Ch. 13

Question 1

Reaction Rates

Answer 1

is the change in concentration of reactants or products with time :
delta [] / Delta t

As the reaction proceeds, concentration of reactants decreases and concentration of products increases

Reaction rates are usually concentration dependent

Concentrations of reactants change as the reaction proceeds, which in turn changes the rate of the reaction, therefore rate of rxn is not constant (unless 1st order rxn)

Faster the rxn = steeper the slope

Question 2

Instantaneous rate

Answer 2

is the slope of the concentration
vs. time curve at a given point.

Over a sufficiently short time interval, average rate approaches instantaneous rate.

Question 3

Rate Laws

Answer 3

For any reaction, we can write a rate law to describe how the instantaneous rate depends on reactant concentration:

Rate = k[reactant]ⁿ

“Rate is equal to the [] of reactant to the power of some exponent multiplied by some constant.”

The rate law does not account for reverse reaction

Question 4

Rate constant (k)

Answer 4

k is the rate constant

– constant for a specific reaction at a specific temperature
– Fast reactions have large k, slow reactions have small k
– increases with temperature

Question 5

Reaction order (n)

Answer 5

n is the reaction order

– an exponent that defines how sensitively the rate depends on reactant concentration
– n is experimentally determined and depends on the reaction mechanism
– n may be different from the stoichiometric coefficient

Ex: exactly how much faster depends on nature of that reaction
–> for some, might double [] of rxn and see twice the rate if rxn. therefore n = 1.
or double the [] of rxn and see rate of rxn increase by 4 times, therefore n = 2.

0 order rxn, rate constant is moles per liter per sec
1st order, rate constant is mols per second to the minus 1
2nd order, rate constant is liters per mol per sec

• the higher the order, the more - the exponent gets in the [] units & rate constant.

Question 6

method of initial rates

Answer 6

An experimental technique designed to measure the instantaneous rate at the beginning of the reaction to make data analysis much easier to interpret.

– Rate is measured immediately after mixing reactants
– The measuring interval is short enough to prevent the concentrations from changing
appreciably
– The experiment is repeated with different concentrations of reactants

However, random experimental error and non-integer multiples of
concentration can make it difficult to determine reaction order by inspection of real experimental data

Question 7

determining reaction order mathematically using method of initial rates with a single reactant

Answer 7

Rate₁ = k[A]₁ⁿ = Rate₁ = k̶[A]₁ⁿ = Rate₁ = [A]₁ⁿ
———————————————————
Rate₂ = k[A]₂ⁿ = Rate₂ = k̶[A]₂ⁿ = Rate₂ = [A]₂ⁿ

“ratio of 2 rates equals the ratio of [] of A to the exponent of the rxn order”

Next, you take the log of both sides.

log (rate₁/rate₂) = log([A]₁ⁿ/[A]₂ⁿ) = log([A]₁/[A]₂)ⁿ = (n)log([A]₁/[A]₂)

“log of ratio of rates = log of ratio [] multiplied by exponent, multiplied by the rxn”

n = log (rate₁/rate₂) / log([A]₁/[A]₂)

Once n is known, k can be calculated from the rate data

if n = 0 –> change of [] will not change rate.
if n = 1 –> double [] will double rate.
if n = 2 –> doubling [] will 4x the rate

Question 8

determining reaction order mathematically using method of initial rates with multiple reactants

Answer 8

Rate = k[A]ⁿ[B]^m

Rate₁ = k[A]₁ⁿ[B]^m = Rate₁ = k̶[A]₁ⁿ[̶B̶]̶^m̶ = Rate₁ = [A]₁ⁿ
divided by
Rate₂ = k[A]₂ⁿ[B]^m = Rate₂ = k̶[A]₂ⁿ[̶B̶]̶^m̶ = Rate₂ = [A]₂ⁿ

Next, you take the log of both sides.

log (rate₁/rate₂) = log([A]₁ⁿ/[A]₂ⁿ) = log([A]₁/[A]₂)ⁿ = (n)log([A]₁/[A]₂)

n = log (rate₁/rate₂) / log([A]₁/[A]₂)

Repeat for all other reactants

Once all reaction orders are known, k can be calculated from the rate data

gives n, rxn order with respect to reactant A and observe the effect on the ratio
if double [A] –> double rate –> know its 1st order with respect to A
if double [A] –> Quadruples rate –> know its 2nd order with respect to A

Question 9

differential rate law

Answer 9

How the rate depends on concentration.

Rate = k[reactant]ⁿ

It tells us how fast the reaction will be if we know the reactant concentrations

It does not (directly) tell us how far the reaction will proceed in a given time

Question 10

integrated rate law

Answer 10

How concentration depends on time.

helps us determine what the concentrations will be at some time in the future.

Ex: rxn proceeds at a certain rate, if let go on longer then did in method of initial rates, to a point where reactant [] are changing the significantly, that’ll feed back and change the rxn rate. therefore you have to integrate that function over time to figure out what happens.

Question 11

Half life

Answer 11

the time required the initial
concentration of reactant to decrease by half

Question 12

Zero Order Reactions

Answer 12

• not dependent on the reactant [].
• proceed at a constant rate until the reactant is depleted, therefore the more initial reaction have, the longer the rxn will run.

Rate Law: Rate = k[reactant]⁰ = k

• Anything to the power of 0 is 1, therefore rate = rate constant

Units of k: M x s^-1
- “moles per liter per second”

Integrated Rate Law: [A] = [A]₀ - kt

• ”[] at time t is equal to initial [] minus kt”

Straight Line Plot:
[A]_t vs time
Slope = -k
y-intercept = [A]₀

Half Life Expression: t_1/2 = [A]₀/2k

• [] dependent, therefore HL is shorter than previous HL (is half the previous HL)
• more in[reactant] –> longer life –> slower rxn
• less in[reactant] –> shorter life –> faster rxn

Question 13

1st Order Reactions

Answer 13

• reactions have a constant half-life that does not depend on concentration
  –> exponential decay y ∝ 1/2ⁿ
• To linearize the exponential function, we plot the (natural) log of the concentration vs. time
• no units in rate constant (dimensionless in integrated rate law)
• any [] units work, so long as same
• rxn proceeds quickly at beginning and slows down as reactant is depleted
• half reactant [], must half rate of rxn
• rate increases when [] of reactant higher
• rate decreases when [] of reactant lower

Rate Law: Rate = k[reactant]¹

• “rate = rate constant times [] of reactant”
• rate linearly dependent on [] of reactant

Units of k: s^-1

• “seconds to the minus 1”

Integrated Rate Law: in[A]_t = kt = in[A]₀

• “natural log of [] at some time t = natural log of in[] minus kt”
• if know in[] & time –> able to calculate final reactant [].
• if know in[] & final [] –> able to calculate how long that takes.
• if know final [] & how long rxn went for –> able to calculate in[].

or
in [A]_t / [A]₀ = - kt

• ” natural log of the ratio of []’s = - kt”
• equation allows us to calculate how much reaction occurs in a particular time interval, or how long it will take for the concentration to change by a particular amount

Straight Line Plot:
in[A]_t vs time
slope = - k
y-intercept = in[A]₀

Half Life Expression: t_1/2 = 0.693/k = 1/k (0.693)

• HL related to rate constant, (is proportional), meaning if know HL can calculate rate constant and vice versa
• wherever you start off, half way to completion will always take the same amount of time proceeding HLs.
• larger rate constant = faster rxn = shorter the HL
• smaller rate constant = slower rxn = longer the HL.

Question 14

2nd Order Reactions

Answer 14

• more dependent on [reactant] b/s it depends on the square of the [], therefore, see more of a rapid decrease in rxn as [reactant] goes down.
• hyperbolic function y ∝ 1/x

Rate Law: Rate = k[reactant]²

• ” rate = rate constant times [] of reactant squared”
• rxn rate proportational to square of [reactant]
• halfing [reactant] cuts rxn rate by a factor of 4 –> will take twice as long.

Units of k: M^-1 x s^-1
or
L x mol^-1 x s^-1

• “liters per mol to the minus 1 times sec to the minus 1”

Integrated Rate Law: 1/ [A]_t = kt + 1/ [A]₀

• “reciprocal of [A] at some time = reciprocal of in[A] + kt”
• To linearize this function, plot the reciprocal of concentration vs. time.

Straight Line Plot:
1/[A]_t vs time
slope = k
y-intercept = 1/ [A]₀
slope goes from bottom (left) up (right), while other two go from top left, to down right.

Half Life Expression: t_1/2 = 1/[A]₀ = 1/k = 1/[A]₀

• depends on []
• Each successive half-life is now twice the previous half life.
• HL is not a constant

Question 15

Determining rxn order from experimental data

Answer 15

• graphically determine the integrated rate law
• Reactant concentration (or a related value) is plotted vs. time
• Each integrated rate law (and thus each reaction order) requires a different plot to produce a straight line.
• Trying all possibilities until a linear graph is obtained will allow the reaction order to be determined
• The rate constant can then be determined from the slope.

if plot [A]_t vs time and get a /, = 0 order. but if get curve, then plot nat log of [] (in[A]_t vs time), if get /, = 1st order. but if get curve, plot reciprocal, if get /, = 2nd order.

Question 16

More Complex Cases

Answer 16

The method of initial rates is readily extended to more complicated
reactions.

More concentrations must be controlled, more trials must be run and more calculations must be done.

Data analysis still follows the same patterns.

Question 17

More Complex Integrated Rate Laws

Answer 17

Using the method of initial rates, we found:

Rate = k[BrO3-][Br-][H+]^2

Integrating a more complex rate law like this is beyond the scope of this course

The problem can be simplified if conditions are chosen such that one of the reagents can be completely consumed without substantially affecting the concentrations of the others

The essentially unchanging concentrations can be regarded as constants.

For example:
* If [Br-]0 and [H+]0 are large relative to [BrO3-]0, we can define:
* k’ = k[Br-]0[H+]02 and Rate = k’[BrO3-]
* This is a pseudo-first order rate law
* A plot of ln[BrO3-] = -kt + ln[BrO3-]0 should be linear
* Given an initial [BrO3-] and the rate constant, we can calculate the [BrO3-] at sometime in the future

Question 18

Summary of Rate Laws

Answer 18

Rate laws are experimentally determined

• Rates are studied under conditions where only the forward reaction is occurring
  to any appreciable extent
• Two types of rate laws:
  – Differential rate law is determined using the method of initial rates.
  – Instantaneous rates are determined as the initial concentrations are varied.
  – Integrated rate law is determined by measuring the concentration of a reactant at varying times.
  – The rate law is determined by curve fitting

Concentrations that don’t change under the experimental conditions can be
combined with the rate constant
– This yields pseudo-order rate law
– Pseudo-order rate laws simplify integration

Question 19

Theoretical Explanation of Rate Laws

Answer 19

Chemical reactions involve breaking of bonds, formation of new bonds and/or transfer of electrons between atoms or molecules

• In the simplest reactions, all of these changes can occur in a concerted fashion (at the same time in a single step)
• The rate laws for these simple reactions correspond to the reaction stoichiometry
• A simple decomposition will be first order
  – A → product(s) rate = k[A]
• An aggregation or exchange reaction will be second order
  – A + B → product(s) rate = k[A][B]

Question 20

Unimolecular Kinetics

Answer 20

• Why are decomposition reactions first order?
  – Only a single molecule is involved
• Under a given set of conditions, there is a certain probability that the molecule will decompose in a given period of time
  – This probability determines the rate constant
• Statistically, the number of decompositions in a period of time will depend on how many precursor molecules there are
  – This results in first order kinetics
• While not a chemical reaction, radioactive decay is a classic example of a first order process

Question 21

Bimolecular Kinetics

Answer 21

• For an aggregation or exchange reaction to occur, two molecules must encounter each other (collide)
• The probability of this encounter increases as both concentrations increase
  – The rate law is first order with respect to each reactant
• What if both reactants are the same?
  – The probability of two identical molecules encountering one another increases as the square of concentration
• In both cases, the overall reaction is second order.

Question 22

Rate determining steps

Answer 22

• Most reactions involve a series of these simple (or elementary) reactions
• The overall rate of a multi-step reaction is limited by the slowest elementary step
• Consider a fluid flow analogy:
  – placing a restriction (such as a valve) anywhere in a large pipe will reduce the rate at which water arrives at the end
  – total flow is controlled by the restriction (opening or closing the valve controls the flow rate)
• The rate law of a multi-step reaction is determined by the rate determining step
• Reactions can be zero order w.r.t a reactant if that reactant is not involved in the rate determining step

Zero-order kinetics also occur in systems that can be saturated with reactant
* Such systems include enzymes and heterogeneous catalysts
– At sufficiently low [reactant], the reaction order changes

Question 23

Reaction Mechanisms

Answer 23

• The sequence of elementary reactions that leads from reactants to products is called a reaction mechanism
• Kinetic data can be used to help discover reaction mechanisms
• A proposed mechanism whose slowest elementary step does not match the observed rate law can be ruled out
• bond breaking is expected to be slower than bond formation

Question 24

higher order kinetics

Answer 24

What about reactions with more than two species in the rate law, or with unusual orders (higher than 2, fractional orders, etc.)?

• Very few reactions involve more than two molecules
  – Termolecular collisions are extremely unlikely
• For many reactions the first step is not rate determining
• Earlier steps in the reaction are often equilibria
  – The equilibrium determines the concentrations of the reactants in the rate determining step

Equilibrium intermediate concentrations depend on both the equilibrium constant and the reactant concentrations

• There may be more than one reactant (or multiple molecules of the same reactant) in the equilibrium
  – This leads to rate laws with an overall order greater than 2nd
• There may be multiple molecules/ atoms of the product of the equilibrium
  – This can lead to rate laws with fractional orders

Question 25

Why do reactions proceed at different rates?

Answer 25

• For many reactions, the reaction rate is much lower than the rate at which the reactant molecules encounter each other
• This implies that the majority of collisions between reactant molecules do NOT result in a
  reaction
• There are two reasons why a collision may not result in reaction:
  – The collision is not energetic enough
  – The collision occurs with the molecules in the wrong orientation

Question 26

Activation Energy

Answer 26

Product cannot form unless
enough energy is added to the
reactant to get to the transition state

Question 27

Collision Model

Answer 27

• Kinetic energy can be harnessed to break bonds when molecules collide
• Only collisions that are energetic enough to overcome the activation barrier can produce a reaction
• In a gas or liquid sample, there is a distribution of velocities
• By extension, there also is a distribution of collision energies
• The distribution of energies varies with temperature (higher temperatures, higher velocities, higher number of collisions)
• The fraction of collisions increases exponentially with temperature

𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛𝑠 𝑤𝑖𝑡ℎ E_a = (total collisions)e^-(E_a/RT)

Question 28

The Effect of Temperature

Answer 28

As temp increases, the fraction of molecules with enough energy to surmount the activation energy barrier also increases.

Question 29

The effect of orientation

Answer 29

Molecules colliding in the wrong orientation cannot from the new bond needed in the products and reaction cannot occur, regardless of energy.

Question 30

Arrhenius Equation

Answer 30

1. Collisions must have at least Ea
   for reaction to occur
2. Orientation of the molecules needs to be taken into account
   * Rate constant can be represented as: k = zpe^-(E_a/RT)

• z is the collision frequency (total number of collisions/s)
• p is the steric factor which reflects the fraction of collisions with the
  appropriate orientation
• Generally this expression is written as:
  𝑘 = 𝐴𝑒^{−(𝐸𝑎/𝑅𝑇)}
• Z and p are combined in the pre-exponential (or frequency) factor, A
• The Arrhenius equation can be rearranged to obtain the following linearized equation:

in 𝑘 = − 𝐸𝑎/𝑅 (1/𝑇) + in𝐴

• The activation energy can be obtained by graphing ln(k) vs. 1/T
• y = mx + b, m = -Ea/R
• Alternatively, one can simply measure the rate constant at two different
  temperatures:
  𝑙𝑛(𝑘2/𝑘1) = 𝐸𝑎/𝑅 (1/𝑇1 − 1/𝑇2)

Question 31

Increasing Reaction Rates

Answer 31

• According to the Arrhenius equation, (k = Ae-(Ea/RT))the rate constant depends on the temp, the frequency factor and the activation energy
• One can easily increase the rate of reaction by increasing the temp, but this is not always a practical option
  – Heating requires energy and more expensive reaction vessels
  – Heating will also speed up side reactions, sometimes by a greater amount than the main reaction
• The other option is to reduce the activation energy and/or increase the frequency factor
• This requires that the reaction proceed via a different pathway
  – Providing this pathway is the job of a catalyst

Question 32

Catalysis

Answer 32

• A catalyst is a substance that
  speeds up a reaction without
  being consumed itself
• A catalyst provides an alternate
  pathway that has a lower Ea
• The energy difference between
  reactants and products remains
  the same when using a catalyst!
  – Catalysts lower Ea, but do not
  change delta H!

Question 33

Heterogeneous Catalysis

Answer 33

• Heterogeneous catalysis occurs when the catalyst and reagents are in different phases
• Involves reagents being adsorbed to a solid catalyst
• Intermediates and transition states are bound to the catalyst surface by covalent and/or non-covalent interactions
• These interactions stabilize the intermediates and/or transition states and lower their energy
• Surface area limitations result in zero-order kinetics at high reactant concentrations
  – Reaction is still first order w.r.t catalyst surface area

Question 34

Adsorption vs. Absorption

Answer 34

• Adsorption refers to the collection of one substance on the surface of another substance
• Absorption refers to the penetration of one substance into another substance

Question 35

Homogeneous Catalysis

Answer 35

• For homogeneous catalysis the catalyst is in the same phase as the reactant molecules
• The destruction of stratospheric ozone can be catalysed by chlorine atoms
  Cl + O3 → ClO + O2
  ClO + O3 → Cl + 2 O2
• Note that while chlorine atoms participate in the reaction, they are not consumed
• Both reactions are first order in O3
  (they do not become zero order at high [O3])
• Both reactions have a much lower activation energy than the uncatalysed process (2 O3 → 3 O2), which does not normally occur to an appreciable extent

Question 36

Enzyme Catalysis

Answer 36

• Enzymes are highly effective catalysts found in biological systems
• All enzymes operate in aqueous solution, and most only work at near-neutral pH and temps below ~45 °C
• Enzymes can catalyse reactions that are otherwise impossible
  – the temperature required to allow the uncatalysed reaction to proceed at a reasonable rate would result in decomposition
• Enzymes are homogeneous catalysts (they are dissolved in the reaction mixture), but they bind their substrates like heterogeneous catalysts
• Substrates are bound in a way that ensures proper orientation of reactants
• Transition states are bound even more tightly than reactants or products
• This reduces the activation energy by lowering the energy of the bound transition state relative to the bound reactants
• Substrate binding by a homogeneous catalyst results in mixed kinetics

Question 37

Enzyme Kinetics

Answer 37

• At low substrate concentration the reaction is 1st order w.r.t. substrate, but at higher concentration the enzyme becomes saturated and the reaction becomes zero order
  w.r.t. substrate
• The reaction is always 1st order w.r.t. enzyme concentration
• This behavior is described by the Michaelis-Menten equation:

V₁ = V_max[S] / {K_m + [S]}

• V₁ is the reaction rate, and Vmax is the
  rate when the enzyme is saturated with substrate
• Km is the substrate concentration at which the rate is ½ of the maximum
• Enzyme reactions may have multiple substrates