Chemical Kinetics

Suggested HW (Ch 17): 6, 12, 18, 22, 42, 47 50

About Kinetics

Chemists ask three questions when studying chemical reactions:1. What happens?2. How fast?3. To what extent?

The first question is addressed by writing balanced chemical reactions, calculating yields, making qualitative observations, etc.

The second question is addressed in an area of chemistry called KINETICS, which will be covered over the next two lectures

The third question will be answered in chapter 19 (Equilibrium)

Reactions Rates

Different reactions proceed with different rates

The rate of a reaction is defined as the decrease in the concentration of reactant per unit time (or the increase in the concentration of product)

The rate of a reaction depends on several factors, including:reactant concentrationtemperaturecatalystssurface area

Today, we will focus exclusively on the relationship between reaction rates and reactant concentration

Intro

Lets take the reaction: A ---> B.

This reaction tells us that as A is consumed, B is formed at an equal rate. We can express this mathematically in terms of changing concentrations by:

Imagine we have 10 moles of A in 1 L of solution. If we can freeze time for an instant, such that the reaction has not yet begin (t=0), the concentration of A is 10M.

−∆ [𝐴 ]∆𝑡

=∆ [𝐵]∆ 𝑡

= 1 mol of A

[A] = 10 M

t = 0

A B

After 10 seconds, 3 moles of B have formed.

[A] = 7 M[B] = 3 M

t = 10

= 1 mol of B

10 more seconds

[A] = 5 M[B] = 5 M

t = 20

20 more seconds

[A] = 4 M[B] = 6 M

40 more seconds

[A] = 3 M[B] = 7 M

t = 40

t = 80

0 10 20 30 40 50 60 70 800

2

4

6

8

10A ---> B

Time (sec)

Co

ncen

trati

on

(M

ol/

L)

A

B

Plotting the data from previous slide

Reactions Follow a Rate Law

The graph in the previous slide shows that the disappearance of A (formation of B) is not linear.

As the reactant concentration decreases, the reaction slows down.

This dependence of rate on concentration suggests that reaction rates follow a rate law, a mathematical expression that ties concentration and rate together

Instantaneous Rates

0 20 40 60 800

2

4

6

8

10 A ---> B

Time (sec)

Con

cen

trati

on

(M

ol/

L)• The rate of the reaction is constantly

changing, but we can determine instantaneous rates (reaction rate at a specific time and concentration)

• Instantaneous rate at t=0 is the initial rate

• We can determine the instantaneous rate by taking the slope of the tangent at the point of interest

• Note: a tangent line is linear and ONLY touches the point in question. It does NOT cross the curve

Instantaneous rate of disappearance of A at t=20 sec

𝑟𝑎𝑡𝑒20=−8−0𝑀0−50 𝑠

=− .160𝑀𝑠

Rates and Stochiometry

In the previous example (A---->B), we had 1:1 stoichiometry. Thus, at any given time, the rate of disappearance of A equals the rate of formation of B. If the stoichiometry is NOT 1:1, we have a much different situation, as shown below:

As you can see, 2 moles of HI are consumed for every 1 mole of H2 and 1 mole of I2 formed. Thus, the disappearance of HI is twice as fast as the appearance of the products.

2𝐻𝐼 (𝑔 )→𝐻2 (𝑔)+ 𝐼2(𝑔)

Example: N2O5(g) ----> 2NO2(g) + ½ O2(g)

6.4 𝑥10−4𝑀𝑚𝑖𝑛

1 .6𝑥 10− 4𝑀𝑚𝑖𝑛

Looking at average ratesaverage rate of disappearance after 10 minutes

average rate of disappearance after 100 minutes

2 .38 𝑥10− 4𝑀𝑚𝑖𝑛

0.59 𝑥10− 4𝑀𝑚𝑖𝑛

N2O5(g) ----> 2NO2(g) + ½ O2(g)

fast

slow

Rate Laws

We see that reducing reactant concentration lowers the reaction rate, but to what extent? What is the mathematical correlation?

The equation that relates the concentration of the reactants to the rate of reaction is called the rate law of the reaction.

We can derive the rate law of a reaction by seeing HOW THE REACTION RATE CHANGES WITH REACTANT CONCENTRATION.

For any reaction aA + bB……. ----> cC + dD…..

In this expression, k is the rate constant, m and n are reaction orders.

𝑹𝒂𝒕𝒆=𝒌¿

Lets go back to the previous reaction: Below is a table of data, showing the initial reaction rate as a function of the

starting concentration of N2O5 (g). We perform multiple experiments to collect enough data to determine our rate law.

Our rate law will be in the form: Rate = k [N2O5]m.

Never include products in a rate law !!!!!

Let’s use the data in the table to obtain the value of m. We must pay attention to how the rate is affected by [N2O5]

Reaction Orders and the Method of Initial Rates

N2O5(g) 2NO2(g) + ½ O2(g)

Experiment [N2O5]o (M) Rate, M/s

1 0.010 .018

2 0.020 .036

3 0.040 .072

Let’s solve for m. We can choose any two sets of data values to set up a system of equations. We’ll choose data from experiments 1 & 2.

Now, we have two equations and two variables (k and m), but the k values can cancel, allowing us to solve for m as shown below:

Rate Laws

0.36𝑀𝑠− 1

0.18𝑀𝑠− 1=𝑘¿ ¿

2=[2 ]𝑚

𝑚=1

Reaction Orders (example continued)

This means that the reaction is FIRST ORDER WITH RESPECT TO [N2O5]

When we double [N2O5]o, the rate also doubles. When we quadruple [N2O5]o, the rate quadruples.

Physically speaking, a reaction order of 1 means that one molecule of the reactant is involved in the formation of products. Thus, the rate is directly proportional to [N2O5]o by the rate constant, k.

We can write the rate law as:

The overall reaction order is the sum of the individual reaction orders. (1st order reaction). We can easily solve for k by plugging in any corresponding rate and concentration (k = 1.8 s-1)

𝑹𝒂𝒕𝒆=𝒌¿

Rate Laws/Reaction Orders

𝑹𝒂𝒕𝒆=𝒌¿

• Reaction orders of non-elementary reactions must be determined experimentally. You can not assume based on the stoichiometry.

• When you have multiple reactants, you must determine the reaction order of each one. • To do this, you must vary the concentration of only one reactant

at a time while holding the others fixed.

• Let’s attempt to determine the rate law for the reaction below:2NO(g) + O2(g) ---> 2NO2

Example: 2NO(g) + O2(g) ---> 2NO2

Using the data below, determine the rate law of this reaction in the form:

𝑹𝒂𝒕𝒆=𝒌¿

Experiment

[NO]o (M) [O2]o (M) Rate (M/s)

1 .0126 .0125 2.82 x 10-2

2 .0252 .0250 1.13 x 10-1

3 .0252 .0125 5.64 x 10-2

• This time, we have two reactants. Lets start by determining the value of ‘m’. To do so, we hold [O2]o fixed and vary [NO]o. This will show how the rate depends on [NO]o.

• In experiments #1 and #3, [O2]o is fixed, so we will use these experiments to find ‘m’.

𝑹𝒂𝒕𝒆=𝒌¿

Experiment [NO]o (M) [O2]o (M) Rate (M/s)1 .0126 .0125 2.82 x 10-2

2 .0252 .0250 1.13 x 10-1

3 .0252 .0125 5.64 x 10-2

• Remember, rate is proportional to [NO] by the power m. The factor of change in the rate is equal to the factor of change of [NO] to the mth power:

factor of rate change factor of change

in [NO]

order

m = 1

• The reaction is 1st order with respect to [NO]

( 5.64 𝑥10− 22.82 𝑥10− 2 )=( .0252.0126 )𝑚

2=2𝑚

𝑹𝒂𝒕𝒆=𝒌¿

Run [NO]o (M) [O2]o (M) Rate (M/s)1 .0126 .0125 2.82 x 10-2

2 .0252 .0250 1.13 x 10-1

3 .0252 .0125 5.64 x 10-2

factor of rate change factor of change in [O2]

order

n = 1𝑹𝒂𝒕𝒆=𝒌¿The reaction is 1st order with respect to [NO], 1st order with respect to [O2] and 2nd order overall.

• Now we can find ‘n’ by varying [O2]o and holding [NO]o fixed. We can use experiments #2 and #3 for this. This will show how the rate depends on [O2]o.

( 5.64 𝑥10− 21.13 𝑥10− 1 )=( .0125.025 )𝑛

( 12 )=(12 )𝑛

k = 179 M-1s-1

Pay Attention to the Units of k, As They Change with Overall Reaction Order

The rate constant, k, is the constant of proportionality between rate and concentration. Higher values of k = faster reactions

It is important to note that the units of k depend on the overall reaction order. Ex:

Rate is always in units of molarity per unit time (sec, hr, etc). Concentrations are always expressed as molarities (M or mol/L). Thus, we have:

Recall for a 1st order reaction:

For zth order (z = any integer)

𝑹𝒂𝒕𝒆=𝒌¿

𝑀𝑠

=𝑘 (𝑀 ) (𝑀 )→𝑘=1𝑀𝑠

=𝑀−1𝑠−1 Units of k for a 2nd order reaction

𝑘=𝑠−1 Units of k for a 1st order reaction

𝑘=𝑀−(𝑧 −1 )𝑠− 1

Group Example

Determine the relative (m & n) and overall (m+n) reaction order of the reaction below. Then, derive the rate law and determine the value of k.

N𝑂2+𝐶𝑂→𝑁𝑂+𝐶𝑂2

𝑹𝒂𝒕𝒆=𝒌¿

Experiment [NO2]o (M) [CO]o (M) Rate (M/s)1 .0300 .200 1 x 105

2 .0900 .200 9 x 105

3 .300 .0400 1 x 107

4 .300 .0800 1 x 107

• Tripling [NO2] causes the rate to increase nine-fold. This means that the rate is squarely proportional to [NO2], so the reaction is second order with respect to NO2 (m=2). Doubling [CO] does nothing. Thus, the rate does not depend on [CO], and is zero order with respect to CO (m=0). Overall 2nd order.

𝑹𝒂𝒕𝒆=𝒌¿ k = 1.11 x 108 M-1s-1

Determining the Overall Rate Order of A Reaction Graphically As we have shown, a first-order reaction depends on the concentration of

a single reactant to the 1st power. For the reaction: A----> products

Using calculus, we can convert this to:

This equation is in y = mx + b form. Therefore, for any 1st order reaction, the plot of the natural log of [A]t vs time will be linear. The slope of the line will be –k.

𝐥𝐧 ¿natural log of concentration at time t natural log of starting

concentrationrate constanty

axism (slope)

b

𝑹𝒂𝒕𝒆=𝒌[𝑨]

time

x axis

Plotting 1st Order Reactions

𝐥𝐧 ¿b

time values on x-axis

slope = -kunits: s-

1

natural log of [A]t on y-axis

Determining the Overall Rate Order of A Reaction graphically

A second-order reaction depends on the concentration of [A] to the 2nd power. For the reaction: A ----> B

Therefore, for any 2nd order reaction, the plot of the inverse of [A]t vs time will be linear. The slope of the line will be k.

𝑹𝒂𝒕𝒆=𝒌¿𝟏¿¿y

m•x

b

Important!!! Natural logs (ln) are exponential terms! To solve for the value of a

term within a natural log function, use the exponential function on your calculator (ex)

example : ln ( x )=0.3 solve for x

e ln (x)=e0.3

e ln ( x )=x

Plotting a 2nd Order Reaction

𝟏¿¿

b

slope = kunits = M-

1 s-1

time values on x-axis

1/[A]t on y-axis

Determining Overall Rate Order From Plotting Time-Dependent Data

• We can determine if a process is first or second order by plotting the data against both equations. Which ever fitting method yields a linear plot gives the overall order.

𝐥𝐧 ¿ 𝟏¿¿

not linear:NOT 1st order

linear!2nd order

Example

From the given data, write the rate law of the reaction, including k. What is the rate of the reaction at t=4 seconds? t=5 seconds? What concentration of A remains at t=10 seconds?

Suggested HW: Ch 18: 38, 41, 44, 47

Ch 18: Kinetics Pt. 2 Temperature Dependence of Rate

Constants

Temperature and Rate

The rates of most chemical reactions increase with temperature.

How is this temperature dependence reflected in the rate expression?

Rates increase with temperature because rate constants increase with temperature. An example is the 1st order reaction:

CH3NC ---> CH3CN

𝒓𝒂𝒕𝒆=𝒌[𝑪 𝑯𝟑𝑵𝑪 ]

Variation in k with temperature.

Collision Model

The collision model makes sense of this. This model is based on kinetic theory

We’ve seen that the thermal energy of a molecule is converted to kinetic energy in order to facilitate motion, and we’ve seen that the velocity of a molecule increases with T.

The central idea of the collision model is that molecules must collide to react. The more collisions per second, the faster the reaction goes.

This model also rationalizes the concentration dependence. The more molecules present, the more collisions you have.

𝑉 𝑉

Activation Energy

Of course, there is more to a chemical reaction than just collisions of molecules.

Molecules must have some minimum energy in order to overcome the energy barrier of reaction

Upon colliding, the kinetic energy of molecules is used to stretch, bend, and break bonds in order to cause a reaction.

But if the molecules don’t have enough kinetic energy, they simply bounce off one another.

This minimum energy that molecules must have is called the activation energy,(Ea)

𝑉 𝑉 𝑉𝑉

E too low

Activation Energy Needed to Reach Transition State

• Lets go back to the reaction CH3NC ---> CH3CN

• The molecule passes through a transition state, a high-energy intermediate in which the CN bond gets rotated 90o.

• This is an unstable configuration and requires an activation energy to be reached.

• Once this transition state is reached, the process is energetically downhill. The transition state is always the highest energy point in the reaction pathway. As drawn, this reaction is exothermic because the energy of the products is less than the energy of the reactants.

Transition state

nitrogen

carbon

ΔH

En

thal

py

Arrhenius noted that reaction-rate data depended on three aspects(1) the fraction of molecules possessing an energy Ea or greater(2) the number of collisions per second(3) the fraction of molecules oriented in the right way for a reaction to

occur

These factors are incorporated into the Arrhenius Equation

This equation relates k to temperature. A is the frequency factor, which is related to criteria 2) and 3) above, and the fraction of particles possessing an energy of Ea or greater is given by

e−E aRT

𝐤=𝐀𝒆−𝑬𝒂

𝑹𝑻

Rearranging the Arrhenius Equation

Taking the natural log of both sides, the Arrhenius equation becomes:

𝐥𝐧𝒌=−𝑬𝒂

𝑹𝑻+𝐥𝐧 𝑨

• As you see, we once again have an equation in y=mx+b form.

y=ln k𝑚 (𝑠𝑙𝑜𝑝𝑒 )=−

𝐸𝑎

𝑅

𝑥=1𝑇

Plotting ln k vs 1/T yields a linear plot with slope –Ea/R and a y-intercept of ln A

Showing Changes in k Graphically

From the data below, determine the activation energy of the 1st order reaction

Temperature, oC K (s-1)189.7 2.52 x 10-5

198.9 5.25 x 10-5

230.3 6.30 x 10-4

251.2 3.16 x 10-3

𝐥𝐧𝒌=−𝑬𝒂

𝑹𝑻+𝐥𝐧 𝑨

Plot ln k vs 1/T to get the slope. We know R, so we can calculate Ea. T must be in Kelvin.

1/T (oK-1) ln K

0.002161 -10.5887

0.002119 -9.8547

0.001987 -7.36979

0.001908 -5.75718

0.0018 0.0019 0.002 0.0021 0.0022

-12

-10

-8

-6

-4

-2

0

f(x) = − 19038.464147378 x + 30.516686342063

1/Tln

k

𝑠𝑙𝑜𝑝𝑒=−19038=−𝐸𝑎

𝑅=−

𝐸𝑎

(8.314 𝐽𝑚𝑜𝑙 𝐾

)

𝑬𝒂=𝟏 .𝟓𝟖 𝒙𝟏𝟎𝟓 𝑱𝒎𝒐𝒍

Calculating Changes in k Mathematically

The Arrhenius equation can be rearranged once again to compare k values at different temperatures

Remember:

ln𝑘2𝑘1

=𝐸𝑎

𝑅 (𝑇 2−𝑇1

𝑇1𝑇2)

𝑒ln (𝑥)=𝑥

Example

The reaction above has an activation energy of 43.5 kJ/mol. The reaction occurs at 298 K with a rate constant of 110 s-1. What will the rate constant be if we increase the temperature to 308 K?

ln𝑘2𝑘1

=𝐸𝑎

𝑅 (𝑇 2−𝑇1

𝑇1𝑇2)

ln𝑘2110

= 43500 𝐽𝑚𝑜𝑙− 1

8.314 𝐽 𝑚𝑜𝑙−1𝐾− 1 ( 308−29891784 )ln

𝑘2110

=.57

𝑒( 𝑙𝑛 𝑘2

110)=𝑒.57𝑘2110

=1.77

𝑘2=194.7 𝑠−1

Reaction Mechanisms

Consider the following reaction:

In looking at this equation, and applying your understanding of collision theory, you would assume that this reaction proceeds when NO2 and CO collideThe activation energy is exceededThe atoms rearrange to form NO and CO2

• Thus, we would expect the rate of the reaction to depend on both [NO2] and [CO]

𝑵 𝑶𝟐 (𝒈 )+𝑪𝑶 (𝒈 )→𝑵𝑶 (𝒈 )+𝑪𝑶𝟐(𝒈)

Reaction Mechanisms

However, according to experimental data, the reaction is 2nd order with respect to NO2 and zero order with respect to CO. The rate law is:

𝑟𝑎𝑡𝑒=𝑘 ¿

• This means that the rate of the reaction does not depend on [CO] at all. How can this be? What does this tell us about the reaction mechanism?

• This particular reaction must proceed through multiple steps, and the rate-determining step must only depend on NO2

𝑵 𝑶𝟐 (𝒈 )+𝑪𝑶 (𝒈 )→𝑵𝑶 (𝒈 )+𝑪𝑶𝟐(𝒈)

Reaction Mechanisms

(1 ) 𝑁𝑂2 (𝑔 )+𝑁𝑂2 (𝑔)→𝑁𝑂3 (𝑔 )+𝑁𝑂 (𝑔)(𝑠𝑙𝑜𝑤)• For this particular reaction, there are two steps:

(3 )𝑁𝑂2 (𝑔)+𝐶𝑂 (𝑔)→𝑁𝑂 (𝑔 )+𝐶𝑂2 (𝑔 )

• This step-by-step description of the molecular pathway is the reaction mechanism. From reaction (1), we can physically understand why the reaction is 2nd order with respect to [NO2].

• The 1st step is substantially slower than the 2nd. Therefore, step (1) acts as the “reaction bottleneck”, and the overall reaction speed can only be as fast as its slowest step.

• Thus, step (1) is the rate determining step.

𝑵 𝑶𝟐 (𝒈 )+𝑪𝑶 (𝒈 )→𝑵𝑶 (𝒈 )+𝑪𝑶𝟐(𝒈)

Each step in a reaction mechanism is called an elementary reaction.

An elementary reaction is any reaction that proceeds in a single step. Mechanisms consist of sums of elementary reactions.

Going back to rate laws, each elementary reaction must have a corresponding rate constant

*** For elementary reactions only, you can assume that the rate law depends directly on the number of species present. Thus, the rate law of an elementary reaction follows the stoichiometry.

(1 ) 𝑁𝑂2 (𝑔 )+𝑁𝑂2 (𝑔) 𝑁𝑂3 (𝑔)+𝑁𝑂 (𝑔 )(𝑠𝑙𝑜𝑤)

k1

k2

𝑟𝑎𝑡𝑒1=𝑘1 [𝑁𝑂2 ] [𝑁𝑂2 ]=𝑘1¿ rate of overall reaction

intermediate (transition step)

Examples

Write the rate law for the following reaction given the elementary reaction steps of the mechanism:

2𝑁𝑂2 (𝑔)2𝑁𝑂 (𝑔 )+𝑂2(𝑔)

(1 ) 𝑁𝑂2 (𝑔 )+𝑁𝑂2 (𝑔 ) 𝑁𝑂 (𝑔 )+𝑁𝑂3 (𝑔)(𝑠𝑙𝑜𝑤)

(2 ) 𝑁𝑂3 (𝑔) 𝑁𝑂 (𝑔 )+𝑂2 (𝑔)( 𝑓𝑎𝑠𝑡 )

k1

k2

Catalysis

The rates of reactions can be increased by using catalysts. Catalysts work by providing a different reaction path (mechanism)

between reactants and products with a lower activation energy The catalyst itself is not consumed in the reaction.

Example: Oxidation of Sulfur Dioxide Uncatalyzed

2SO2(g) + O2(g) 2SO3(g)

(1) SO2(g) ---> S(s) + O2(g) ΔHorxn (1) = 296.8 kJ/mol (slow step)

(2) 2S(s) + 3O2(g) ---> 2SO3(g) ΔHorxn (2) = -791. 4 kJ/mol

2SO2

Formation of S(s)

combustion of S to SO3

ΔHorxn

(Ea)ΔHo1 ΔHo2

Example: Oxidation of Sulfur Dioxide Pt-Catalyzed

2SO2(g) + O2(g) 2SO3(g)

(1) O2(g) ---> 2O ΔHorxn (1) = 93.4 kJ/mol (slow step)

(2) SO2(s) + O(g) ---> SO3(g) ΔHorxn (2) = -791. 4 kJ/mol

Pt

2SO2

Formation of O

ΔH

Ea

Ea,cat

Example: Oxidation of Sulfur Dioxide Pt-Catalyzed