Kinetics 3 OUT OF 75 M/C QUESTIONS FREE RESPONSE—ALMOST EVERY YEAR Chapter 12

Kinetics

3 OUT OF 75 M/C QUESTIONSFREE RESPONSE—ALMOST

EVERY YEAR

Chapter 12

Chemical Kinetics

Study ofReaction ratesReaction mechanism—series of steps by which a reaction takes place

Rate of Reaction the change in concentration of a

reactant or product per unit time

Rate = [A]t2 – [A]t1 / (t2 – t1)

OR Rate = D[A] / Dt Usually considered positive even if

concentration is decreasing.

The square brackets indicate concentration of a substance.

Example Problem

Nitrogen dioxide decomposes to nitrogen monoxide and oxygen gas. At the beginning of the reaction, a pressurized 1-L flask contains 2.0 mol of nitrogen dioxide. At the end of 20 minutes, the flask contains 1.5 mol of nitrogen dioxide and 0.5 mol of nitrogen monoxide. What is the rate of the reaction in terms of M/s of nitrogen dioxide?

Example Problem Solution• Important information:

2.0 mol/ L nitrogen dioxide initially After 20 minutes, 1.5 mol/L of nitrogen dioxide• Change in concentration: 2.0 mol – 1.5 mol = 0.5 mol• Change in time: 20 min• Rate: 0.5 mol / 20 min = 0.025 mol/min

Average vs. Instantaneous Rate

Rate is usually not constant, but decreases over time

Average rate—total change over total time (like the previous example)

Instantaneous rate—rate at a particular instant in time

Instantaneous Rate

Calculated by line’s slope tangent to the point in time on the rate curve Page

558

Think about a slope: change in "y” over change in “x”—It’s the same thing as the average rate if y is the concentration and x is the time.

Rate Law

A mathematical expression which describes the dependence of reaction rate on concentration of reactants

Rate = k[A]n

k = proportionality constant

This is a very important concept!

The [A] means concentration of substance A. The exponent “n” is known as the order of the reaction. The order must be determined experimentally from data.

The lower case k is the rate constant. It must be lower case because upper case K is an equilibrium constant.

Rate Law[A] = concentration of

reactant(s)n = order of reactantBoth k and n must be

experimentally determinedCannot determine from

balanced equation!!!

Write the rate law

2NO2 2NO + O2 if NO2 is second order

Rate = k [NO2]2

2H2 + 2NO N2 + 2H2O if NO is second order and H2 is first order

Rate law always begins with “rate = k….”

Next, put the formula for each reactant in brackets and raise it to the power of the order.

You don’t need to include zero order reactants since anything raised to the zero power is equal to one.

Combining Rate Laws

Rate = D[A] / DtRate = k[A]n

D[A] / Dt = k[A]n

More product terms can be added

Since both of these are equal to the rate, they are equal to each other.

Use this equation to solve for either rate, the rate constant or a concentration.

Example

Rate = k[A]n[B]m[C]p

(Greater value of exponent means greater effect on rate.)

To find exponents, determine what happens if concentration of a reactant doubles.

Determining orders of reactants

Find two experiments in which one concentration remains the same while the other changes.

Rate 2Rate 1

= (D concentration)n

Example Calculation

TrialInitial Concentration of

Reactants (M)[A] [B] [C]

Initial Rate

(M/min)

1 0.10 0.10 0.10 0.01

2 0.10 0.10 0.20 0.01

3 0.10 0.20 0.10 0.02

4 0.20 0.20 0.10 0.08

To determine orders of reactions, by method of initial rates: 1. Look at the data and find two trials

that start with the same concentrations in all but one reactant.

2. See trials 1 and 2. The concentration of reactant C was doubled while A and B stayed the same. There was no effect on the rate, so reactant C is zero order. Leave it out of the rate law.

Continued on next slide

Example Calculation

TrialInitial Concentration of

Reactants (M)[A] [B] [C]

Initial Rate

(M/min)

1 0.10 0.10 0.10 0.01

2 0.10 0.10 0.20 0.01

3 0.10 0.20 0.10 0.02

4 0.20 0.20 0.10 0.08

3. Compare trials 1 and 3. While the concentrations of A and C stayed the same, the concentration of B was doubled and the rate doubled. Since the effect was the same, B is a first order reactant. (2 x B = 2 x rate; 21 = 2)—The exponent is the order.

4. Compare trials 3 and 4. The concentrations of B and C are the same while the concentration of A was doubled. The rate increased by a factor of 4 so A is a second order reactant. (2 x A = 4 x rate; 22 = 4)—The exponent is the order.

5.Rate = k [A]2 [B]1

Example

Overall Rate:

Rate = k[A]2[B]1[C]0

Overall order of reaction:2 + 1 + 0 = 3

Applying Rate Laws

Once the rate law is established, use data to solve for k and to find rates at different conditions.

Example

Calculate the rate constant k for the previous reaction.

Rate = k[A]2[B]Trial

Initial Concentration of Reactants (M)

[A] [B] [C]

Initial Rate

(M/min)

1 0.10 0.10 0.10 0.01

ExampleSince [C] has no effect on rate, we

can leave it out when solving for k.

Rate = k[A]2[B]k = Rate / [A]2[B]k = .01M/sec / (.10M)2 (.10M)k = 10 M-2sec-1

To find the rate constant, k, choose any of the trials and substitute the concentrations and initial rate into the equation. Solve for k.

Pay attention to units and include them for k.

Write the Rate Law Equation

Look at data sets from different trials:

NH4+ + NO2

- N2 + 2 H2O

Experiment #

Initial Conc. Of

NH4+

Initial Conc. Of

NO2-

Initial Rate (mol/L *s)

1 0.100 M 0.0050 M 1.35 x 10-7

2 0.100 M 0.010 M 2.70 x 10-7

3 0.200 M 0.010 M 5.40 x 10-7

Answer

Both reactants are first order, so Rate = k [NH4

+]1[NO2-]1

Solve for k.

Answer

Rate = k [NH4+]1[NO2

-]1

Using information from trial #1:1.35 x 10-7 M/s = k (0.100 M)(0.0050

M)k = 2.7 x 10-4 M-1s-1

2 ClO2 + 2 OH- ClO3- +

H2O

[ClO2] in mol/L [OH-] in mol/L Initial Rate in M/s

0.0500 0.100 5.75 x 10-2

0.100 0.100 2.30 x 10-1

0.100 0.0500 1.15 x 10-1

Write the rate law equation and find the rate constant.

Answer

Rate = k [ClO2]2 [OH-]k = 230 M-2 s-1

Types of Rate Laws

Differential Rate Law—expression of how rate depends on concentration (previous problems)

Integrated Rate Law—expression of how concentration depends on time

When one is determined, the other can be calculated.

Integrated Rate LawShows how concentration of A depends on time

Usually given [A]0 and kEquation (whichever form) is in the form y = mx + b so it will be a straight line when graphed.

Calculating Integrated Rate Law—First Order

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

ln = natural logarithmt = time[A] = conc. at time t[A]0 = conc. at time 0

Integrated Rate Law Equation

Equation can also be expressed as a ratio:

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

Half LifeThe time required for a reactant to reach half its original concentration

t1/2

Calculating Half LifeWhen t = t1/2, then [A] = [A]0/2Use equation ln ([A]0 / [A]) = kt

ln ([A]0 / [A]0/2) = kt1/2

So…ln(2) = kt1/2

Notice, half life does not depend on concentration.

Calculating Integrated Rate Law—Second Order

Rate = k[A]2

Integrated Rate Law

1/[A] = kt + 1/[A]0

Integrated Second Order

Plot of 1/[A] versus time is a straight line with slope = k

Half life for second order:

t1/2 = 1/k[A]0

For 2nd order, half life does depend on initial concentration.

Zero Order Rate LawsOften encountered when a

reaction rate is limited because a surface (such as a platinum catalytic converter) is required.

When surface is covered, increasing concentration of a reactant has no effect.

If the reaction takes place on a surface, increasing concentration will not affect reaction rate.

Zero Order Rate LawsRate = k[A]0 = k(2) = k

Integrated Rate Law:

[A] = -kt + [A]0Half Life:

t1/2 = [A]0 / 2k

More Complex Rate LawsMany reactions have several

reactants—each affects rateTo study them, high concentrations of all but one reactant are used.

Highly concentrated reactants stay practically constant, so the remaining reactant can be studied.

Pseudo-First-OrderIf conc. of B & C are large,Rate = k[A]n[B]m[C]p =

k’[A]Calculate k’ and solve for k

Good Summary—Table 12.6

ImportanceHelps determine reaction mechanism

Find rate determining step in a series of reactions so total process can be sped up

Reaction MechanismsSum of elementary steps must give overall balanced equation

Must agree with experimentally determined rate law

Reaction MechanismsMost reactions are much

more complicated that they appear from their equations.

Intermediate—a species that is neither a reactant nor a product but that appears and then is consumed in the course of a reaction

Reaction MechanismsEach reaction is called an

elementary step.Rate of each elementary step can

be written from its molecularity.Molecularity = the number of

species that must collide to cause the reaction

MolecularityUnimolecular—depends on one

molecule; Rate = k[A]Bimolecular—depends on two

molecules; Rate = k[A]2 or Rate = k[A][B]

Termolecular—depends on 3 molecules; Rate = k[A]2[B] (etc)

Rate Determining StepThe slowest step in a

reaction mechanismDetermines overall rate much like pouring water through a funnel—limiting factor

Rate Law

Comes from the slow step and every step prior to it.

Only consider species that are in the overall reaction.

Only consider reactants.

Species Not in Overall Reaction

Intermediate—a species that is produced and then consumed in a reaction—appears in mechanism but not in overall reaction—appears as a product and then a reactant

Catalyst—a species that is used during one step of a mechanism but is reproduced later—appears as a reactant and then a product

Determining Mechanism See if rates of elementary

steps agree with observed rate law

If yes, it is an acceptable mechanism

Never proven—only possibly correct

Example Problem

2NO2 + F2 2NO2F

Rate = k[NO2][F2]Possible Mechanism??

NO2 + F2 NO2F + F (slow)

F + NO2 NO2F (fast)

ExampleOverall Reaction: I2 + H2 2HI

Step 1: I2 2I

Step 2: I + H2 H2I

Step 3: H2I + I 2HIDoes this give the overall equation?If step one is rate determining, what is

the rate law? Step 2?Identify any catalysts or intermediates.

Example2 NO2 (g) + F2 (g) 2NO2F (g)

Rate Law: Rate = k [NO2] [F2]Possible Mechanism:

NO2 + F2 NO2F + F (slow)

F + NO2 NO2F (fast)

Is this a possible mechanism?

Identify any catalysts or intermediates.

ExampleNO2(g) + CO(g) NO (g) + CO2(g)

Step 1: NO2 + NO2 NO3 + NO (slow)

Step 2: CO + NO3 NO2 + CO2 (fast)If the mechanism is correct, what is

the rate law?Identify any catalysts or intermediates.

ExampleWrite the rate law and overall reaction:

A + A + B C + D (fast)C + E D + B (slow)Identify any catalysts or intermediates.

Collision ModelMolecules must collide in order to react

Anything that increases frequency or energy or effectiveness of collisions increases reaction rate.

Effective CollisionsOnly a small fraction of collisions produce reactions.

Many ineffective collisions occur because energy is too low or orientation is wrong.

Some orientations for a collision between BrNO molecules. Orientations (a) and (b) can lead to a reaction, but (c) cannot.

Activation EnergyThe minimum amount (threshold) of energy that a system must have to produce a reaction

Using k to Calculate Activation Energy

k = zpe-Ea/RT

k = rate constantz = collision frequencyp = steric factor (fraction of

collisions with proper orientation)

e-Ea/RT = fraction of collisions with

enough energy to produce reaction

Arrhenius Equationk = Ae-Ea/RT

zp replaced by AA is the frequency factor for the

reactionEa is activation energyR is universal gas constant (8.3145)Taking ln of both sides gives a

y = mx +b form line equation

Arrhenius Equation

ln(k) = -Ea/RT+ lnA

In slope intercept form:Slope = Ea

Intercept = A

To Find Activation Energy

Measure rate constant (k) at several temps

Plot ln(k) versus 1/T

To Find Activation Energy

Use equation:

ln(k2/k1) = Ea/R (1/T1 – 1/T2)

Example Problem—p. 588

To Speed up a Reaction:

1. Increase temperature2. Increase pressure3. Increase concentration4. Increase surface area5. Add a catalyst

CatalystA substance that speeds up

a reaction without being consumed in the reaction

Enzymes—biological catalysts

Works by providing the reaction with a new pathway with lower activation energy

Effect of a CatalystNotice that energy difference between products and reactants is unchanged.

Effect of a Catalyst-

Heterogeneous Catalysis

Usually gaseous reactants adsorbed on a metal surface

Example—hydrogenation of ethylene

Breaking H-H bond requires lots of energy, but metal-H interactions weaken bond and allow bonds to break at lower energy

Homogeneous Catalysis

Catalyst and reactants exist in the same phases

Example: Catalysis of ozone by nitric oxide and freons (CCl2F2)