KINETICS�Chapter 14

Chemical kinetics is the study of how fast chemical reactions occur.

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• The speed of the chemical reaction is called its reaction rate

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• A change that occurs per unit time.
• concentration/time (M/s)
• Mole/time (mol/s)
• Pressure/time (atm/s)

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Generally, the more frequently collisions between reaction particles occur, the faster the reaction.

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Factors that affect reaction rates:�

1. Physical state of the reactants

• Homogeneous mixtures (same phase) react faster than heterogeneous mixtures. (different phases)

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• Heterogeneous Mixtures:

more surface area 🡪 faster reaction

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Factors that affect reaction rates:�

2. Concentration of the reactants

• more reactant molecules in the same amount of space
• Better chance for collisions to happen🡪 faster reaction

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Factors that affect reaction rates:�

3. Temperature of the reaction

• higher temperature molecules move faster and collide more frequently
• Bond breaking requires energy. More energy is available at higher temps

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4. Catalysts

A substance that changes the mechanism of the reaction (the way it happens to increase speed)

(More on this later!)

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Calculating the Rate of Reaction

Reaction rates are always positive values!

• Suppose A reacts to form B. A → B

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• We can uses this information to find the average rate of the reaction.

• For the reaction A → B, there are 2 ways of measuring rate:

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• The rate of appearance of product B.

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• The rate of disappearance of reactant A

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The isomerization of methyl isonitrile CH₃NC to acetonitrile, CH₃CN, was studied in the gas phase at 215^oC and the following data was obtained:

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Time (s) 0 2000 5000 8000 12000 15000

[CH₃NC] (M) 0.0165 0.0110 0.00591 0.00314 0.00137 0.00074

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(a) Calculate the average rate of disappearance of CH₃NC in M/s, for the time interval from 2000 to 5000 seconds.

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0.014

0.023

0.029

0.034

600s

1800s

Change of Rate with Time

The rate is usually not constant and generally decreases with time

Instantaneous Rate

• The rate at any instant in time is called the instantaneous rate.

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• It is the slope of the straight line tangent to the curve at that instant.

Tangent

straight line that goes through a point on a curve and only touches the object at that single point

Example:

Using the data in table above, calculate the average rate of disappearance of C₄H₉Cl over the time interval from 50.0 to 150.0 seconds. (b) Using the figure, estimate the instantaneous rate of disappearance of C₄H₉Cl at t = 0 (the initial rate).

b. pick two points on the

tangent line that passes through the initial time.

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(0, .100) and (200, .060)

Reaction Rates and Stoichiometry

For the reaction: aA + bB → cC + dD

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For the reaction: 2HI(g) → H₂(g) + I₂(g)

Consider the combustion of H₂(g),

2 H₂(g) + O₂(g) → 2 H₂O(g)

If hydrogen is burning at the rate of 0.85 mol/s, what is the rate of consumption of oxygen? What is the rate of formation of water vapor?

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O₂ is consumed: H₂O is produced:

ΔP_{TOT =}

- 23

-11.5

+ 23

-11.5 torr/min

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Part 2�Rate Law

Rate Law

• The rate law (rate expression) is an equation that relates the rate of a reaction to concentration.

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For a general reaction the rate law is: A + B 🡪 C

Rate = k [A]^m [B]ⁿ

[A] and [B] are concentrations (molarity) or pressures of the reactants

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The constant, k, is called the rate constant.

(different for each reaction and conditions – more later!)

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Rate = k [A]^m [B]ⁿ

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The exponents m and n are called reaction orders

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The rate is “m” order in A

The rate is “n” order in B

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We commonly encounter reaction orders of 0, 1 or 2.

But, fractional or negative values are possible.

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The overall reaction order is the sum of the reaction orders.

The overall order of a reaction is m + n + ….

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Note that reaction orders must be determined experimentally.

They DO NOT necessarily correspond to the stoichiometric coefficients in the balanced chemical equation! (but they can and sometimes do)

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Example 8:

What are the individual and overall reaction orders for the reactions described in the following equations:

1. 2 N₂O₅(g) → 4 NO₂(g) + O₂(g) Rate = k [N₂O₅]

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1^st order in N₂O₅

1^st order overall.

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2. CHCl₃(g) + Cl₂(g) → CCl₄(g) + HCl (g) Rate = k [CHCl₃][Cl₂]^1/2

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1^st order in CHCl₃

½ order in Cl₂

3/2 order overall (add the exponents).

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We have to find m and n for the rate law from experimental data:

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In general, rates:

• Increase when reactant concentration is increased.

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• We examine the effect of concentration on reaction rate by measuring the way in which reaction rate at the beginning of a reaction depends on different starting conditions.

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• We measure initial reaction rates.
• The initial rate is the instantaneous rate at time t = 0.
• We find this at various initial concentrations of each reactant.

A + B 🡪 C

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Using Initial Rates to Determine Rate Laws

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• To determine the rate law, we observe the effect of changing initial concentrations.

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• A reaction is nth order if multiplying the concentration by x causes a xⁿ increase in rate.

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x increase in concentration = xⁿ increase in rate

(n is the order)

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• Most Common Scenarios:

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change in conc. of reactant change in rate order

doubled no change - rate is multiplied by 1 or 2⁰

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doubled doubled - rate is multiplied by 2 or 2¹

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doubled quadrupled - rate is multiplied by 4 or 2²

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tripled no change - rate is multiplied by 1 or 3⁰

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tripled tripled - rate is multiplied by 3 or 3¹

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tripled rate is multiplied by 9 or 3²

Zero order rxn: ∆ concentration = no change in rate

1^st order rxn: ∆ concentration = ∆ rate

2^nd order rxn: (∆ concentration)² = ∆ rate

Xth order: (∆ concentration)^x = ∆ rate

Another way to look at order of reaction vs. change in concentration:

0

1

2

0

1

2

A🡪 B

Rate = k [A]²

A🡪 B

Rate = k

Using Initial Rates to Determine Rate Laws

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A + B 🡪 C

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Rate = k [A]²[B]

Rate = k [NO₂]²

The iodide ion reacts with hypochlorite ion (the active ingredient in bleach) in the following way: OCl^-1 + I^-1 → OI^-1 + Cl^-1. This rapid reaction gives the following rate data:

Experiment [OCl^-1] (M) [I^-1] (M) Rate (M/s)

1 0.0015 0.0015 1.36 × 10^-4

2 0.0030 0.0015 2.72 × 10^-4

3 0.0015 0.0030 2.72 × 10^-4

1. What is the rate law for the reaction?

rate = k [OCl^-1][I^-1]

(b) What is the value of the rate constant?

solving the rate law for k and using the data of experiment 1:

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(c) Calculate the rate when [OCl^-1] = 0.0020 M and [I^-1] = 0.00050 M

using the rate law (from a) and the value of k (from b):

60. M^-1s^-1

A reaction A + B → C, obeys the following rate law: rate = k [B]².

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1. if [A] is doubled, how will the rate change?

It won’t!

1. if [B] is doubled, will the rate constant change?

Nope!

1. What are the reaction orders for A and B?

zero order in A

2^nd order in B

1. What is the overall reaction order?

2^nd order overall

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Consider the gas-phase reaction between nitric oxide and bromine at 273^oC:

2 NO + Br₂ → 2 NOBr. The following data for the initial rate of NOBr were obtained:

Experiment [NO] (M) [Br₂] (M) Initial Rate (M/s)

1 0.10 0.20 24

2 0.25 0.20 150

3 0.10 0.50 60

4 0.35 0.50 735

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1. Determine the rate law.

rate = k [NO]²[Br₂]

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(b) Calculate the value of the rate constant (with units) for the appearance of NOBr.

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24 M/s = k (0.10M)²(0.20M)

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k=12000 M^-2s^-1

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Consider the gas-phase reaction between nitric oxide and bromine at 273^oC:

2 NO + Br₂ → 2 NOBr. The following data for the initial rate of NOBr were obtained:

Experiment [NO] (M) [Br₂] (M) Initial Rate (M/s)

1 0.10 0.20 24

2 0.25 0.20 150

3 0.10 0.50 60

4 0.35 0.50 735

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3. How is the rate of appearance of NOBr related to the rate of disappearance of Br₂?

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(d) What is the rate of disappearance of Br₂ when [NO] = 0.075 M and [Br₂] = 0.25?

rate = k [NO]²[Br₂]

rate = (12000)(0.075)²(0.25)

rate = 17 M/s

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Consider the reaction of peroxydisulfate ion, S₂O₈^-2 , with iodide ion, I^-1 , in aqueous solution:

S₂O₈^-2 + 3 I^-1 → 2 SO₄^-2 + I₃^-1

At a particular temperature, the rate of disappearance of S₂O₈^-2 varies with reactant concentrations in the following manner:

Experiment [S₂O₈^-2] (M) [I^-1] (M) Initial Rate (M/s)

1 0.018 0.036 2.6 × 10^-6

2 0.027 0.036 3.9 × 10^-6

3 0.036 0.054 7.8 × 10^-6

4 0.050 0.072 1.4 × 10^-5

1. Determine the rate law for the reaction.

(a) from experiments 1 and 2, increasing [S₂O₈^-2] by 1.5 increases the rate by 1.5 so [S₂O₈^-2] is 1^st order

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Experiment [S₂O₈^-2] (M) [I^-1] (M) Initial Rate (M/s)

1 0.018 0.036 2.6 × 10^-6

2 0.027 0.036 3.9 × 10^-6

3 0.036 0.054 7.8 × 10^-6

4 0.050 0.072 1.4 × 10^-5

So we know that the rate law is: rate = k [S₂O₈^-2] [I^-1]^x and we need to find x

Choose 2 experiments: I used 1 & 3 because the rate goes up by a factor of 3 so it’s an easy number to work with.

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EXP 3: Rate k [S₂O₈^-2] [I^-1]^x

-------- = --------------------

EXP 1: Rate k [S₂O₈^-2] [I^-1]^x

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2

2

rate = k [S₂O₈^-2] [I^-1]

x=1

You CAN, but shouldn’t NEED to use this method on quizzes/tests

Kinetics Part 3

Integrated Rate Laws

• Goal: Convert the rate law into a convenient equation that gives concentration as a function of time.
• This takes some calculus!

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• We will only look at zero, 1^st, and 2^nd order

Zero Order Reactions

Rate = k

After some calculus………

y = mx + b

[A]_t

Time (s)

(slope= −k)

On AP equation sheet (rearranged)

•A plot of [A]_t versus t is a straight line with slope -k and intercept [A]₀

First-Order Reactions

• Rate must depend on only one reactant and be raised to the 1st power.
• For a first-order reaction, the rate doubles as the concentration of a reactant doubles.

Rate = k[A]

After some calculus………

•A plot of ln[A]_t versus t is a straight line with slope -k and intercept ln[A]₀

y = mx + b

On AP equation sheet (rearranged)

Second-Order Reactions

• For a 2nd order reaction with just one reactant:

calculus

A plot of 1/[A]t versus t is a straight line with slope k and intercept 1/[A]₀.

y = mx + b

On AP equation sheet (rearranged)

Half-life

• Half-life, t_½ , is the time required for the concentration of a reactant to decrease to half its original value.

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• That is, half life, t_½, is the time taken for [A]₀ to reach ½ [A]₀.

Half Life Summary

Radioactive decay

is 1^st order.

Sample Exercise #1

• The first order rate constant for the decomposition of a certain insecticide in water at 12^oC is 1.45 yr^-1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 x 10^-7 g/cm³ of water. Assume that the effective temperature of the lake is 12^oC.

(a) What is the concentration of the insecticide on June 1 of the following year?

ln [A]_t = - k t + ln [A]₀

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ln [A] _t = - (1.45 yr^-1)(1.00 yr) + ln(5.0 x 10^-7)

ln [A] _t = -15.96

e^{ln [A] t} = e^-15.96

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[A] _t = 1.2 x 10^-7 g/cm³

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(note: conc. units for [A] and [A]₀ must be the same)

(b) How long for the conc. of the insecticide to drop to 3.0 x 10^-7 g/cm³?

ln [A]_t = - k t + ln [A]₀

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ln (3.0 × 10^-7) = - (1.45)(t) + ln (5.0 × 10^-7)

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t = 0.35 yr

The decomposition of sulfuryl chloride (SO₂Cl₂) is a first-order process. The rate constant for the decomposition at 660 K is 4.5 × 10^-2 s^-1. (a) if we begin with an initial SO₂Cl₂ pressure of 375 torr, what is the pressure of this substance after 65 s? (b) at what time will the pressure of SO₂Cl₂ decline to one-tenth its initial value?

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(a) ln P_t = - kt + ln P_o

ln P₆₅ = - 4.5 × 10^-2 s^-1 (65) + ln(375)

P₆₅ = 20 torr

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(b) P_t = 0.10 P₀ = (.10)(375) = 37.5 torr

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ln P_t = - kt + ln P_o

ln (37.5) = - (4.5 x 10^-2 s^-1 )t + ln (375)

t = 51 s

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• The initial value of [C₃H₉Cl] is 0.100 M.
• The half-life for this reaction is the time for [C₃H₉Cl] to be 0.050 M.
• This point occurs at approx. 340 s.

Using the figure to the right, estimate the half-life of the 1^st order reaction of C₄H₉Cl with water.

The reaction SO₂Cl₂ → SO₂ + Cl₂ is first order in SO₂Cl₂. Using the following kinetic data, determine the magnitude of the first order rate constant:

Time (s) Pressure SO₂Cl₂ (atm) ln Pressure SO₂Cl₂

0 1.000 0

2500 0.947 - 0.0545

5000 0.895 - 0.111

7500 0.848 - 0.165

10000 0.803 - 0.219

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Graph ln P vs. time (first order)

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Sample Exercise #5

The following data was obtained for the gas phase decomposition of NO₂(g) at 300^oC:

2 NO₂(g) → 2 NO(g) + O₂(g)

Time (s) [NO₂] (M)

0.0 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

What is the rate law for this reaction?

• To test whether the reaction is first or second order, we can construct plots of ln [NO₂] and 1 / [NO₂] against time.

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Time (s) [NO₂] ln [NO₂] 1/[NO₂]

0.0 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

-4.610

-4.845

-5.038

-5.337

-5.573

100

127

154

208

263

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From the graphs above, only the plot of 1/[NO₂] is a straight line. Thus, the reaction is second order in NO₂.

Rate = k [NO₂]²

From the slope of the straight line graph, we can get:

k = 0.543 M^-1 s^-1

Consider the data presented

1. By using appropriate

graphs, determine

whether the reaction

is first or second order.

2. What is the value of the

rate constant for the

reaction?

3. What is the half life for

the reaction?

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1. make both first and second order plots to see which is linear

the plot of 1/[A] vs. time is linear, so the reaction is second order in [A]

2. the slope of the line in the graph is the rate constant

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(c) use the half-life formula for a 2^nd order reaction:

t_1/2 = 1 / k[A]_o = 1 / [(0.043 M^-1 min^-1) (0.65 M)] = 36 min

1/[A]

Temperature and Rate

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• Most reactions speed up as temperature increases

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• The rate law has no temperature term in it, so the rate constant (k) must depend on temperature.
• We see an approximate doubling of the rate with each 10^oC increase in temperature.

The Collision Model

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• Rates of reactions are affected

by concentration and temperature.

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In order for molecules to react they must collide.

• The greater the number of collisions the faster the rate.

- The more molecules present, the greater the probability of collision 🡪faster rate

- The higher the temperature, the faster molecules move and more often they collide 🡪 faster rate

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However, not all collisions lead to products.

• In order for a reaction to occur the reactant molecules must collide in the correct orientation AND with enough energy to form products.

• The Orientation Factor
• The orientation of a molecule during collision can have a profound effect on whether or not a reaction occurs.
• Consider the reaction between Cl and NOCl:
• If Cl collides with Cl of NOCl, the products are Cl₂ and NO.
• If the Cl collides with the O of NOCl, no products are formed.

Activation Energy

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• Arrhenius: molecules must posses a minimum amount of energy to react. In order to form products, bonds must be broken in the reactants.

• Bond breakage requires energy.

(ΔH_rxn = bonds broken – bonds formed)

• Molecules moving too slowly, with too little kinetic energy, don’t react when they collide.

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• Activation energy, E_a is the minimum energy required to initiate a chemical reaction (varies with the reaction).

H₃C-N≡C 🡪 H₃C-C≡N

• Energy is required to stretch and break the bond between the CH₃ group and the N≡C group

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• The species at the top of the barrier is called the activated complex or transition state.

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• Energy is released when the C-C bond is formed.

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• The change in energy for the reaction is the difference in energy between CH₃NC and CH₃CN.

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• The activation energy is the difference in energy between reactants, (CH₃NC) and the transition state.

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The rate depends on the

magnitude of the E_a.

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In general, the lower the E_a,

the faster the rate.

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• Notice that if a forward reaction is exothermic (CH₃NC → CH₃CN), then the reverse reaction is endothermic (CH₃CN → CH₃NC).

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Solution

The lower the activation energy, the faster the reaction. The value of ∆E does not affect the rate.

Hence the order is (2) < (3) < (1).

SAMPLE 1:

Consider a series of reactions having the following energy profiles:

Assuming that all three reactions have nearly the same frequency factors, rank the reactions from slowest to fastest.

The gas phase reaction Cl(g) + HBr(g) 🡪 HCl(g) + Br(g) has an overall enthalphy change of - 66 kJ. The activation energy for the reaction is 7 kJ.

(a) Sketch the energy profile for the reaction and label E_a and ΔE.

(b) What is the activation energy for the reverse reaction?

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(b) the reverse reaction has an E_a value of 73 kJ

How does this relate to temperature?

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• At any particular temperature, the molecules present have an average kinetic energy.
• Some molecules have less energy than the average while others have more than the average value.

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• Molecules that have an energy equal to or greater than E_a have sufficient energy to react.

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At a higher temperature, more molecules have the minimum E_a needed to react and the rate is faster.

Maxwell-Boltzman Distribution (we looked at this in our gas unit too!)

The Arrhenius Equation

• Arrhenius discovered that most reaction-rate data obeyed an equation based on three factors

1. The number of collisions per unit time.

2. The fraction of collisions that occur with the correct orientation.

3. The fraction of the colliding molecules that have an energy equal to or greater than E_a.

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Arrhenius equation:

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• Where k is the rate constant, E_a is the activation energy(J), R is the ideal-gas constant (8.314 J/K•mol) and T is the temperature (K).
• A is the frequency factor.
• It is related to the frequency of collisions and the probability that a collision will have a favorable orientation.

More Useful Version of the Arrhenius Equation

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y = mx + b

• A graph of ln k vs 1/T will be a line with slope of –E_a/R and a y-intercept of ln A.

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The rate of the reaction CH₃COOC₂H₅ + OH^-1 → CH₃COO^-1 + C₂H₅OH was measured at several temperatures and the following data were collected. Using the data, construct a graph of ln k versus 1/T and determine the value of E_a

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.00347

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.00336

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.00325

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.00314

E_a= 47,000 J/mol or 47 kJ/mol

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14.6 Reaction Mechanisms

• A chemical equation provides information about substances present before and after a reaction.

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2Na + Cl₂ 🡪 2NaCl

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• The reaction mechanism is the process of how the reactants become products.

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• Mechanisms provide a picture of which bonds are broken and formed during the course of a reaction.

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• Most reactions occur as a series of small steps or changes.

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Notice that ions are formed in the intermediate steps.

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Na + small energy 🡪 Na⁺¹ + e^-1

Cl + e^-1 🡪 Cl^-1 + large energy

Na⁺¹ + Cl^-1 🡪 NaCl + small energy

Overall reaction:

Na + Cl 🡪 NaCl + large energy

metal nonmetal ionic compound

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Elementary steps are processes that occur in a single step.

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• The number of molecules present in an elementary step is the molecularity of that elementary step.

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Unimolecular: one molecule in the elementary step.

Bimolecular: two molecules in the elementary step.

Termolecular: three molecules in the elementary step.

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Rate Laws for Elementary Steps

• The rate laws of the elementary steps determine the overall rate law of the reaction.
• For elementary processes, we DO NOT need to find the reaction order by experimentation – it matches the stoichiometry of the equation

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• The rate law of an elementary step is determined by its molecularity.
• Unimolecular processes are first order.
• Bimolecular processes are second order.
• Termolecular processes are third order.

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What is the molecularity of each of the following elementary processes? Write the rate law for each.

(a) Cl₂ → 2 Cl

unimolecular

rate = k [Cl₂]

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(b) OCl^-1 + H₂O → HOCl + OH^-1

bimolecular

rate = k [OCl^-1] [H₂O]

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(c) NO + Cl₂ → NOCl₂

bimolecular

rate = k [NO] [Cl₂]

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Rate Laws for Multistep Mechanisms

• When a reaction occurs by mechanisms with more than one step, the slowest step limits the overall reaction rate.
• This is called the rate-determining step (rds) of the reaction
• This step governs the overall rate law for the overall reaction.

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Step 1: NO₂(g) + NO₂(g) → NO₃(g) + NO(g) slow step

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Step 2: NO₃(g) + CO(g) → NO₂(g) + CO₂(g) fast step

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• Intermediates - these are species that appear in an elementary step but are neither a reactant nor product.
• They are formed in one elementary step and consumed in another.
• They are NOT found in the balanced equation for the overall reaction.

Step 1: NO₂(g) + NO₂(g) → NO₃(g) + NO(g) slow step

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Step 2: NO₃(g) + CO(g) → NO₂(g) + CO₂(g) fast step

Consider the reaction:

NO₂(g) + CO(g) → NO(g) + CO₂(g)

Here is a proposed a mechanism for the reaction:

Step 1: NO₂(g) + NO₂(g) → NO₃(g) + NO(g) slow step

Step 2: NO₃(g) + CO(g) → NO₂(g) + CO₂(g) fast step

(NO₃ is an intermediate.)

The overall reaction rate will depend on the slowest step:

Rate = k[NO₂]²

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Supporting our mechanism:

1. If we do an experiment and find that the experimentally derived rate law is: Rate = k[NO₂]² Then our theoretical rate law is in agreement with the experimental rate law. This supports (but does not prove) our mechanism.
2. Detection of the intermediates during the course of the reaction is another common way to support a reaction mechanism.

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The following mechanism has been proposed for the reaction of methane gas with chlorine gas. All species are in the gas phase.��Step 1                  Cl₂  2 Cl              fast equilibrium��Step 2        CH₄ + Cl  🡪 CH₃ + HCl          slow��Step 3        CH₃ + Cl₂  🡪 CH₃Cl + Cl        fast��Step 4      CH₃Cl + Cl  🡪 CH₂Cl₂ + H        fast��Step 5            H + Cl  🡪 HCl                    fast�

�Write the rate law for this reaction.

The overall rate law depends on the rate law of the slow step:

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step 2: rate = k[CH₄][Cl]

The following mechanism has been proposed for the gas-phase reaction of H₂ with ICl:

H₂ + ICl HI + HCl

HI + ICl → I₂ + HCl

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H₂ + 2 ICl → I₂ + 2 HCl

(a) Write the balanced equation for the overall reaction

(b) Identify any intermediates

HI

(c) If the first step is slow and the second one is fast, what rate law do you expect to be observed for the overall reaction?

rate = k [H₂] [ICl]

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14.7 Catalysis

• A catalyst is a substance that changes the speed of a chemical reaction without any permanent change to itself

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• Works by providing an alternate mechanism for the reaction:

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• Lowering activation energy of an elementary step by stabilizing a transition state
• Lowering the activation energy allows for more molecules with E_a large enough to react – more collisions that can make products – higher rxn rate (without changing temperature)

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• Forming a new intermediate to change the mechanism of the reaction.

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Catalysis

• Homogeneous catalyst – catalyst is in the same phase as reactants
• Heterogeneous catalyst – catalyst is in different phase than reactants

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• Acid-Base Catalysis:

- A reactant gains or loses a proton

• Surface Catalysis:
• New intermediate is formed or probability of successful collisions is increased. (catalytic converter in automobile)
• Enzymes:
• Catalyst in biological systems

• Much research to find new catalysts AND, much research to find “inhibitors”(substances that slow down an undesired reaction)
• Example: corrosion

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Effect of a catalyst on a reaction

A catalyst provides an alternate mechanism that has a lower E_a than the original mechanism – this allows the reaction to occur at a faster rate

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(a) Based on the following reaction profile, how many intermediates are formed in the reaction A → D?

This is a three-step mechanism: A → B, B → C, C → D

2 intermediates: B and C

2. How many transition states are there?
3. Which step is the fastest?

(d) Is the overall reaction exothermic or endothermic?

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Step 3 : C → D

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endothermic

The oxidation of SO₂ to SO₃ is catalyzed by NO₂. The reaction proceeds as follows:

NO₂(g) + SO₂(g) → NO(g) + SO₃(g)

2 NO(g) + O₂(g) → 2 NO₂(g)

(a) Show that the two reactions can be summed to give the overall oxidation of SO₂ by O₂ to give SO₃. Hint: the top reaction must be multiplied by a factor so the NO and NO₂ cancel out.

(a) 2 [NO₂ + SO₂ → NO + SO₃] = 2 NO₂ + 2 SO₂ → 2 NO + 2 SO₃

2 NO + O₂ → 2 NO₂ __

2 SO₂ + O₂ → 2 SO₃

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(b) Why do we consider NO₂ a catalyst and not an intermediate in this reaction?

NO₂ is a catalyst because it is present at the beginning and end of the reaction. (NO is an intermediate because it is produced and then consumed.)

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(c) Is this an example of homogeneous catalysis or heterogeneous catalysis?

homogeneous catalysis

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• This process works for the two mechanism situations described:
• 1^st step is slow
• 1^st step is fast and 2^nd step is slow.
• There are many more complex mechanisms that we won’t worry about!

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Note: Textbook p 588:

“In general, whenever a fast step precedes a slow one, we can solve for the concentration of the intermediate by assuming that an equilibrium is established in the first step.”

If the 1^st step is fast and not equilibrium just assume it is in equilibrium!

• Mathematically, the half life of a first-order reaction is:

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Note that the half-life of a first-order reaction is independent of the initial concentration of the reactant.

Radioactive decay

is 1^st order.

On AP equation sheet

• We can show that the half-life of a second order reaction is:

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• Note that the half-life of a second-order reaction is dependent on the initial concentration of reactant.
• The half life of a second order reaction is NOT constant!!

Consider the following reaction: CH₃Br + OH^-1 → CH₃OH + Br^-1 . The rate law for this reaction is first order in CH₃Br and first order in OH^-1. When [CH₃Br] is 0.0050 M and [OH^-1] is 0.050 M, the reaction rate at 298 K is 0.0432 M/s.

1. What is the value of the rate constant? (with units)

rate = k [CH₃Br] [OH^-1]

0.0432 M/s = k(0.0050M)(0.050M)

k = 170 M^-1s^-1

(b) What would happen to the rate if the concentration if OH^-1 were tripled?

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Since the rate law is first order in [OH^-1],

if [OH^-1] is tripled, the rate triples.

The decomposition of N₂O₅ proceeds as follows:

2 N₂O₅ → 2 NO₂ + O₂.

The rate law is first order in N₂O₅. At 64^oC, the rate constant is 4.82 × 10^-3 s^-1.

1. Write the rate law for the reaction.

rate = k [N₂O₅]

2. What is the rate of reaction when [N₂O₅] = 0.0240 M?

rate = (4.82 × 10^-3 s^-1) (0.0240 M) = 1.16 × 10^-4 M/s

(c) What happens to the rate when the concentration of N₂O₅ is doubled to 0.0480 M?

one way: rate = 4.82 × 10^-3 s^-1 (0.0480 M) = 2.31 × 10^-4 M/s

quicker way:

We know that the rxn is 1^st order in N₂O₅, so doubling the conc. will double the rate

Not sure on this….

• Add an example there there is no equilibrium:

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• A + B🡪C +D (fast)
• A+D🡪E (slow)

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• Rate = k[A]2B
• Use all reactants except intermediats up to and including the slow step

Taken out…

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• Useful when we have 2 different k values at 2 different temperatures

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Calculate the fraction of atoms in a sample of argon gas at 400 K that have an activation energy of 10.0 kJ or greater.

A certain first order reaction has a rate constant of 2.75 × 10^-2 s^-1 at 20^oC. What is the value of k at 60^oC if E_a = 75.5 kJ/mol

T₁ = 293 K T₂ = 333 K

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The activation energy of a certain reaction is 65.7 kJ/mol. How many times faster will the reaction occur at 50^oC rather then 0^oC assuming equal initial concentrations?

Since we are considering one reaction and temp only affects k which directly affects rate…the ratio of k values equals the ratio of rates at the two temperatures.

T₁ = 323 K ; T₂ = 273 K

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The reaction will occur about 88 times faster at 50^oC

Using Initial Rates to Determine Rate Laws

• To determine the rate law, we observe the effect of changing initial concentrations.

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• A reaction is nth order if multiplying the concentration by x causes a xⁿ increase in rate.

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x increase in concentration = xⁿ increase in rate

(n is the order)

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For our example – the concentration was multiplied by 2 and the rate was also multiplied by 2 (2¹) so the rate is 1^st order with respect to both NH₄⁺(aq) and NO₂^– (aq)

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2 (increase in concentration) = 2¹ (increase in rate)

(1^st order)

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• As [NH₄⁺] doubles with [NO₂^–] constant the rate doubles.
• We conclude the rate is proportional to [NH₄⁺].
• As [NO₂^–] doubles with [NH₄⁺] constant the rate doubles
• We conclude that the rate is proportional to [NO₂^–].

• For the reaction in our example:

NH₄⁺(aq) + NO₂^– (aq) → N₂(g) + 2H₂O(l)

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The rate law is:

Rate = k [NH₄⁺] [ NO₂^–]

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• The reaction is said to be first order in [NH₄⁺], first order in [NO₂^–], and second order overall.

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Units of Rate Constants

• Units of the rate constant depend on the overall reaction order.
• Second order overall:
• Rate = k [A]² or Rate = k[A][B] or etc

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• First order overall:
• Rate = k [A]

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Units of Rate Constants

• Units of the rate constant depend on the overall reaction order.
• Second order overall:
• Rate = k [A]² or Rate = k[A][B]

M/s = k (M)²

(M)² (M)²

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• First order overall:
• Rate = k [A]

M/s = k (M)

(M) (M)

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= M^-1s^-1

= s^-1

How does this relate to temperature?

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• At any particular temperature, the molecules present have an average kinetic energy.
• Some molecules have less energy than the average while others have more than the average value.

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• Molecules that have an energy equal to or greater than E_a have sufficient energy to react.

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• The fraction of molecules with an energy equal to or greater than E_a is given by:

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(R is 8.314 J/mol-k , Temp in K, E_a in J)

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