [A] = mol/L of A at time t. k = specific rate constant. t = time elapsed since beginning of reaction. a = stoichiometric coefficient of A in balanced overall equation.

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Second order reactions also have a half-life. – Using the second order integrated ratelaw as a starting point. At the half-life, t1/2 [A] = 1/2[A]0.

1 ln 2 ak 0.693 t1/2  ak

Example 16-4: Cyclopropane, an anesthetic, decomposes to propene according to the following equation. H2 C

CH2

CH (g)

H3 C

CH2

(g)

The reaction is first order in cyclopropane with k = 9.2 s-1 at 10000C. Calculate the half life of cyclopropane at 10000C. Example 16-5: Refer to Example 16-4. How much of a 3.0 g sample of cyclopropane remains after 0.50 seconds? – The integrated rate laws can be used for any unit that represents moles or concentration.

For reactions that are second order with respect to a particular reactant and second order overall, the rate equation is:

1 1  akt A A0

t1/2 

CH2

 k t  ln A 0  ln A   k t

A  e1.052  2.86 g  2.9 g or 97% unreacted

Define the half-life, t1/2, of a reactant as the time required for half of the reactant to be consumed, or the time at which [A]=1/2[A]0. At time t = t1/2, the expression becomes:

t1/2 

A0 A

ln(3.0) - ln A   (0.00101 hr -1 )( 48 hr) 1.10 - ln A   0.048 ln A   -(0.048 - 1.10)  1.052

Solve the first order integrated rate equation for t.

t

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ln

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2 1   a k t1/2 or A 0 A 0 1  a k t1/2 A 0

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If we solve for t1/2:

t1/2  •

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1 a k A 0

Note that the half-life of a second order reaction depends on the initial concentration of A. Example 16-7: Acetaldehyde, CH3CHO, undergoes gas phase thermal decomposition to methane and carbon monoxide.  CH3CHOg    CH4g  + COg 

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The rate-law expression is Rate = k[CH3CHO]2, and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 527oC? Example 16-7: The rate-law expression is Rate = k[CH3CHO]2, and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 527oC?  CH3CHOg    CH4g  + COg 

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For the first order reaction 1  A   Rate = -   a A → products a  t  the rate can be written as:

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For a first-order reaction, the rate is proportional to the first power of [A].

1  A   -    kA  a  t 

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1  d A   -    kA  a  dt  •

(b) How many moles of CH3CHO remain after 200 hours? (c) What percent of the CH3CHO remains after 200 hours? Example 16-8: Refer to Example 16-7. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 10.0 L vessel at 527oC? – Note that the vessel size is increased by a factor of 10 which decreases the concentration by a factor of 10! (b) How many moles of CH3CHO remain after 200 hours? (c) What percent of the CH3CHO remains after 200 hours? Let us now summarize the results from the last two examples.

In calculus, the rate is defined as the infinitesimal change of concentration d[A] in an infinitesimally short time dt as the derivative of [A] with respect to time.

Rearrange the equation so that all of the [A] terms are on the left and all of the t terms are on the right.

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d A   a k  dt A

Express the equation in integral form.

d A -   a k   dt A 0 A 0 A 

t

This equation can be evaluated as:

- ln A  0t  a k t 0t or

- ln A t  ln A 0  a k  t - a k  0 which becomes - ln A t  ln A 0  a k t

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Which rearranges to the integrated first order rate equation.

ln

A0 At

akt

– •

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Derive the rate equation for a reaction that is second order in reactant A and second order overall. The rate equation is:

d A   k dt 2 a A 

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Then integrate the equation over the limits as for the first order reaction. A 



A 0

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d A t  a k dt A2 0

Which integrates the second order integrated rate equation.

1 1  akt A A0 •

For a zero order reaction the rate expression is:

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d A  k adt

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Which rearranges to:

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Then we integrate as for the other two cases:

 d A  a k d t

A 

t

A 0

0

  d A   a k d t

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Which gives the zeroeth order integrated rate equation.

A  A0  -a k t or A  A0 - a k t

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ln A0  ln A  a k t

or ln A  a k t  ln A0

d A 2   kA  adt

Separate the variables so that the A terms are on the left and the t terms on the right.

Enrichment - Rate Equations to Determine Reaction Order Plots of the integrated rate equations can help us determine the order of a reaction. If the first-order integrated rate equation is rearranged.

This law of logarithms, ln (x/y) = ln x - ln y, was applied to the first-order integrated rate-equation.

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The equation for a straight line is:

y  mx  b •

Compare this equation to the rearranged first order rate-law.

y  mx  b ln A  a k t  ln A0

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Now we can interpret the parts of the equation as follows:  y can be identified with ln[A] and plotted on the y-axis.  m can be identified with –ak and is the slope of the line.  x can be identified with t and plotted on the x-axis.  b can be identified with ln[A]0 and is the y-intercept. Example 16-9: Concentration-versus-time data for the thermal decomposition of ethyl bromide are given in the table below. Use the following graphs of the data to determine the rate of the reaction and the value of the rate constant.  C2 H5Brg    C2 H4g   HBrg  at 700K