tro chapter 13 kinetics spring 2015 (1)

CHAPTER 13

Chemical Kinetics

Definitions/Terminology

Kinetics is the study of rates of chemical reactions and the mechanisms (pathways) by which they occur.

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• The rate is the increase in concentration of a product per unit time or decrease in concentration of a reactant per unit time.

• Rates have units of +/-[conc/time] (e.g., Molar/sec).

• A mechanism is the series of molecular steps by which a reaction occurs.

Reaction Rates

Thermodynamics determines whether or not a reaction can occur.

Kinetics describes how quickly a reaction occurs. Some reactions are thermodynamically favorable (G0

< 0), but are kinetically so slow as to be imperceptible at room temperature:

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CH4(g) + O2(g) CO2(g) + H2O(g) G0 = -800 kJ/mol

H2(g) + O2(g) H2O(g) G0 = -230 kJ/mol

• These reactions do not occur at room temperature and pressure without activation (e.g., a spark or a catalyst).

Reaction Rates

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• Rates define the velocity at which reactants disappear or products appear

• Rates have units of [concentration/time].

Reaction Rates

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• in many reactions, the coefficients of reactants and products in the balanced equation are not all the same, for example:

H2 (g) + I2 (g) 2 HI(g)

For every 1 mole of H2 consumed, 1 mole of I2 will also be consumed, and 2 moles of HI will be produced.

The rate of disappearance of [H2] and [I2] will, therefore, be ½ the rate of appearance of [HI].

• The rate of the overall reaction is, thus, the change in the concentration of each substance multiplied by

1/[coefficient in the balanced equation]:

Reaction Rates

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For [H2], the instantaneous rate at 50 s is:

For [HI], the instantaneous rate at 50 s is:

sM 0 .0 0 7 0 R a te

s 4 0

M 2 8.0 R a te

s

M 0 .0 0 7 0 R a te

s 4 0

M 5 6.0

2

1 R a te

Reaction Rates Mathematically, the rate of a generic reaction:

aA + bB --> cC + dD can be written as:

For X = A, B, C or D, [X] = [X]t - [X]0, where[X]t is concentration at time, t after the start of the reaction, and [X]0 is the initial concentration at the beginning of the reaction, t = 0.Minus sign indicates that the reactants concentrations decrease with time.

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• For example, for a reaction 2A B:Rate = -(1/2)[A]/t = [B]/t

• Rates are often approximated as instantaneous rates, and the notation of calculus is used:

Rate = -1/2d[A]/dt = d[B]/dt

The rate equation

Reaction Rates The rate of a simple one-reactant, one-step reaction

is directly proportional to the concentration of the reacting substance:

Rate has units of conc/unit time (e.g., Ms-1) [A] is the concentration of A, e.g., in molarity (M). The proportionality constant, k is called the rate constant. For this simple expression, k must have units of inverse

time (e.g., s-1). Rate constants are always positive numerical quantities. 9

A B + Crate [A] or rate = k[A] the rate law expression

(different from the rate eqn!)

Reaction Rates

For the simple expression, rate = k[A]: If the initial concentration of A is doubled, the

initial rate of the reaction is doubled. If the initial concentration of A is halved, the initial

rate of reaction is cut in half. The initial rate is, thus, directly proportional to

the initial [A].

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Reaction Rates If more than one reactant molecule appears in

the chemical equation for a one-step reaction:

2A B + C

The experimentally determined rate law is:

rate = k[A]2

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• This relationship means that, If [A] is doubled, the rate increases by a factor of 4 (= 22).

• If initial [A] is halved, the initial rate decreases by a factor of 4 (=(1/2)2).

Reaction Rates Rate = k[X] or rate = k[X]2 are examples of a rate

law.

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Rate laws can only be determined experimentally. The rate law cannot be determined from the balanced

chemical equation because: most chemical reactions are not one-step reactions.

Reaction Rates

The order of a reaction expresses: the dependence on the concentrations of each reactant in the

rate law (= the exponent for each concentration). the sum of the orders for each reactant.

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• For the reaction:N2O5(g) 2NO2(g) + 1/2O2(g)

The experimentally determined rate law is: rate = k[N2O5]

• The reaction is said to be first order in N2O5 and first order overall.

Reaction Rates First order reactions are dependent on the

concentration of only a single reactant. First order reactions are common for many

chemical reactions and all simple radioactive decays.

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Two examples of first order reactions:

2N2O5(g) 2 N2O4(g) + O2(g)

238U 234Th + 4He

Rate = k[N2O5]

Rate = k[238U]

Reaction Rates

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• For the reaction:2NO(g) + O2(g) 2NO2(g)

• The experimentally determined rate law is:rate = k[NO]2[O2]

• This reaction is second order in NO, first order in O2 and third order overall.

Reaction Rates

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• For the reaction:(CH3)3CBr(aq) + OH-

(aq) (CH3)3COH(aq) + Br -(aq)

The experimentally determined rate law is:rate = k[(CH3)3CBr]

• This reaction is first order in (CH3)3CBr and zero order in OH- and first order overall.

Concentrations of Reactants: The Rate-Law Expression

Example 1: The following rate data were obtained at 25oC for the reaction: 2A(g) + B(g) ® 3C(g)

What are the rate law and rate constant for this reaction?

ExperimentNumber

Initial [A](M)

Initial [B](M)

Initial rate (M/s)

1 0.10 0.10 2.0 x 10-4

2 0.20 0.10 4.0 x 10-4

3 0.10 0.20 2.0 x 10-4

22


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• Compare experiments 1 and 3.• When [A] is held constant and [B] is doubled, the

rate does not change. • The reaction is zeroth order in [B].

• The rate law can be written generically as:Rate = k[A]x[B]y

• The rate law requires experimental determination of the exponents, x and y.

• Therefore, y = 0 and the rate law reduces to:Rate = k[A]x


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• Next compare experiments 1 and 2. • If [A] is doubled, the rate increases by a

factor of 2.• Therefore, (2)x = 2 and x = 1.

• The rate law reduces to:rate = k[A]

• The reaction is first order in [A] and first order overall.


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Concentrations of Reactants: The Rate-Law ExpressionExample 2: The following data were obtained for the

reaction:

2 A(g) + B(g) + 2 C(g) ® 3 D(g) + 2 E(g)

ExperimentInitial [A]

(M)Initial [B]

(M)Initial [C]

(M)Initial rate

(M/s)

1 0.20 0.10 0.10 2.0 x 10-4

2 0.20 0.30 0.20 6.0 x 10-4

3 0.20 0.10 0.30 2.0 x 10-4

4 0.60 0.30 0.40 1.8 x 10-3

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From these initial rate data we can determine the rate-law expression and the rate constant for this reaction.


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• Comparing experiments 1 and 3:• [A] and [B] are held constant, but [C] is

increased by 3-times. • The rate does not change.

• Therefore, the exponent, z = 0, and the rate law simplifies to:

Rate = k[A]x[B]y

Rate = k[A]x[B]y[C]z


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• Comparing experiments 1 and 2:• [A] is held constant, but [B] is increased by 3-

times. • The rate also increases by 3-times.

• Therefore, y = 1, and the rate law further simplifies to:

Rate = k[A]x[B]


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• Comparing experiments 2 and 4:• [B] is held constant, but [A] is increased by 3-

times. • The rate also increases by 3-times.

• The reaction is, thus, first order in A, first order in B and second order overall.

• Therefore, x = 1, and the rate law further simplifies to:

Rate = k[A][B]


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Example 3: A reaction between compounds A and B is determined to be first order in A, first order in B, and second order overall. From the information given below, fill in the blanks.

ExperimentInitial Rate

(M/s)Initial [A]

(M)Initial [B]

(M)1 4.0 x 10-3 0.20 0.050

2 1.6 x 10-2 ? 0.050

3 3.2 x 10-2 0.40 ? 31

Rate = k[A][B]


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33


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Concentration vs. Time: Integrated Rate Laws

The integrated rate equation relates time and concentration for chemical reactions. The integrated rate equation can be used to predict the

amount of product that is produced in a given amount of time.

Provides an alternative to the initial rate method for obtaining rate constants.

Initially we will look at the integrated rate equation for first order reactions. These reactions are 1st order in one reactant and 1st order

overall. For a generic first order reaction with one reactant: A products, rate = k[A]

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or -d[A]/dt = k[A]


where:

t = time elapsed since beginning of reaction.

[A]0= mol/L of A at t = 0 (sometimes called initial conc., [A]i)

[A] = mol/L of A at time t ([A] sometimes written as [A]t)

k = rate constant

Take the integral of this rate law from the beginning (t = 0, [A]0) to [A] at any later reaction time, t.

The integrated rate equation for a 1st order reaction:

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ln([A]/[A]0) = -kt,

MEMORIZE and be able to interconvert these equations!

or[A] = [A]0e-kt

ln([A]0/[A]) = kt, y = mx + bor ln[A] = -kt + ln[A]0

or ln[A]0 - ln[A] = kt,

2N2O5(g) 2 N2O4(g) + O2(g)


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• The reaction is found experimentally to be first order in N2O5 and first order overall.

• Rate = k[N2O5]

plots of the first order integrated rate equations for this reaction…


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[N2O5] = [N2O5]0e-kt

ln[N2O5] = -kt + ln[N2O5]0


Define the half-time, t1/2, of a reaction as the time required for half of the reactant to be consumed, i.e., the time, t, at which [A]=1/2[A]0, then at t1/2, [A]0/[A] = 2.

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ln 2 = 0.693 = kt1/2

• So, the half-time for any overall first order reaction is given by:

t1/2 = 0.693/k MEMORIZE (or take ln 2)

• The half-time is independent of [A]0!

[A]0/[A] = 2 = ekt at t = t1/2


Example: Cyclopropane decomposes to propene according to the following equation:

The reaction is found to be first order in cyclopropane with k = 9.2 s-1 at 1000 0C. What is the half time for disappearance of cyclopropane at 1000 0C?

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t1/2 = 0.693/9.2 s-1 = 0.075 s


Example: Refer to the previous example. How much of a 3.0 g sample of cyclopropane remains after 0.50 seconds? The integrated rate laws can be use for any unit

that’s proportional to moles or concentration. This example uses grams rather than mol/L.

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ln [A]0 - ln [A] = kt

[A]0 = 3, t = 0.50 s, k = 9.2 s-1, solve for [A]:

ln 3 - ln [A] = 9.2 s-1(0.5 s)

1.1 - ln [A] = 4.6

ln [A] = -(4.6-1.1) = -3.5

[A] = e-3.5 = 0.03 g ~ 1% remains

Use the integrated first order rate equation:


Example: The half-time for the first order reaction:

CS2(g) ® CS(g) + S(g)

is 688 hours at 1000 0C. Calculate the rate constant, k, at 1000 0C and the amount of a 3.0 g sample of CS2 that remains after 48 hours.

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k = 0.693/688 hr = 0.00101 hr-1.

Concentration vs. Time: The Integrated Rate Laws

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ln [A]0 - ln [A] = kt

ln (3.0) - ln [A] = (0.00101 hr-1)(48 hr)

1.1 - ln [A] = 0.048

ln [A] = -(0.048 - 1.1) = 1.052

[A] = e1.052 = 2.86g ~ 2.9 g or 97% unreacted

Concentration vs. Time: The Integrated Rate Laws For reactions that are second order with respect to

a single reactant, A, and second order overall, with

rate = k[A]2 = (1/a)(d[A]/dt), the integrated rate law is:

Where a = stoichiometric coefficient of A in the balanced overall equation. If A P, then a = 1. If 2A P, then a = 2

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MEMORIZE!

• An alternative form of the 2nd order integrated rate law:

[A] = [A]0/(1 + akt[A]0).


Half-time, t1/2, for second order reactions: Use the second order integrated rate-law as a starting

point. At the half-time, t1/2 is [A] = 1/2[A]0.

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Concentration vs. Time: The Integrated Rate Equations

If we solve for t1/2:

So the half-time of a second order reaction depends on [A]0. Less [A]0, longer half-time Half-time increases as the reaction proceeds (unlike an

overall first order reaction, where t1/2 remains constant).

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Example: Acetaldehyde, CH3CHO, undergoes gas phase thermal decomposition to methane and carbon monoxide.

The rate law was found to be:

rate = k[CH3CHO]2, and k = 2.0 x 10-2 M-1hr-1 at 527 oC.

(a) What is the half-time (t1/2) for disappearance of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 527 oC? 48


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(b) How many moles of CH3CHO remain after 200 hours in the 1 L vessel?

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(c) What percent of the CH3CHO remains after 200 hours in the 1 L vessel?

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(a) For the same reaction as the previous example, what is the half-time (t1/2) for disappearance of CH3CHO if 0.10 mole is injected into a 10.0 L vessel at 527 oC?

●So, if [A]0 decreases by 10-times, t1/2 must increase by 10-times,

●10 x 5.0 x 102 hr = 5.0 x 103 hr

●Note that the vessel size is increased by a factor of 10 compared to the previous example, which decreases the initial concentration by a factor of 10.

●Recall the equation for half time of this 2nd order reaction (in which the coefficient, a = 1) is:

t1/2 = 1/k[A]0)

For an overall zeroth order reaction the rate expression is:

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Which gives the zeroth order integrated rate equation:

Using Integrated Rate Laws to Determine Reaction Order

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y = mx + b

ln [A] = -kt + ln [A]0

Plots of linear form of the integrated rate equations can determine the reaction order and rate constant.

1st order:

zeroth order: [A] = -kt + [A]0

2nd order:


Example: For the thermal decomposition of ethyl bromide:

C2H5Br(g) C2H4(g) + HBr(g)

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Time (min) 0 1 2 3 4 5

[C2H5Br] 1.00 0.82 0.67 0.55 0.45 0.37

ln [C2H5Br] 0.00 -0.20 -0.40 -0.60 -0.80 -0.99

1/[C2H5Br] 1.0 1.2 1.5 1.8 2.2 2.7

D

Using Rate Integrated Laws to Determine Reaction Order

Make an x,y plot for each of the three sets of values in the table:

1. [C2H5Br] (y-axis) vs. time (x-axis)

If the plot is linear, then the reaction is zeroth order with respect to [C2H5Br].

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2. ln [C2H5Br] (y-axis) vs. time (x-axis) If the plot is linear, then the reaction is first order

with respect to [C2H5Br].

3. 1/ [C2H5Br] (y-axis) vs. time (x-axis) If the plot is linear, then the reaction is second

order with respect to [C2H5Br].

Using Integrated Rate Laws to Determine Reaction Order Plot of [C2H5Br] versus time.

62Is it linear?

[C2H5Br]

Using Rate Laws to Determine Reaction Order

Plot of ln [C2H5Br] versus time.

Is it linear?

ln [C2H5Br]


Plot of 1/[C2H5Br] versus time.

64Is it linear?

1/[C2H5Br]

Using Integrated Rate Laws to Determine Reaction Order ln[C2H5Br] vs. time is the only linear plot. The reaction is, therefore, first order with respect to

[C2H5Br] and first order overall. From the integrated rate law for a first order reaction:

ln[A] = -kt + ln[A]0, slope = -k.

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●Slope = y2 - y1/x2 - x1.

• So k = 0.2 min-1

Using Rate Equations to Determine Reaction Order

Example: Concentration-versus-time data for the reaction:

2NO2(g) --> 2NO(g) + O2(g),

are given in the table below. Plot each of these time functions to determine the rate of the reaction and the value of the rate constant.

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Time(min) 0 1 2 3 4 5[NO2] 1.0 0.53 0.36 0.27 0.22 0.18

ln [NO2] 0.0 -0.63 -1.0 -1.3 -1.5 -1.7

1/[NO2] 1.0 1.9 2.8 3.7 4.6 5.5


Plot of [NO2] versus time.

69Is it linear?

[NO2]

Using Rate Equations to Determine Reaction Order Plot of ln [NO2] versus time.

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Is it linear?

ln [NO2]


Plot of 1/[NO2] versus time.

71Is it linear?

1/[NO2]


1/[NO2] vs. time is the only linear plot. The reaction is, therefore, second order in

[NO2] and second order overall.

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●Determine the value of the rate constant from the slope of the line.


From the equation for a second order reaction the slope = a k In this reaction a = 2.

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Slope = 0.90 = 2 k

k = 0.45 M-1min-1

Collision Theory of Reaction Rates

Three basic events must occur for a bimolecular reaction to occur.

The atoms, molecules or ions must:1. Collide.

2. Collide with enough energy to break and form bonds.

3. Collide with the proper orientation for a reaction to occur.

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For a simple second order reaction between two different gaseous atoms, A and B:

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A B

A B

A B BA B

A BA BA B

4 different possible A-B collisions



A(g) + B(g) products, where rate = k[A][B]

Increasing the number of atoms per unit volume (equivalent to increasing their concentrations) increases the probability of A/B collisions per unit time and, thereby increases the reaction rate.


An example of an effective collision is:

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X Y

X Y

X Y

X Y

X Y +

X Y

Colliding molecules (i.e., individual reactants that contain more than one atom) must be oriented properly for reaction.

X2(g) + Y2(g) 2 XY(g)


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X

X

Y YY

Y

X X X X Y Y

Some possible ineffective collisions are :

Collision Theory of Reaction Rates Effective and ineffective molecular collisions for:

NO + N2O --> NO2 + N2

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Effective

reactants collision

products

Ineffective (wrong orientation)

reactants collisionreactants

(blue spheres = N, red spheres = O)

Reaction Mechanisms and the Rate-Law Expression Many chemical reactions occur by more than one step. The sequence of individual chemical steps by which the

reaction occurs is called the reaction mechanism. Each step consists of either a unimolecular or bimolecular

chemical reaction. The sum of the individual steps must equal that of the overall

balanced reaction. Rate laws can be used to propose reaction mechanisms.

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Reaction Mechanisms and the Rate-Law Expression The slowest step (the “bottleneck”) in a reaction mechanism

is the rate-determining step (r.d.s.) (sometimes called the rate-limiting step).

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Rules for a mechanism to be consistent with the experimentally determined rate law.

Reaction orders indicate the number of molecules involved in: The r.d.s. only

or The r.d.s. and any steps preceding the r.d.s.

Any step after the r.d.s. cannot be part of the rate law. Sum of the steps in the mechanism must equal the

balanced equation for the overall reaction. Intermediates in a mechanism cannot appear in the rate

law.

Reaction Mechanisms and the Rate-Law Expression

Use the experimental rate law to postulate a mechanism.

Consider the iodide ion-catalyzed decomposition of hydrogen peroxide to water and oxygen.

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2H2O2(aq) ---> 2H2O(l) + O2(g)

I-


This reaction has been experimentally determined to be first order in H2O2, first order in I- , and second order overall, i.e. Rate = k[H2O2][I-]

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Step 1, slow: H2O2 + I- --> IO- +H2O

Step 2, fast: IO- + H2O2 --> H2O + O2 + I-

Overall reaction: 2H2O2 --> 2H2O + O2

A mechanism consistent with this rate law is:


According to this mechanism: One hydrogen peroxide molecule and one iodide

ion react in the slow step (the r.d.s.).

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The iodide ion functions as a catalyst: it is consumed in step 1 and regenerated in step 2.

Hypoiodite (IO-) should not accumulate appreciably because it is produced in the r.d.s., then rapidly consumed in the subsequent fast reaction with hydrogen peroxide. (IO- has been detected in very small amounts as a

short-lived reaction intermediate).


Ozone, O3, reacts very rapidly with nitric oxide, NO, in a reaction that is first order in each reactant and second order overall.

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Reaction Mechanisms and the Rate-Law Expression One possible mechanism that is consistent with the

rate law is:

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k1

k2

Rate = k1[O3][NO]

• k2 >> k1, so the slow step is the “bottleneck” which limits the overall reaction rate

• k1 = k, i.e., the rate constant for the overall reaction


A mechanism that is inconsistent with the rate-law is:

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• Experimentally determined rate laws can, thus, DISPROVE, but cannot, by themselves, PROVE a particular mechanism


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• Consider the reaction:

2NO(g) + Br2(g) 2NOBr(g)

• The experimentally determined rate law is:

Rate = k[NO]2[Br2]• One possible mechanism consistent with the rate law is a simultaneous collision and reaction of two NO molecules and one Br2 molecule. However, simultaneous three-body collisions are highly improbable.

• More likely mechanisms would consist of a sequence of bimolecular steps.


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NO + Br2 NOBr2 fast pre-equilibrium

NOBr2 + NO ---> 2NOBr2NO + Br2 --> 2NOBr overall reaction

• According to this mechanism, the rate law for the slow step is:

Rate = k2[NOBr2][NO]

How to reconcile with the experimental rate law, rate = k[NO]2[Br2]?• For the fast pre-equilibrium, the rates of the forward and reverse reactions must be equal: k1[NO][Br2] = k-1[NOBr2].

• Rearrange to [NOBr2] = k1/k-1[NO][Br2], substitute into the rate law for the r.d.s.:

• Rate = k2(k1/k-1[NO][Br2])[NO] = k[NO]2[Br2], where k = k2k1/k-1

k2

k-1

k1

slow

Transition State Theory

Transition state theory postulates that the reactants form a high energy intermediate, the transition state, which then rearranges into the lower energy products.

For a reaction to occur, the reactants must acquire sufficient energy to form the transition state. This energy is called the activation energy, Ea.

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95

Energy changes during a chemical reaction.

Ea = activation energy

transition state

ΔE ~ ΔH

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


96

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Transition State TheoryNO2(g) + CO(g) NO(g) + CO2(g)

experimental rate law: Rate = k[NO2]2

• Step 1) is the r.d.s (k1 << k2).

• Step 1) is slower than step 2) because Ea1 > Ea2.

• The rate law of step 1) is the same as the rate law of the overall reaction.

A mechanism consistent with the rate law:

step 1: NO2(g) + NO2(g) NO3(g) + NO(g) Rate1 = k1[NO2]2 slow

step 2: NO3(g) + CO(g) NO2(g) + CO2(g) Rate2 = k2[NO3][CO] fast

Transition State Theory The relationship between the activation energy

for forward and reverse reactions is: Forward reaction activation energy = Ea

Reverse reaction activation energy = Ea + DE

difference = DE (~DHrxn, a thermodynamic quantity)

Higher Ea, slower reaction, lower rate constant, k. In order for a reaction to reach the transition state and

convert to products, the combined kinetic energy of the colliding molecules must be ≥ Ea.

One way to increase kinetic energy of molecules is to raise the temperature.

Therefore, the rates of most chemical reaction increase with increasing temperature.

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Temperature Effects: Transition State Theory

Molecular kinetic energies increase as the temperature is increased.

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Temperature Effects: Transition State Theory One method to increase the energy necessary to

break and reform bonds is to heat the molecules. Recall the reaction of methane with molecular

oxygen is exothermic (DH0 < 0) but slow at 25 0C:

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The reaction can be started with a spark or match. This provides the initial activation energy necessary to break

the first few bonds. Afterwards the reaction is self-sustaining, i.e., heat

generated by reaction of the first few molecules provides the activation energy to break bonds in other molecules.

CH4(g) + O2(g) CO2(g) + H2O(g) H0 = -880 kJ/mol

Temperature Effects: The Arrhenius Equation

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MEMORIZE and be able to use the Arrhenius equation!

• k is the rate constant

• T is the absolute temperature

• Ea is the activation energy

• R is the gas constant (8.314 J/mol∙K)

• A is the “pre-exponential factor” or “frequency factor” • The number of approaches to the activation barrier per unit time.

• e(-Ea/RT) is the “exponential factor” • the fraction of molecules that convert to product after approaching

the activation barrier.

or

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Temp, K k, M-1∙s-1 Temp, K k, M-1∙s-1

600 3.37 x 103 1300 7.83 x 107

700 4.83 x 104 1400 1.45 x 108

800 3.58 x 105 1500 2.46 x 108

900 1.70 x 106 1600 3.93 x 108

1000 5.90 x 106 1700 5.93 x 108

1100 1.63 x 107 1800 8.55 x 108

1200 3.81 x 107 1900 1.19 x 109


Example: Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data:

105

Plot these data as lnk vs I/T:


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Ea = m(-R)

solve for Ea

A = ey-intercept

solve for A 11-11

118.26

sM 1036.4

1036.4

A

eA


y = mx + b, where m = -(Ea/R) and b = ln(A)


If the Arrhenius equation is written for two temperatures, T2 and T1 with T2 > T1.

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2

a2

1

a1

-Aln ln

and

-Aln ln

RT

Ek

RT

Ek

Temperature Effects: The Arrhenius EquationSubtract one equation from the other:

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Temperature Effects: The Arrhenius Equation Consider the rate of a reaction for which Ea = 50

kJ/mol, at 20 oC (293 K) and at 30 oC (303 K). How much do the two rates differ?

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Temperature Effects: The Arrhenius Equation Many reactions have Ea »50 kJ/mol.

For these reactions the rate approximately doubles for every 10 0C rise in temperature (near room temperature).

This is a good “rule of thumb” for dependence of reaction rates (strictly speaking, rate constants) on temperature.

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Catalysts

Catalysts increase reaction rates by providing an alternative reaction pathway with a lower activation energy.

There is no net consumption of the catalyst during the reaction it catalyzes.

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Catalysts

Homogeneous catalysts exist in same phase as the reactants.

Heterogeneous catalysts exist in different phases than the reactants. Solid catalysts often catalyze solution reactions.

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2H2O2 --> O2(g) + 2H2OMnO2

Catalysts

Examples of commercial catalyst systems include:

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Catalysts

This movie shows catalytic converter chemistry on the molecular scale

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Catalysts

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Catalytic converters also catalyze oxidation of CO to CO2

Overall reaction

2 CO(g)+ O2(g) ® 2CO2(g)

Adsorption

CO(g) ® CO(surface)

O2(g) ® O2(surface)

Activation

O2(surface) ® 2 O(surface)

Reaction

CO(surface) +O(surface) ® CO2(surface) Desorption

CO2(surface) ® CO2(g)

Catalysts The Haber process is used industrially for “fixing” nitrogen

by producing ammonia, NH3, from N2 and H2:

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N2(g) + 3H2(g) --------> 2NH3(g)

Fe,Fe2O3

• A good illustration of the interplay between kinetics and thermodynamics.

• This reaction is actually an equilibrium that lies far towards the right at 25 0C and is also exothermic.

N2(g) + 3H2(g) 2NH3(g) Kc = 108 at 25 0C, H = -92 kJ/mol

• However, the rate of this reaction is negligible at 25 0C.• The reaction is carried out at 500 0C with catalyst to increase the rate.

• But high temperature favors the reverse reaction (because it’s exothermic), so high pressure (200-1000 atm) is used to shift the equilibrium to the right (fewer moles of gaseous product than reactants).

The Haber Process: Balancing Factors that Affect Equilibria

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Enzymes

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Catalase: 2H2O2 O2(g) + 2H2O

Enzymes are biological catalysts (usually proteins) that: increase reaction rates provide specificity, optimized to bind only particular

substrates and convert them to single sets of products.

After studying Tro Chapter 13 text and lecture and working the problems, you should be able to:

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1. Determine reaction orders, rate laws, and values of rate constants given a set of initial rates and initial concentrations.

2. Identify overall orders of reactions and order in each reactant given a rate law.

3. Know units for zeroth, 1st and 2nd order rate constants.4. Write the rate laws for simple overall zeroth, 1st and 2nd order

reactions.5. Write and use the integrated rate equations for zeroth, 1st and 2nd

order rate laws with one reactant.6. Describe how concentrations change with time or how the rate will

change for a given concentration change for each of the rate laws in #3.

7. Use these rate laws to determine: half-times, concentrations remaining after a given reaction time or from a given starting concentration and rate constants.

8. Identify or draw and describe equations for linear plots of zeroth, 1st and 2nd order reactions.

9. Determine order of reaction given plots in #8.

After studying Tro Chapter 13 text and lecture and working the problems, you should be able to:10. Write reasonable mechanisms given rate laws and vice versa. 11. Define the rate-determining step of a reaction mechanism.12. Effects of collisions and molecular orientations on reactions.13. Diffusion-controlled reactions are limited only by collision rates.14. Define activation energy draw reaction progress diagrams and

label activation energy, transition state and DHrxn.15. Qualitatively describe temperature dependence of reactions in

terms of molecular kinetic energies. 16. Qualitatively draw and/or interpret Maxwell-Boltzmann distributions

of molecular kinetic energies and label areas equal to and exceeding activation energy.

17. Memorize and use Arrhenius equation(s) to calculate temperature dependence of reaction rates and activation energies.

18. Define catalyst and its effect on activation energy and reaction rate.19. Not responsible for enzyme catalysis or chain reactions.

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Mid-term Exam 4

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Chapter 18 (Electrochemistry) including:

1. Use the Nernst equation (memorize) to calculate Ecell under non-standard conditions given half-reaction E0 values, initial concentrations and the Nernst equation.

2. Calculate values of K and Grxn from E0cell using the Nernst equation and

G = -nFEcell (memorize).

3. Describe and calculate Eo for a concentration cell.

4. Identify components in an alkaline dry cell given the half reactions and a diagram of the cell. Show direction of current flow, and explain why voltage stays fairly constant during discharge.

5. Draw a diagram of a lead acid car battery given the half reactions, and describe the discharge and recharge cycles.

6. Describe and draw a diagram of a H2/O2 fuel cell.

7. Given E0’ values (pH 7), describe and calculate how energy can be derived from biological redox reactions.

All of Chapter 13 (kinetics)

tro chapter 13 kinetics spring 2015 (1)

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