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Chapter 11

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Chapter 11

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.

- 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:

- 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:

- 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.

- 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?

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

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.

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

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

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.

- 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.

- 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

- 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.

- 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

- 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.

- 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:

- 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…

- 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

- 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

- 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

- 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.

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

- Rxn: A + 2B C

- 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

- 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].

- 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

- 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

- 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

- 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.

- 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.

- 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

- 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

- 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

- 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

- 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

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

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

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

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

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 = ?

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

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

- 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)

- 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

- 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)

- 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

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

- 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)

- 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.

- 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 – 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.
- 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

- 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 – 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

Exothermic Reaction

- 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.

- 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.

- 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.

- 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.

- 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.

- 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:

- 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.

- This reaction takes place in three steps

First step is fast

Low activation energy

Ea

Ea

Second step is slow

High activation energy

Ea

Third step is fast

Low activation energy

Second step is rate determining

Intermediates are present

Activated Complexes or Transition States

- 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

- 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

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

- 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

- Examples

Use the ex key times 0.282

- Examples

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

- 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

- Rearrange to solve for Ea

- Rearrange to solve for Ea

- 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

- 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.

- 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

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

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

Rate

Concentration of reactants