Rates of ReactionSuroviec

Spring 2014

Chapter 13

I. Reaction Rates

Chemical Kinetics: the investigation of the rate at which reactions occur

Rate: how fast a quantity changes with time

I. Reaction Rates

Rate = change in quantity / time elapsed

A. Average Rate Vs. Instantaneous Rate

B. Writing Rate Reactions

Consider the rate reaction: 2NO2 (g) 2NO (g) + O2 (g)

General form

aA + bB cC + dD

Example

Give the relative rates of disappearance of the reactants and formation of products for each of the following reactions:

2O3 (g) 3O2 (g)

2HOF (g) 2HF (g) + O2 (g)

Example

A 2B

Rate increases or decreases as there is less A to react

How is rate of change of [A] related to rate of change of [B]?

Find rate of change of [A] for time internal from 10.0 to 20.0 s

What is the instantaneous rate when [B] = 0.750M

time [B] (mol/L) Rate ([B]/t)

0.00 0.000

10.0 0.326

20.0 0.572

30.0 0.750

40.0 0.890

Example

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

What are the relative rates?If –[H2]/t = 4.5X10-4 M/min,

What is –[N2]/t?

What is [NH3]/t?

II. Rate Laws

Why look at initial rates?

Ex. 2N2O5 4NO2 + O2

[N2O5] Rate (M/s)

0.010 0.0180

0.020 0.0360

0.040 0.0720

0.060 0.1080

0.100 0.1800

General Form of the Rate Law

aA + bB cC + dD

Rate = k[A]x[B]y

A. Method of Initial Rates

Initial rate of reaction is measured with several different starting amount of reactants

Example

The reaction between ozone and nitrogen dioxide at 231 K is first order in both [NO2] and [O3].

2 NO2 (g) + O3 (g) N2O5 (g) + O2(g)

Write the rate equation for the reaction If the concentration of NO3 is tripled,

what is the change in the reaction rates? What is the effect on the reaction rate

if the concentration of the O3 is halved?

Example

2NO(g) + O2 (g) 2NO2 (g)1. Determine the

order of the reaction for each reactant

2. Write the rate equation for the reaction

3. Calculate the rate constant

4. Calculate the rate in M/s at the instant when [NO] = 0.015 M and [O2] = 0.0050M

[NO] (mol/L) [O2] (mol/L) Rate of [NO] disappearing

(mol/L.s)

0.010 0.010 2.5 X 10-5

0.020 0.010 1.0 X 10-4

0.010 0.020 5.0 X 10-5

Example

A + B CWhat is the rate equation?

Run # [A]i [B]i Initial Rate (M/s)

1 0.10 0.10 2.00X10-4

2 0.20 0.10 8.00X10-4

3 0.40 0.20 2.56X10-2

Base 10 log and Natural Log

Base 10 log: 102 = 100 log 100 = _______103 = 1000 log 1000 = _______log 45 = 1.65 101.65 =_________log x = y 10y =x

Natural Log: base is e = 2.718281828…..e1 = 2.71828ln(2.71828) = 1ln(e)=_______ln(45) = 3.81 e3.81 =_________ln x = y ey =x

Important properties of both log and ln

log ab = log a + log b

log a/b = log a – log b

log ab = b(log a)

Example

Rate = k[A]2[B]x

III. Concentration-time relationships

Useful to know How long a reaction is going to proceed to reach concentration of interest for product or reactant

After a given time what are the concentrations of the reactants and the products?

A. 1st order reactions

A B Rate = k[A] Integrated rate eqn for 1st order reactions:

Useful for 3 reasons Can calculate k if we know [A]/[A]o

if k and [A]o are known then we can determine the amount of material expected after time t ([A])

If k is known than after time t we can calculate the fraction [A]/[A]o

B. 2nd order reactions

Rate = k[A]2

Integrated rate eqn for 2nd order reactions:

1/[A] = kt + 1/[A]o OR

1/[A] – 1[A]o = kt

Example

Ammonium cyanate, NH4NCO, rearranges in

water to give urea, (NH2)2CO.

NH4NCO (aq) (NH2)2CO (aq)

The rate equation for this process is: Rate = k [NH4NCO]2

where k = 0.0113 L/mol.min.

If the original concentration of NH4NCO in solution is 0.229M how long will it take for the concentration of decrease to 0.180M?

time (min) [sucrose] ln [sucrose] 1/[sucrose]

0 0.316 -1.1520131 3.164557

39 0.274 -1.2946272 3.649635

80 0.238 -1.4354846 4.2016807

140 0.190 -1.6607312 5.2631579

210 0.146 -1.9241487 6.8493151

Example

Sucrose breaks down to form glucose and fructose.C12H22O11(aq) +H2O (l) 2C6H12O6 (aq)

1. Plot ln sucrose versus timeand 1/[sucrose] versus time.Find the order of the reaction2. Write the rate constant for the reaction and calculate k3. Find concentration of sucrose After 175 min

C. Zero Order Reactions

Rate =k[A]0

Integrated rate eqn for 0th order reactions:

[A]o – [A] = kt

Table 15-1, p.719

D. Half-Life

Time required for half of a reactant to be consumed

1. 1st order

Fig. 15-9, p.720

Example

The rate equation for the decomposition of producing NO2 and O2 is

–[ N2O5]/t = k[N2O5]

The value of k = 5.0 X 10-4 s-1 for the reaction at a known temperature.

Calculate the half-life of N2O5

How long does it take for the N2O5 concentration to drop to 1/10th of its original value?

2. 2nd order and 0th order

A B Rate = k[A]2

A BRate = k

IV. Activation Energy

NO2 (g) + F2 (g) FNO2 (g) + F (g)

What makes this reaction occur?

What makes this reaction not occur?

A. Collision Theory

Reactant Molecules must collide with each other

Reactant molecules must collide with sufficient energy

Reactant molecules must have proper orientation

2 questions:What makes a reaction go faster?

What does sufficient energy mean?

A. What makes a reaction go faster?

1. 2.

B. What does sufficient energy mean?

RXN: AB + C A + BC

Mechanism:

C. How can we measure Ea?

From collision theory

Rate of reaction = (collision freq)X(fraction with proper orientation)X (fraction of collisions with enough energy)

k = Ae-Ea/RT

Arrhenuis equation

Example

When heated to a high temperature, cyclobutane, C4H8, decomposes to ethylene.

C4H8 (g) 2C2H4 (g)

The activation energy, Ea for this reaction is 260 kJ/mol. At 800K the ate constant k = 0.0315 s-1. Determine the value of k at 850K.

D. What is a catalyst?

A substance added to a reaction to speed it up, but not consumed in the reaction

MnO2 catalyzes decomposition of H2O2

2 H2O2 ---> 2 H2O + O2

Uncatalyzed reactionUncatalyzed reaction

Catalyzed reactionCatalyzed reaction

V. Reaction mechanisms

AB + C A + BC

Possible mechanisms

V. Reaction Mechanisms

Molecularity

Elementary step

A. Mechanisms and Rate Laws

1. For net reactions, no simple relationship can be assumed between reaction coefficients and reactant orders

2. For each elementary step reactant coefficients DO equal reactant orders

A. Mechanisms and Rate Laws

3. The rate of a net reaction is essentially that of the slowest step (the rate determining step)

A. Mechanisms and Rate Laws

4. When a fast step proceeds a slow step the fast step can be assumed to be at equilibrium

RXN: H2 (g) + CO (g) H2CO (g)

Step 1. H2 (g) 2 H (g) fast

Step 2. H (g) HCO (g) slow

Step 3. H (g) H2CO (g) fast

A. Mechanisms and Rate Laws

5. A catalyst can provide a new mechanistic pathway for a reaction

2Ce4+ (aq) + Tl+ (aq) 2Ce3+ (aq) + Tl3+ (aq)

Mechanism

Rate equation

Add Mn2+ catalyst

In General

1. Measure reaction rates1. Through initial rates

2. Graphically concentraion vs. time

2. Formulate the rate law

3. Postulate the mechanism1. Cannot prove, can only disprove