Kinetics
Reaction Rates, Rate Law, Collision
Theory and Activation Energy (PLN 710)
PLN 7
• Important Concepts:
– Reactions can occur at different rates
– Factors that help determine the reaction rate
– Reaction characteristics:
• Mechanism of reaction (PLN 11)
• Rate of Reaction
• Rate Law (PLN 8)
Basic Kinetics
• Reaction Rate – Speed that reactants
disappear and products form
– How fast reactants become/form products
Examples:
• Very Fast Rates (Almost Instantaneous):
– Most Acid-Base Reactions
• 𝐻𝐶𝑙(𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝐻2 𝑂(𝑙) + 𝐻2 (𝑔)
– Some Precipitation Reactions
• Slower Reaction Rates:
– Rusting
• 3𝐹𝑒(𝑠) + 𝑂2 (𝑔) + 2𝐻2 𝑂(𝑙) → 𝐹𝑒3 𝑂4 (𝑠) + 2𝐻2 (𝑔)
What Determines the Rate?
•
•
•
•
Temperature
Pressure
Concentration
Catalyst (PLN 12)
– Lowers activation energy
• Surface Area
– Not going to be covered on this test
Mechanism of Reaction
• Lists the individual steps of a reaction
• Describe reactions at a molecular level
• Not all reactions occur in one step or all at
once
• Chemical equation is overall summary of the
reaction
Rate of Reaction
• The calculated rate at which reactants are
used up/disappear or products are
formed/appear
• For general reaction:
𝑎𝐴 + 𝑏𝐵 → 𝑐𝐶 + 𝑑𝐷
Where a, b, c and d are coefficients,
𝑅𝑎𝑡𝑒 = −
1∆ 𝐴
1∆ 𝐵
1∆ 𝐶
1 ∆[𝐷]
=−
=
=
𝑎 ∆𝑡
𝑏 ∆𝑡
𝑐 ∆𝑡
𝑑 ∆𝑡
Rate Law
• Mathematic expression for the rate of
reaction
– Expressed in terms of the concentrations of the
reactants
• For a reaction:
A+BC+D
• 𝑅𝑎𝑡𝑒 = 𝑘[𝐴]𝑚 [𝐵]𝑛
Reaction Rates
• Definition:
– The rate of a reaction is the change in molar
concentration of a reactant or product per unit of
time in a reaction
• Example: 2𝑁2𝑂5 (𝑔) → 4𝑁𝑂2 (𝑔) + 𝑂2 (𝑔)
∆[𝑁2 𝑂5 ]
𝑁
𝑂
=
−
• Rate of decomposition of 2 5
∆𝑡
• However, this gives the average rate over the
period of time Δt
• The instantaneous rate can be calculated as the
slope of the tangent line at a given point
Overall Rate of Reaction
• The rate of reaction is more commonly
described in terms of the equation
• For the reaction: 2𝑁2𝑂5 (𝑔) → 4𝑁𝑂2 (𝑔) + 𝑂2 (𝑔)
• For every 2 moles of N2O5 lost:
– 4 moles of NO2 is formed
– And 1 mole of O2 is formed
𝑅𝑎𝑡𝑒 = −
1 ∆ 𝑁2 𝑂5
1 ∆ 𝑁𝑂2
1 ∆[𝑂2 ]
=
=
2 ∆𝑡
4 ∆𝑡
1 ∆𝑡
Note: The negative sign placed in front of the reactants is to count for
the fact that their concentrations are decreasing
PLN 8
• Important Concepts:
– Rate Laws
– Rate Constant (k)
– Order of Reaction
– Initial Rate Method
Rate Laws for Chemical Reactions
• Rates depend on concentrations of certain
reactants and the concentration of the
catalyst, if there is one
• Definition:
– A Rate Law is an equation that relates the rate of a
reaction to the concentrations of the reactants
(and catalyst, if used) raised to various powers, or
exponents.
𝑚
𝑅𝑎𝑡𝑒 = 𝑘[𝐴] [𝐵]
𝑛
• Rate – Expressed in mol/L/time or M/time
• k – Rate constant
– Specific to a certain reaction at a specific temperature
– Units depend on the overall reaction order (explained later)
• [A] & [B] – Concentrations of reactants as
mol/L or M
• m & n – Orders of reaction with respect to
reactants
k
• The reaction constant, k, is called the rate
constant and is dependent on the particular
reaction as well as the specific temperature at
which the reaction takes place
• The units of k depend on the order of reaction
Orders of Reaction
• The rate law exponents are determined using
experimental data
• Examples:
2𝑁𝑂(𝑔) + 2𝐻2 (𝑔) → 𝑁2 (𝑔) + 2𝐻2 𝑂(𝑔)
𝑅𝑎𝑡𝑒 = 𝑘[𝑁𝑂]2 𝐻2
• The overall order of reaction is the sum of all orders
with respect to each reactant
• So for the example, where the rate is 2nd order with
respect to NO and first order with respect to H2:
• The overall order of reaction is 2 + 1 = 3, or a 3rd order
reaction
Determining the Rate Law
Experimentally
• The Initial Rate Method
– Uses the relationship between the measured
initial rate of a reaction and the concentrations of
each reactant
• The Integrated Rate Law Method
– Uses the relationship between reactant or product
concentration and its changes over time
The Initial Rate Method
• By determining the ratio of Δrate to
Δ[reactant] between 2 experiments
•
𝑟𝑎𝑡𝑒2
𝑟𝑎𝑡𝑒1
=
𝑘[𝐴]2𝑛 [𝐵]2𝑚
𝑘[𝐴]1𝑛 [𝐵]1𝑛
=
𝑘 [𝐴]2 𝑛 [𝐵]2 𝑚
𝑘 [𝐴]1
[𝐵]1
• Solve for the exponents for each reactant
The Initial Rate Method – Data
Collection
1. Determine the initial rate of reaction, fixing
the concentration of all reactants except one
2. Repeat step 1, fixing the concentration of
each reactant in turn
Example: 𝑁𝐻4 𝑁𝐶𝑂(𝑎𝑞) → 𝐶𝑂(𝑁𝐻2 )2 (𝑎𝑞)
Experiment
[NH4NCO]
M
Rate of loss of NH4NCO
M/min
1
0.14
2.2 × 10-4
2
0.28
8.8 × 10-4
The Initial Rate Method – Calculations
•
•
•
•
𝑟𝑎𝑡𝑒2
𝑟𝑎𝑡𝑒1
=
𝑘[𝐴]2𝑛
𝑘[𝐴]1𝑛
=
8.8×10−4 𝑀/𝑚𝑖𝑛
2.2×10−4 𝑀/𝑚𝑖𝑛
4 = 2𝑛
ln 4 = 𝑛 ln 2
ln 2
=
𝑘 [𝐴]2 𝑛
𝑘 [𝐴]1
=
ln 4
ln 2
𝑘0.28𝑛
𝑘0.14 𝑛
=
𝑘 0.28 𝑛
𝑘 0.14
=𝑛
• 𝑛=2
• The reaction is 2nd order with respect to NH4NCO
• The reaction is also 2nd order overall, since NH4NCO is the
only reactant and there is no catalyst
Initial Rate Method – Rate Law and k
• Given NH4NCO is 2nd order, we can now determine the
rate law to be:
𝑅𝑎𝑡𝑒 = 𝑘[𝑁𝐻4 𝑁𝐶𝑂]2
• Using this rate law and the experimental data, the rate
constant can also be calculated:
• 2.2 × 10−4 𝑀/𝑚𝑖𝑛 = 𝑘(0.14𝑀)2
•
•
1
× 2.2
0.0196𝑀2
0.011
=𝑘
𝑀×𝑚𝑖𝑛
×
−4 𝑀
10
𝑚𝑖𝑛
2
= 𝑘(0.0196𝑀 ) ×
𝑅𝑎𝑡𝑒 = 0.011[𝑁𝐻4 𝑁𝐶𝑂]2
1
0.0196𝑀2
PLN 9
• Important Concepts:
– Integrated Rate Law Method
• 0th, 1st and 2nd order reactions
– Half-Life
• 0th, 1st and 2nd order reactions
– Units for k
• 0th, 1st and 2nd order reactions
The Integrated Rate Law Method
• Initial Rate Method describes change of rate
as we change initial reactant concentrations
• Using integral calculus, we can convert Rate
Laws into equations that can give us
concentrations of the reactant(s) or product(s)
at anytime during the reaction
• The Integrated Rate Law Method fits
experimental data to a mathematical
relationship
First Order Reactions
• Basic Example:
𝑎𝐴 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
∆𝐴
𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑅𝑎𝑡𝑒 = −
=𝑘 𝐴
∆𝑡
• Which can be rewritten as:
𝑑𝐴
𝑅𝑎𝑡𝑒 = −
= 𝑘[𝐴]
𝑑𝑡
• And simply by cross-multiplying, you can get:
𝑑𝐴
−
= 𝑘 𝑑𝑡
[𝐴]
First Order Reactions (cont.)
• This setup allows integration of both sides:
[𝐴]𝑡
−
[𝐴]0
1
𝑑 𝐴 =𝑘
[𝐴]
𝑡
1 𝑑𝑡
0
Note: The k can be pulled out of the integral since it is a constant.
• ln[𝐴]0 − ln[𝐴]𝑡 = 𝑘𝑡
• Rearranging: ln
[𝐴]0
[𝐴]𝑡
= 𝑘𝑡
Note: The next two slides detail all the steps in the integration process and may be
skipped if you already understand the integration done here.
Explaining the Integration: 1st Order
[𝐴]𝑡 1
[𝐴]0 [𝐴]
𝑡
1
0
•
Beginning with: −
•
The integral of 𝑥 on the interval from 𝑎 to 𝑏 is written as:
𝑑 𝐴 =𝑘
1
𝑏
•
𝑑𝑡
𝑎
1
𝑑𝑥
𝑥
And is solved as:
𝑏
𝑎
1
𝑏
𝑏
𝑑𝑥 = ln 𝑥 = ln 𝑏 − ln 𝑎 = ln
𝑥
𝑎
𝑎
Note: In calculus, the notation “… 𝑥 𝑎𝑏 ” means that you plug in b for x and plug in a for x and subtract the second equation (one
with the a’s) from the first (one with the b’s)
•
So, using our equation, the integration of just the left side looks like this:
𝐴𝑡
−
𝐴0
1
𝑑 𝐴 = − ln 𝐴
𝐴
𝐴
𝐴
𝑡
0
= ln 𝐴
= − ln 𝐴 𝑡 − ln 𝐴
0 − ln 𝐴
𝑡
= ln
0
[𝐴]0
[𝐴]𝑡
= − ln 𝐴
𝑡
− − ln[𝐴]0
Explaining the Integration: 1st Order
(cont.)
[𝐴]0
[𝐴]𝑡
𝑡
1
0
•
Beginning with: ln
•
The integral of 1 on the interval from a to b is written as:
=𝑘
𝑑𝑡
𝑏
1 𝑑𝑥
•
𝑎
And is solved as:
𝑏
1 𝑑𝑥 = 𝑥
•
𝑎
So, using our equation, the integration of just the right side looks like this:
𝑡
𝑘
•
𝑏
= 𝑏−𝑎
𝑎
1 𝑑𝑡 = 𝑘 𝑡
0
𝑡
= 𝑘 𝑡 − 0 = 𝑘𝑡
0
The final equation:
ln
[𝐴]0
= 𝑘𝑡
[𝐴]𝑡
Half-Life of a Reaction
• Definition
– The half-life of a first order reaction is the time taken for the reactant
amount to reach one-half of its initial (or previous) value
• This is saying that [𝐴]𝑡 = half the value of [𝐴]0
1
[𝐴]𝑡 = [𝐴]0
2
• Using substitution into the integrated rate law for 1st order
reactions, we get time of half life 𝑡1 2 :
[𝐴]0
1
ln
= ln
= ln 2
1
1
[𝐴]0
2
2
ln 2 = 𝑘𝑡
ln 2
𝑡1 =
2
𝑘
Rate Law, k, Integrated Rate Law and
Half-Life for 1st, 2nd and 3rd Order
• For only 1st order reactions: The half-life
doesn’t depend on the initial concentration
Order
Rate Law
k
0
Rate = k
𝑚𝑜𝑙
𝐿 × 𝑡𝑖𝑚𝑒
[𝐴]𝑡 − 𝐴
1
Rate = k [B]
1
𝑡𝑖𝑚𝑒
ln
2
𝑅𝑎𝑡𝑒 = 𝑘 [𝐶]2
𝐿
𝑜𝑟
𝑚𝑜𝑙 × 𝑡𝑖𝑚𝑒
𝑅𝑎𝑡𝑒 = 𝑘 𝐶 [𝐷]
Integrated Rate Law
0
= −𝑘𝑡
[𝐵]0
= 𝑘𝑡
[𝐵]𝑡
1
1
−
= 𝑘𝑡
[𝐶]𝑡
𝐶 0
Half-Life
1 [𝐴]0
𝑡1 =
2
𝑘 2
𝑡1
𝑡1
2
2
=
ln 2
𝑘
=
1
𝑘 [𝐶]0
n (number of half lives)
2−𝑛 = (𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑙𝑒𝑓𝑡)
PLN 10
• Important Concepts:
– Collision Theory
• Pre-exponential constant (A)
• fKE
– Importance of Correct Orientation
– Arrhenius Equation
– Activation Energy (EA)
– Transition State Theory
– Potential Energy Diagrams
What Affects Reaction Rates, Again?
• Reaction rates are dependent upon:
– Temperature
– Pressure
– Concentration
– Catalyst
– Surface Area
How Temperature influences Reaction
Rates
• Sometimes the influence temperature has on
the rate of reaction can be quite dramatic, for:
𝑁𝑂(𝑔) + 𝐶𝑙2 (𝑔) → 𝑁𝑂𝐶𝑙(𝑔) + 𝐶𝑙 ∙(𝑔)
• Data:
At 25°C: 𝑘 = 4.9 × 10−6 𝐿 𝑚𝑜𝑙 −1 𝑠 −1
At 35°C: 𝑘 = 1.5 × 10−5 𝐿 𝑚𝑜𝑙 −1 𝑠 −1
Collision Theory
• The Collision Model says that, in order to react,
molecules have to collide, both:
– With enough energy
– And with correct orientation
• In the Collision Model, k depends on 3 factors:
– Z = Collision frequency
– 𝑓𝑜𝑟𝑖𝑒𝑛𝑡 = fraction of collisions that occur with the
molecules properly oriented
– 𝑓𝐾𝐸 = fraction of molecules having or exceeding the
required activation energy
Changes in Temperature
• Z and forient are generally combined into one
– Pre-exponential constant = A
• A is essentially independent of any temperature
change, so fKE is the critical factor of k when it
comes to changes in temperature
• 𝑓𝐾𝐸 = 𝑒 −𝐸𝐴 /𝑅𝑇 where:
– R = ideal gas constant in terms of 𝐽 𝑚𝑜𝑙 −1 𝐾 −1
• 8.314 𝐽 𝑚𝑜𝑙 −1 𝐾 −1
– T = temperature in kelvin
– EA = Activation Energy for the process
Importance of Correct Orientation
• For the reaction: 𝑁𝑂2 𝐶𝑙 + 𝐶𝑙 → 𝑁𝑂2 + 𝐶𝑙2
• Consider two possible ways for the reactant
molecules to collide:
Arrhenius Equation
𝑘 = 𝑓𝑜𝑟𝑖𝑒𝑛𝑡 × Z × 𝑒 −𝐸𝐴 /𝑅𝑇 = 𝐴 𝑒 −𝐸𝐴 /𝑅𝑇
• Taking the natural log of both sides:
• ln 𝑘 = ln 𝐴 −
𝐸𝐴
𝑅𝑇
• Rearranged: ln 𝑘 =
−𝐸𝐴
𝑅
1
𝑇
+ln 𝐴
looks somewhat similar to: 𝑦 = 𝑚𝑥 + 𝑏
• In fact, it is a linear equation if you plot ln 𝑘 𝑣𝑠.
– Where the 𝑠𝑙𝑜𝑝𝑒 =
−𝐸𝐴
𝑅
– And the 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = ln 𝐴
1
𝑇
Calculating EA for an Equation
• By subtracting: ln 𝑘2 = ln 𝐴 −
• From: ln 𝑘1 = ln 𝐴 −
• We get:
𝐸𝐴
𝑅𝑇1
𝐸𝐴
𝑅𝑇2
−𝐸𝐴 −𝐸𝐴
ln 𝑘1 − ln 𝑘2 = ln 𝐴 − ln 𝐴 +
−
𝑅𝑇1 𝑅𝑇2
• (ln A – ln A) = 0, simplify ln x – ln y and combine like
−𝐸𝐴
terms
:
𝑅
𝑘1
𝐸𝐴 1
1
ln
=
−
𝑘2
𝑅 𝑇2 𝑇1
Example
• For: 𝐶𝐻4 (𝑔) + 2𝑆2 (𝑔) → 𝐶𝑆2 (𝑔) + 2𝐻2 𝑆(𝑔)
k (L mol-1 s-1)
T (°C)
1.1
550
6.4
625
• Plug in values for k1, k2, T1 and T2 into the equation:
• ln
1.1𝐿𝑚𝑜𝑙 −1 𝑠 −1
6.4𝐿𝑚𝑜𝑙 −1 𝑠 −1
=
𝐸𝐴
1
8.314𝐽𝑚𝑜𝑙 −1 𝐾 −1 898𝐾
−1 −1
• Solve for EA: −1.76𝐿𝑚𝑜𝑙 𝑠 =
𝐸𝐴
−1 −1 −0.000101𝐾
8.314𝐽𝑚𝑜𝑙
𝐾
𝐸𝐴 = 140,000 𝐽 = 140 𝑘𝐽
−
1
823𝐾
Transition State Theory
• Transition State Theory describes what
happens to the reactant molecules as a
reaction proceeds
• When the reactants collide, they form a
temporary “substance” composed of a
combination of the two reactants
– This temporary “substance” is called the
transition state or activated complex
Transition State
• For:
𝑁𝑂2 𝐶𝑙 + 𝐶𝑙 → 𝑁𝑂2 ⋯ 𝐶𝑙 ⋯ 𝐶𝑙
• Where a temporary complex:
– 𝑁𝑂2 ⋯ 𝐶𝑙 ⋯ 𝐶𝑙
•
•
•
•
‡
Is formed, before the
𝑁 ⋯ 𝐶𝑙 bond is broken
and the 𝐶𝑙 ⋯ 𝐶𝑙 bond
is completely formed
‡
→ 𝑁𝑂2 + 𝐶𝑙2
Transition States (cont.)
• The double dagger (‡) indicates a transition
state
• The transition state is a step in-between
forming a bond and breaking another
– Can be thought of as one half broken bond and
one half formed bond
• The partially formed and partially broken
bonds are denoted with (⋯)
Potential Energy Diagrams
• A graph of Potential Energy vs. the Reaction
Coordinate
– The reaction coordinate is essentially the progress
of the reaction
Exothermic Reaction
Endothermic Reaction