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### Chapter 11

Chemical Kinetics

Chemical Kinetics

- Chemical Kinetics – the study of the rates of chemical rxns, the factors that affect rxn rates, and the mechanisms (the series of steps) by which the rxns occur.
- Rates of reaction – describes how fast reactants are used up and products are formed.
- In thermodynamics we learned to assess whether a particular rxn will occur.
- Kinetics determines if a substantial rxn will occur in a certain time period.

Chemical Kinetics

- If a rxn is not thermodynamically favored it will not occur under the given condition.
- Or even if it is thermodynamically favored it might not occur at a measurable rate.
- Examples:
- C(diamond) + O2(g) CO2(g) ΔGorxn = -397 kj/mol
- Does not occur at an observable rate at RT.

- C(graphite) + O2(g) CO2(g) ΔGorxn = -394 kj/mol
- Once started occurs rapidly at RT.
Kinetics explains this!

- Once started occurs rapidly at RT.

- C(diamond) + O2(g) CO2(g) ΔGorxn = -397 kj/mol

- Examples:

Chemical Kinetics

- The rate of reaction
- Rate of Rxn usually expressed in moles/L·time or M/time
- With the balanced equation the rate of the rxn can be determined by following the change in concentration of any product or reactant that can be detected quantitatively.
- Examples:
- Measuring color change with spectroscopy
- If a gas is produced, measure the change in pressure

- Examples:

Chemical Kinetics

- Calculating Rate of Rxn:
Sample Rxn: aA + bB cC + dD

Rate of rxn = -1 Δ[A] = -1 Δ[B] = 1 Δ[C] = 1 Δ[D]

aΔt bΔt cΔt dΔt

Coefficients from balanced equation

Reactants – negative b/c they are being used up

Products – positive because they are being created

* We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.

Chemical Kinetics

- Calculating Rate of Rxn:
Rxn: 2N2O5(g) 4 NO2(g) + O2(g)

NO2 is forming at a rate of 0.0072 mol/L·s

a. What is the rate change of [O2] at this time?

b. What is the rate change of [N2O5] at this time?

c. What is the rate of reaction, at this time?

Chemical Kinetics

Rxn: 2N2O5(g) 4 NO2(g) + O2(g)

NO2 is forming at a rate of 0.0072 mol/L·s

- What is the rate change of [O2] at this time?
Mole ratio 1 mole O2: 4 moles NO2

Rate change of [O2] = Δ[O2] = 0.0072 mol NO2 x 1 mol 02

Δt L·s 4 mol NO2

= 0.0018 mols O2/L·s

*The rate of change is positive b/c it is being produced

Chemical Kinetics

Rxn: 2N2O5(g) 4 NO2(g) + O2(g)

NO2 is forming at a rate of 0.0072 mol/L·s

b. What is the rate change of [N2O5] at this time?

Mole ratio 2 mole N2O5: 4 moles NO2

Rate change of [N2O5] = -Δ[N2O5] = 0.0072 mol NO2 x -2 mol N205

Δt L·s 4 mol NO2

= - 0.0036 mols N2O5/L·s

*The rate of change is negative b/c it is being used up.

Chemical Kinetics

Rxn: 2N2O5(g) 4 NO2(g) + O2(g)

c. What is the rate of rxn, at this time?

You can use any of the rate changes for the reactants or products to calculate the rate of rxn.

Using Rate change of [N2O5]

Rate of rxn = -1 Δ[N2O5] = -1 (-0.0036 mol/L·s)

2 Δt 2

= 0.0018 mols /L·s

Chemical Kinetics

Rxn: 2N2O5(g) 4 NO2(g) + O2(g)

c. What is the rate of rxn, at this time?

You can use any of the rate changes for the reactants or products to calculate the rate of rxn.

Using Rate change of [NO2]

Rate of rxn = 1 Δ[NO2] = 1 (0.0072 mol/L·s)

4 Δt 4

= 0.0018 mols /L·s

Chemical Kinetics

Rxn: 2N2O5(g) 4 NO2(g) + O2(g)

c. What is the rate of rxn, at this time?

You can use any of the rate changes for the reactants or products to calculate the rate of rxn.

Using Rate change of [O2]

Rate of rxn = 1 Δ[O2] = 1 (0.0018 mol/L·s)

1 Δt 1

= 0.0018 mols /L·s

Notice that no matter which rate of change you use you get the same rxn rate.

Factors that Affect Reaction Rates

- Four Factors
- Nature of the reactants
- Concentration of reactants
- Temperature
- Presence of a catalyst

- Nature of the Reactants
A. Physical states of reacting substances are important in determining their reactivity’s.

Factors that Affect Reaction Rates

- Examples:
- Gasoline liquid can burn smoothly vs. gasoline vapor which can burn explosively.
- Immiscible liquids react slowly at their interface, but if mixed the reaction speeds up.
- K2SO4(s) + Ba(NO3)2(s) no rxn for years
K2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2 KNO3(aq)

occurs rapidly

Factors that Affect Reaction Rates

- Examples:
- White phosphorus and red phosphorus are allotropes of elemental phosphorus. White phosphorus ignites when exposed to oxygen while red phosphorus can be kept exposed to oxygen for long periods of time.
- Allotropes – different forms of the same element in the same physical state
- Examples:
- O2 (oxygen) & O3 (ozone)
- C (diamond) & C (graphite)

- Examples:

- Allotropes – different forms of the same element in the same physical state

- White phosphorus and red phosphorus are allotropes of elemental phosphorus. White phosphorus ignites when exposed to oxygen while red phosphorus can be kept exposed to oxygen for long periods of time.

Factors that Affect Reaction Rates

- Nature of the Reactants
B. Chemical Identities of Elements & Compounds Affect Rxn Rates

- Examples:
- Na(s) metal + H2O(l) at RT reacts rapidly low ionization energy
- Ca(s) metal + H2O(l) at RT reacts slowly higher ionization energy

Factors that Affect Reaction Rates

- Examples:
- Acids & Bases react rapidly
- Reactions involving electrostatic interactions b/t ions in aqueous solution react more rapidly than rxns involving the breaking of covalent bonds

- Acids & Bases react rapidly

C. Reaction rates depend on surface area & degree of reactant subdivisions

- Example:
- Large chunks of metal do not burn but powdered metals with large surface areas (more atoms exposed to O2) burn easily.

Factors that Affect Reaction Rates

- The ultimate degree of subdivisions would make all reactant molecules (or atoms or ions) accessible at the same time. This situation can be achieved when the reactants are in the gaseous state or in solution.
- Examples:
- Medicines – rapid relief vs. extended release

- Examples:

The Rate Law Expression

- Concentration of Reactants – The Rate Law Expression
- As the concentration of reactants change at constant temperature the rate of rxn changes.
- Rate Law Expression (AKA - Rate Law)
- For Rxn A + B products
- Rate = k[A]x[B]y…

The Rate Law Expression

- Rate Law Expression
- For Rxn A + B products
- Rate = k[A]x[B]y…
- k = specific rate constant at a particular temperature
- x & y = the order of the rxn with respect to A & B; can be integers, zero, fractions or negative; must be determined experimentally has nothing to do with coefficients from the balanced equations.
- x + y = the overall order of the rxn

The Rate Law Expression

- Important Points Regarding Specific Rate Constant (k)
- Once the rxn orders are known, experimental data must be used to determine the value of k for the rxn at appropriate conditions.
- The value is for a specific reaction, represented by a balanced equation
- Units of k depends on the overall order of the rxn
- k does not change w/ concentration of either reactants or products
- k does not change with time
- k refers to the rxn at a particular temperature & changes if we change the temperature
- k value depends on whether a catalyst is present

The Rate Law Expression

- Example:
- Rxn: A + B + C products
- Rate law = k[A][B]2
- What happens to the rxn with the following concentration changes:
- Double concentration A; B & C stay the same
- Double concentration B; A & C stay the same
- Double concentration C; A & B stay the same
- Double concentration A, B & C simultaneously

The Rate Law Expression

- Example:
- Rxn: A + B + C products
- Rate law = k[A][B]2
- What happens to the rxn with the following concentration changes:
- Double concentration A; B & C stay the same.
- rate is directly proportional to [A]; so rate would increase by 2.

- Double concentration B; A & C stay the same
- rate is directly proportional to [B] to the power of 2; so the rate would increase by 22 = 4.

- Double concentration C; A & B stay the same
- rate is independent of [C]; so changing C causes no change in rxn

- Double concentration A, B & C simultaneously rate
- rate increases by a factor of 2 due to change in [A]; and by a factor of 4 due to the change in [B]; and is unaffected by change in C. The result: the rxn rate increase by 2 x 4 = 8.

- Double concentration A; B & C stay the same.

The Rate Law Expression

- Determine the Rate Law from Initial Rates
- Example:
- Rxn: A + 2B C
Rate = k[A]x[B]y

- Rxn: A + 2B C

The Rate Law Expression

- Determine the Rate Law from Initial Rates
- Example:
- Rxn: A + 2B C
- Rate = k[A]x[B]y
Look at experiment 1 & 2 [A] stays the same, so the change

in Initial Rate of Formation of C must be due to the change in [B].

a. How has [B] changed?

2.0 x 10-2 = 2.0 = [B] ratio

1.0 x 10-2

b. How has the rate changed?

3.0 x 10-6 = 2.0 = rate ratio

1.5 x 10-6

The Rate Law Expression

- Determine the Rate Law from Initial Rates
- Example:
c. How to get y.

rate ratio = ([B]ratio)y

2.0 = (2.0)y so y = 1

That means the reaction is first order in [B].

So far: rate = k[A]x[B]1

- To figure out x; we do the same for [A] using experiments 1 & 3.
Since there is no change in [B] in experiment 1 & 3 then the rate difference observed must be solely to do with [A].

The Rate Law Expression

- Determine the Rate Law from Initial Rates
- Example:
- Rxn: A + 2B C
- Rate = k[A]x[B]y
a. How has [A] changed?

2.0 x 10-2 = 2.0 = [A] ratio

1.0 x 10-2

b. How has the rate changed?

6.0 x 10-6 = 4.0 = rate ratio

1.5 x 10-6

The Rate Law Expression

- Determine the Rate Law from Initial Rates
- Example:
c. How to get x.

rate ratio = ([A]ratio)x

4.0 = (2.0)x so x = 2

That means the reaction is second order in [A].

So far: rate = k[A]2[B]1

- You can now solve for k by plugging in the values from any of the 3 experiments and solving for k.
rate = k[A]2[B]1

k = rate

[A]2[B]1

The Rate Law Expression

- Determine the Rate Law from Initial Rates
- Example:
k = rate

[A]2[B]1

Let’s use the values from experiment 2

k = 3.0 x 10-6 M/s

[1.0 x 10-2M]2 [2.0 x 10-2M]1

k = 1.5/M2·s or k = 1.5 L2/mol2·s

The Rate Law Expression

- Determine the Rate Law from Initial Rates
- Example:
rate = 1.5 [A]2[B]1

M2·s

We can say the reaction is 2nd order in [A] and the reaction is 1st order in [B] and the overall reaction is 3rd order.

Integrated Rate Equation

- Concentration vs. Time – Integrated Rate Equation
- The integrated rate equation (IRE) relates concentration and time and can be used to calculate the half-life (t1/2) of a reactant.
- Half-Life (t1/2) – the time it takes for the reactant to be converted to product.
- The IRE & t1/2 are different for rxns of different order.

Integrated Rate Equation

- First Order Rxn:
- Rxn: aA products

Conditions [A] is 1st order Rxn is 1st order overall

Represents coefficients from balanced equation

Rate constant

IRE is ln [A]o = akt (first order)

[A]

time

coefficient

Integrated Rate Equation

- First Order Rxn:
- Rxn: aA products

Conditions [A] is 1st order Rxn is 1st order overall

IRE is ln [A]o = akt (first order)

[A]

[A]o is initial concentration of reactant A

[A] is concentration at some time, t, after rxn begins

Integrated Rate Equation

- First Order Rxn:
- Solving for t:
- t = 1 ln [A]o
[A]

For t1/2:

Definition: [A] = ½ [A]o at t = t1/2

t1/2 = 1 ln [A]o = 1 ln 2 = ln 2 = 0.693

ak ½[A]o ak ak ak

ak

First order

Integrated Rate Equation

- First Order Rxn:
- Example:
1. Half Life: 1st Order Reaction

Rxn: A B + C

k = 0.045/s

What is the half life at 25oC?

Use t1/2 equation from earlier.

a = 1; k is given

- t1/2 = ln 2 = 0.693 = 15.4 s
ak 1(0.045/s)

- Example:

Conditions [A] is 1st order Rxn is 1st order overall

Integrated Rate Equation

- First Order Rxn:
- Example:
1. 1st Order Reaction using IRE

Rxn: 2 N2O5(g) 2 N2O4(g) + 02(g)

Rate = k[N2O5]

k = 0.00840/s

- If 2.50 mol N2O5 were placed in a 5.00L container at that temperature, how many moles of N2O5 would remain after 1.00 min?
- How long would it take for 90% of the original N2O5 to react?

- Example:

Conditions [N2O5] is 1st order Rxn is 1st order overall

Integrated Rate Equation

Example:

Rxn: 2 N2O5(g) 2 N2O4(g) + 02(g)

Rate = k[N2O5]

k = 0.00840/s

- If 2.50 mol N2O5 were placed in a 5.00L container at that temperature, how many moles of N2O5 would remain after 1.00 min?
Use IRE 1st Order Equation:

ln [N2O5]o = akt

[N2O5]

Conditions [N2O5] is 1st order Rxn is 1st order overall

Integrated Rate Equation

Use IRE 1st Order Equation:

ln [N2O5]o = akt

[N2O5]

Step 1: Determine [N2O5]o (initial molar concentration)

[N2O5]o = 2.50 mol = 0.500 M

5.00 L

Step 2: Solve for [N2O5]after 1.00 min.

a = 2k = 0.00840/s t = 1.00 min = 60.0 s

[N2O5] = ?

Note: Be sure

t & k are the same units

Integrated Rate Equation

Solve for unknown [N2O5]:

Definition: ln x = ln x – ln y

ln [N2O5]o = ln [N2O5]o – ln [N2O5] = akt

[N2O5]

ln [N2O5] = ln [N2O5]o – akt

= ln(0.500) – (2)(0.00840/s)(60.0s)

= - 0.693 – 1.008

ln [N2O5] = -1.701 (use the ex key times -1.701)

[N2O5] = 1.82 x 10-1 M = .182 M after 1.00 min

y

Integrated Rate Equation

Step 3: Calculate the number of moles of N2O5 left in the container.

5.00 L x 0.182 mol = 0.910 mol N2O5

L

- IRE involves ratios of concentration, we don’t need the numeric value s for the concentration of N2O5 when 90% has reacted 10.0% remains.
Step 1: [N2O5] = (0.100) [N2O5]o

10.0% original concentration

IRE - First Order Reactions

Step 2: Substitute into IRE:

ln [N2O5]o = akt

[N2O5]

ln [N2O5]o = (2)(0.00840/s)t

(0.100) [N2O5]o

ln(10.0) = (0.0168/s)t

2.302 = (0.0168/s)t

t = 2.302 = 137 seconds

0.0168/s

Remember

a=2

k = 0.00840/s

t = ?

IRE - Second Order Reaction

Conditions

[A] is 2nd order

Rxn is 2nd overall

- Rxn: aA product
2nd Order IRE is:

1 - 1 = akt

[A] [A]o

Initial concentration of reactants

IRE - Second Order Reaction

Conditions

[A] is 2nd order

Rxn is 2nd overall

- Rxn: aA product
Half-life 2nd Order Equation:

t1/2 = 1

ak[A]o

Examples:

Rxn: A + B C + D

rate = k[A]2

k = 0.622 L/mol·min at 30oC

Conditions

[A] is 2nd order

Rxn is 2nd overall

IRE - Second Order Reaction

- What is the half life of A when 4.10 x 10-2 M A is mixed with excess B
- Look at the rate equation: rate = k[A]2
- The rate is independent of B, and as long as there is some of B present to react with A the reaction proceeds at the given rate.
- So use the 2nd Order Half Life Equation:
- t1/2 = 1 = 1
ak[A]o (1)(0.622/M·min)(4.10x10-2M)

= 39.2 min

Initial concentrationof A (given)

Coefficient from balanced equation

k (given)

IRE - Second Order Reaction

- Concentration vs. Time: 2nd Order Rxn – IRE
- Rxn: 2NOBr(g) 2NO(g) + Br2(g)
- rate = k[NOBr]2
- k = 0.810/m·s at 10.0oC
- [NOBr]o = 4.00x10-3 M at 10.0oC
- How many seconds does it take to use up 1.50x10-3M of NOBr
- Step 1: Determine [NOBr] after 1.50x10-3M is used up.
- ? M NOBr remaining = (0.00400 – 0.00150)M
= 0.00250 M NOBr

- ? M NOBr remaining = (0.00400 – 0.00150)M

IRE - Second Order Reaction

- Step 2: Use 2nd Order IRE:
- 1 - 1 = akt (solve for t)
[NOBr] [NOBr]o

t = 1 1 - 1

ak [NOBr] [NOBr]o

= 1 1 - 1

(2)(0.810/M·s) 0.00250M 0.00400M

= 1 (400/M – 250/M)

1.62/M·s

= 92.6 seconds

- 1 - 1 = akt (solve for t)

IRE - Second Order Reaction

- Example 2
- Same Rxn: 2NOBr(g) 2NO(g) + Br2(g)
- rate = k[NOBr]2
- k = 0.810/m·s at 10.0oC
- [NOBr]o = 2.40x10-3 M
- What is the concentration of NOBr after 5.00 min.
- Step 1: Use 2nd Order IRE t = 5.00 min
- 1 - 1 = akt
[NOBr] [NOBr]o

IRE - Second Order Reaction

What is the concentration of NOBr after 5.00 min.

- Step 1: Use 2nd Order IRE t = 5.00 min
- 1 - 1 = akt
[NOBr] [NOBr]o

1 - 1 = (2)(0.810/M·s)(5.00min)(60s/min)

[NOBr] 2.40x10-3M

1 - 4.17x102 = 486

[NOBr] M M

1 = 486 + 417 = 903/M

[NOBr] M M

[NOBr] = 1 = 1.11x10-3M

903/M

IRE - Zero Order Reaction

- Rxn: aA products Zero order reaction
- Therefore rxn rate is independent of concentration
- IRE for Zero Order Rxn:
- rate = k
- [A] = [A]o – akt (Zero Order)

- Half Life Equation for Zero Order Rxn:
- t1/2 = [A]o (Zero Order)
2ak

- t1/2 = [A]o (Zero Order)

Problem Solving Tips: Which Equation Do You Use?

- Step 1: You must decide if you use the Rate Law Expression (RLE) or the Integrated Rate Equation (IRE)
- The RLE relates rate and concentration
- The IRE relates time and concentration
- So if the problem asks you to make calculations involving rxn rate and concentration use RLE
- If the problem asks you to make calculations involving time you have to use the IRE.

Problem Solving Tips: Which Equation Do You Use?

- Step 2: You must decide which form of the RLE or IRE (Zero, First or Second Order) that is right for the given equation.
- Hints:
- The problem may state the order of the rxn
- The RLE may be given; so you can tell the order of the rxn from the exponents in the expression
- The units of the specific rate constant (k) may be given; you can interpret the stated units to figure out the order

- Hints:

Collision Theory of Rxn Rates

- Collision Theory of Rxn Rates – for a rxn to occur, molecules, atoms or ions must first collide.
- Increasing concentration of reacting substances results in a greater number of collisions per unit of time.

Collision Theory of Rxn Rates

- Collision Theory of Rxn Rates.
- Collisions must occur for a chemical reaction to proceed, but they do not guarantee that a rxn will occur.
- Collisions must be effective collisions
- Possess at least a certain minimum energy necessary to rearrange outer electrons in breaking bonds and forming new bonds
- Increasing temperature (↑ KE) creates more molecules with sufficient energy to react.

- Have the proper orientations toward one another at the time of collision.
- Even if molecules collide & have sufficient energy they still might not be in the proper orientation to react.

- Possess at least a certain minimum energy necessary to rearrange outer electrons in breaking bonds and forming new bonds

Transition State Theory

- Chemical Rxn involve making and breaking chemical bonds
- The energy associated with a chemical bond is a form of potential energy
- Therefore, rxns are accompanied by changes in potential energy
- Frequently covalent bonds must be broken so that others can be formed
- This can only occur if molecules collide with enough kinetic energy to overcome the potential energy stabilization the bond provides

Transition State Theory

- Transition State Theory – reactants pass through a short lived, high energy intermediate state called the transition state, before products are formed.
- Activation Energy Ea is the kinetic energy that reactant molecules have to reach the transition state.
- Reactants must possess enough Ea to overcome the “energy barrier” to reach the transition state.
- When the atoms go from the transition state to productsenergy is released.
- If the rxn results in a net release of energy, more energy than Ea is returned to the surroundings, the rxn is exothermic.
- If the rxn results in a net absorption of energy, an amount of energy less than Ea is released to the surroundings then the rxn is endothermic.

Transition State Theory

- The net release of energy = ΔErxn
- ΔErxn relates product energy to reactant energy regardless of pathway
- ΔErxn is negative when energy is given off (exothermic)
- ΔErxn is positive when energy is absorbed (endothermic)

- Increasing temperature changes the rate by altering the fraction of molecules that can get over the energy barrier.
- Introducing a catalyst increases the rate by providing a different pathway that has a lower activation energy.

Rxn Mechanisms & the Rate Law Expression

- Mechanism – the step by step pathway by which a rxn occurs
- Some rxns occur in a single step, but most follow a series of elementary steps.
- The rxn order for a single step elementary step is equal to the coefficients for that step.
- A rxn can never occur faster than its slowest step called the rate determining step.
- The balanced equation for the overall rxn is equal to the sum of all the individual steps.
- Note Again: The rate law exponents do not necessarily match the coefficients of the overall balanced equation.

Rxn Mechanisms & the Rate Law Expression

- Example 1: aA + bB cC + dD
- rate = k [A]x[B]y
- The values of x and y are related to the coefficients of the reactants in the slowest step.

Rxn Mechanisms & the RLE

- Example 2: NO2(g) + CO(g) NO(g) + CO2(g)
- rate = k[NO2]2
- The balanced equation for the overall rxn shows the stoichiometry but does not mean that the rxn is simply the rxn between one NO2 & one CO.
- If the rxn took place in one step then the rxn would be 1st order for both NO2 & CO and the rate would be:
- rate = k [NO2]1[CO]1

- Since the coefficients of the balanced equation do not match the exponents in the rate expression then you know the rxn does not occur in one step.

Rxn Mechanisms & the RLE

- Example 2: NO2(g) + CO(g) NO(g) + CO2(g)
- The current accepted steps for the rxn:
- NO2 + NO2 NO3 + NO (slow)
- NO3 + CO NO2 + CO2 (fast)
NO2 + CO NO + CO2 (overall)

NO3 is an intermediate – a substance formed in one step, but totally consumed in a later step.

Rxn Mechanisms & the RLE

- You should be able to distinguish between various species that appear in a rxn mechanism.
- So far we have three species:
- Reactant – more is consumed than formed
- Product – more is formed than consumed
- Rxn intermediate – formed in earlier steps, then consumed completely in later steps.

- So far we have three species:

Mechanisms and rates

- There is an activation energy for each elementary step.
- Activation energy determines k.
- k = Ae- (Ea/RT)
- k determines rate
- Slowest step (rate determining) must have the highest activation energy.

Temperature: The Arrhenius Equation

- The average kinetic energy of a group of molecules is proportional to the absolute temperature
- At T1: a defined amount of reactant will have enough KE to overcome Ea to form products
- At T2: a larger defined amount of reactant will have enough KE to overcome Ea to form products at a faster rate.
The Arrhenius Equation relates Activation energy, absolute temperature and specific rate constant (k) of a reaction

Temperature: The Arrhenius Equation

- Arrhenius Equation:
- ln k = ln A – Ea/RT
- A = a constant that has the same units as k
- Represents the fraction of collisions with proper orientation when all reactant concentrations are one molar.

- R = universal gas constant
- 8.314 J/mol·K (if Ea is in units of J/mol)

- A = a constant that has the same units as k
- ↑ Ea = ↓ k = ↓ rxn rate
- B/C fewer collisions have enough energy to get over the higher energy barrier.

- ln k = ln A – Ea/RT

Temperature: The Arrhenius Equation

- Arrhenius Equation:
- Predicts: ↑T = ↑rxn rate for the same Ea and concentration
- The Arrhenius Equation for 2 different Temps.

Temperature: The Arrhenius Equation

- Examples
- Rxn: N2O5 NO2 + NO3 Conditions: 1st order rxn
- k = 9.16 x10-3/s @ 0.0oC
- Ea = 88.0 kJ/mol
- Determine k @ 2.0oC
Step 1: Make sure all units agree

Ea = 88,000 J/mol R = 8.314 J/mol·K

k1 = 9.16x10-3/s @ T1 = 0.0oC + 273 = 273 K

k2 = ? @ T2 = 2.0oC + 273 = 275 K

Step 2: Use the 2 Temp Arrhenius Equation

Temperature: The Arrhenius Equation

- Examples

So just raising the temperature 2oC increases the rxn rate significantly

Temperature: The Arrhenius Equation

- Example 2: Activation Energy
- Rxn: C2H5I C2H4 + HI
- k1 = 1.60x10-5/s @ 600.0 K
- k2 = 6.36x10-3/s @ 700.0 K
- What is the Ea
Step 1: Use the 2 Temp Arrhenius Equation to solve for Ea

Temperature: The Arrhenius Equation

- Rearrange to solve for Ea

Temperature: The Arrhenius Equation

- Rearrange to solve for Ea

Temperature: The Arrhenius Equation

- Determining Ea using the Arrhenius 2 Temp equation can have a large degree of error since it depends on the measurement of k at only 2 temperatures.
- To improve the calculation it is better to use many measured values for the same reaction then graph the results.
- ln k = – (Ea/R)(1/T) + ln A
↓ ↓ ↓ ↓

y = m x + b

- ln k = – (Ea/R)(1/T) + ln A

Temperature: The Arrhenius Equation

- ln k = – (Ea/R)(1/T) + ln A
↓ ↓ ↓ ↓

y = m x + b

- A (collision frequency factor) is nearly constant over moderate ΔT.
- Therefore, can be interpreted as the constant term in the equation (the intercept)
- Then the slope of the straight line obtained by plottin
ln k vs. 1/T = – Ea/R

- Then you can determine the value of Ea from the slope.

How Catalysts Work

- Catalysts allow reactions to proceed by a different mechanism - a new pathway.
- New pathway has a lower activation energy.
- More molecules will have this activation energy.
- Do not change E

Catalysts

- Speed up a reaction without being used up in the reaction.
- Enzymes are biological catalysts.
- Homogenous Catalysts are in the same phase as the reactants.
- Heterogeneous Catalysts are in a different phase as the reactants.

Catalysts and rate

- Catalysts will speed up a reaction but only to a certain point.
- Past a certain point adding more reactants won’t change the rate.

Catalysts and rate.

- Rate increases until the active sites of catalyst are filled.
- Then rate is independent of concentration

Rate

Concentration of reactants

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