Chemistry 2 Flashcards
Give some examples of slow and fast reactions
Value of k (rate constant). When k is large the reaction is fast. True or false
Under what conditions will the reaction occur spontaneously?
How far does the reaction proceed?
Kinetics: Important Questions
• How fast/slow do chemical reactions occur? • How can I control a chemical rate?
• What factors affect reaction rates?
• How does the reaction occur (what is the reaction mechanism )?
Slow
rusting (oxidation of Fe metal) – corrosion of Fe
smoldering of peat (slow, low-temperature combustion)
browning of fruit and vegetables (enzymatic oxidation of polyphenols)
• Fast
explosions:
detonation of nitroglycerin in dynamite alkaline metals + water
chemiluminescence:
A + B➡️[AB]➡️[AB] + hv
light produced by a chemical reaction: glowsticks, fireflies etc.
Under what conditions will the reaction occur spontaneously?
– Thermodynamics (we’ll explore this in Ch. 19)
– Enthalpy (ΔH), entropy (ΔS), Gibbs free energy (ΔG)
• How far does the reaction proceed?
– Equilibrium (Principles I material: review Ch. 15)
– Value of K or Keq
(special cases: Keq – acid dissociation; Ksp – precipitation).
What factors affect the speed of a reaction
More about methanol combustion: http://www.youtube.com/watch?v=FSdBB1vBDKY
For platinum catalyst oxidation of
Methanol Combustion (oscillating flame reaction)
2CH3OH + 3O2➡️2CO2 + 4H2O
What fast reaction occurs when b methanol and oxygen are available at room temperature and K for a combustion in dish
Both reactants have to be present for the reaction to occur.
Once the oxygen inside the flask is used up ___________________ will occur until more O2 molecules flow into the flask for combustion in a flask
The flask has been cooled down on
ice. The reaction will ___________________ at low T.
For cold methanol
Pt Wire + Flask + Methanol
Pt catalyzes the reaction as long as ____________ reactants are present.
What factors affect the speed of a reaction?
- Physical State
eg. fine powder vs solid block vs gas vs liquid - Reactant Concentration eg. dilute vs concentrated
- Reaction Temperature eg. hot vs cold
- Catalysts eg. alcohol sensitivity related to aldehyde dehydrogenase
These are macroscopic variables with submicroscopic causes. Objectives: We need to be (1) quantitative and (2) to develop
insight and understanding
Combustion
What are reaction rates
What are the units
What is the formula for average rate for formation of products
What is the formula for average rate of loss of reactants
Calculate the average rate of the hypothetical reaction,
Red → 2Blue
for the first 10.0 seconds of the reaction.
time = 0.0 s 10.0 s
[Red]= 0.08M 0.04M
[Blue] = 0.00 M 0.08 M
Reaction rate - change in the concentration of reactants or products per unit time.
Reaction Rate = the change in concentration per unit time taking into account stoichiometry and reaction direction. – Units are typically in M/s (molarity per second)
For a generic reaction:
aA+bB→cC+dD
Rate =-1/a x delta A/delta t= -1/b x delta B/delta t=1/c x delta C/delta t=1/d x delta D/delta t
(These are related to the slopes for the graph)
Δ[ ] = change in concentration
Δt = change in time
a, b, c, d = stoichiometric coefficients Note:
• minus sign for reactants
• plus sign for products so that the rate of the reaction is always positive.
• Consider also the reaction
A→B
average rate of appearance of B= Δ[B]Δt
average rate of disappearance of A=
− (Δ[A]/Δt)
Formation of products:
Average Rate = Δc/Δt
Δc/Δt > 0
(increasing concentration with time)
Average Rate for Loss of Reactants
Average Rate = -Δc/Δt
Δc/Δt < 0
(concentration decreases with time; negative sign ensures that the rate is positive)
Average rate [A]final - [A]initial/tfinal – tinitial
= Δ[A]/Δt
[A]initial is not necessarily [A] at time 0
Instantaneous rate slope of the tangent to the curve
(rate at an arbitrarily short time) Initial rate slope near time = 0
At Δ[A]➡️d[A], the average rate➡️instantaneous rate.
Reactant)Red → 2Blue(product)
red] = 0.08M 0.04M 0.02M 0.01M
blue] = 0.00 M 0.08 M 0.12 M 0.14 M
Objectives:
1. Understand that rate is associated with the slope of the
concentration vs time plots.
2.Appreciate how stoichiometry can influence the rate and must be
accounted for in defining the rate of the reaction.
3. Be able to discuss how rate changes with time.
Rate=−1/1multiplied by Δ[Red]/Δt
=−1/1 multiplied by (0.04−0.08)/ (10.0−0.0)s
=0.004M/s
Rate=+1/2 multiplied by Δ[Blue]/Δt =+1/2(0.08−0.00)M/(10.0−0.0)s
=0.004M/s
What is the form of the rate law?
O3(g) + 2NO2(g) → O2(g) + N2O5(g) Rate = k[O3][NO2]
This reaction is 1st order with respect to O3, 1st order with respect to NO2, and 2nd order overall.
2 NO2(g) → 2 NO(g) + O2(g) Rate = k[NO2]to the power 2
• ___________ with respect to NO2, and 2nd order overall.
(CH3)3CCI + OH- → (CH3)3COH + Cl- Rate = k[(CH3)3CCI]
• This reaction is ___________with respect to (CH3)3CCI, 0th order with respect to ________, and ______________ overall.
What is the unit of the rate constants for first order reaction and second order reaction
What is the Strategy for Determining Rate Law Using Initial Rates
Consider the reaction and the kinetic data in the table: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) (in CCl4 solvent)
Exp. No. 1 2 [N2O5]o (mol/L) 0.45 M 0.90 M Initial Rate (molL-1⋅s-1) 2.7 × 10-4 5.4 × 10-4 Using the experimental data in the table, determine the rate law. Rate = k[N2O5]to the power m What is “m”? Question 2 Consider the reaction and the kinetic data in the table: 2 NOCl(g) → 2 NO(g) + Cl2 (g) Exp. No. 1 2 [NOCl]o (mol/L) 0.0033 M 0.0066 M Initial Rate (mol⋅L-1⋅s-1) 2.7 × 10-4 1.08 × 10-3
For a generic chemical reaction,
aA + bB → cC + dD
the rate law generally has the form: Rate = k[A]raised to the power m[B]raised to the power n
• k is the rate constant (k depends on the temperature).
• The exponents define the order of the reaction:
• m is the order of the reaction with respect to A.
• n is the order of the reaction with respect to B.
• (m + n) is the overall reaction order.
• m and n do not necessarily equal a and b!
• m and n are usually integers.
Units of rate = (units of rate constant)multiplied by (units of concentration)raised to the power the order
Examples:
First order reaction
Units of rate = rate constant x concentration
Units of rate constant = units of rate /concentration
=M/s/M
= s raised to the power -1
Second order reaction
Units of rate = rate constant x (concentration) to the power 2
Units of rate constant = Units of Rate/concentration to the power 2
= M /s/M to the power 2
= M to the power −1 s to the power −1
How are rate laws determined?
Remember: Rate laws must be determined experimentally.
method of initial rates = determines reaction orders from the effect of changing a reactant’s concentration on the initial rate of the reaction
Rate = k[A]to the power m[B]to the power
Solution:
When concentration doubles, the rate doubles so m = 1 and the reaction is first order.
Rate = k [N2O5]
Exmaple 2 solution
Rate 2/Rate 1= 1.08 × 10-3/2.7 × 10-4
= 4
[NOCl]to the power m exp1/ [NOCl]to the power m exp2
=[0.0066]to the power m / [0.0033]to the power m
=(2) to the power m
m=2
Question 3 on using method of initial rates
Deduce the rate law for the reaction, BrO3-(aq) + 5 Br-(aq) + 6 H+(aq) → 3 Br2(l) + 3 H2O(l) from the following experimental data: Exp. 1 2 3 4 [BrO3-]o 0.10M 0.20M 0.20M 0.10M [Br-]o 0.10M 0.10M 0.20M 0.10M
[H+]o
- 10M
- 10M
- 10M
- 20M
Initial Rate (mol⋅L-1⋅s-1) 8.0 × 10-4 1.6 × 10-3 3.2 × 10-3 3.2 × 10-3 Assume a rate law of the form: Rate = k[BrO3-]n[Br-]m[H+]p • What is your strategy? • What do you notice about the experimental design, i.e. the pattern of concentrations?
Determine n by examining data from Experiments 1 & 2 Exp. 1 2 [BrO3-]o 0.10M 0.20M [Br-]o 0.10M 0.10M [H+]o 0.10M 0.10M Initial Rate (mol⋅L-1⋅s-1) 8.0 × 10-4 1.6 × 10-3
Rate 2/Rate 1=
16×10−4 M/s /8.0×10−4 M/s
= 2.0
If, Rate = k [BrO3-]n [Br-]m [H+]p then
Rate 2/Rate1 = k(0.20 M)n (0.10 M) m(0.10 M)p divided by k(0.10 M)n (0.10 M) m(0.10 M)p
=(2.0)raised to the power n
2.0 is = to (2.0)raised to the power n
n=1
Determine p and m from Experiments 1 & 4 and 2 & 3 respectively Exp. 1 2 3 4 [BrO3-]o 0.10M 0.20M 0.20M 0.10M [Br-]o 0.10M 0.10M 0.10M 0.20M [H+]o 0.10M 0.10M 0.10M 0.20M Initial Rate (mol⋅L-1⋅s-1) 8.0 × 10-4 1.6 × 10-3 3.2 × 10-3 3.2 × 10-3
Solution: Assume a rate law of the form, Rate = k[BrO3-]n[Br-]m[H+]p Exp 1 & 2 ⇒ n = 1 Exp2&3⇒ m=1 Exp1&4⇒ p=2 Rate =\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
What are integrated rate laws
Solve this At 500°C, cyclopropane forms propene in a first-order reaction with rate constant 6.7 × 10-4 s-1. If initially the concentration of cyclopropane is 0.25 M, what is its concentration after 10.0 minutes?
Equations expressing the relationship between concentration and time can be derived using calculus.
• The resulting equations depend on the rate law used and are called integrated rate laws.
First-Order Reaction: A → products
Rate = delta [A]/delta t
=k[A]
Using calculus,
Integrated rate law is gotten :
In[A]subscript t -In[A]subscriptnaught = -kt
Or In[A]subscript t/In[A]subscriptnaught
=-kt
ln[A]t = natural log of the molarity of A after t seconds k⋅t = the rate constant times the elapsed time ln[A]o = natural log of the molarity of A at t = 0.0
InB/A=InA − l n B
ln(A to the power n ) = n ln A
ln( AxB) = ln A + ln B
ln(eraised to the powerx) =x
For First Order Reactions:
ln[A]t =−kt+ln[A]0
In[A]subscript t/In[A]subscriptnaught =-kt
In[A]subscript t=[A]subscript naught e superscript -kt
Solution: ln[A]t = -k⋅t + ln[A]o
ln[A] = -(6.7 × 10-4 s-1)(600. s) + ln(0.25) ln[A] = -1.788
[A] = antiloge(-1.788) = e-1.788 = 0.17 M
Use eto the power x key
What is the physical meaning of the rate law
Rate = k[A]m[B]n
• Connection - rate constant, k
larger k => faster reaction
• Order of the reaction (m and n): how ‘sensitive’ the rate of reaction is to the concentrations of A and B, respectively.
In Example 3:
BrO3-(aq) + 5 Br-(aq) + 6 H+(aq) → 3 Br2(l) + 3 H2O(l)
Rate = k[BrO3-][Br-][H+]2
• Out of the 3 reactants, increasing [H+] will have the _________ effect on the rate of reaction (2nd order in H+) .
What other method can be used to determine the order of a reaction and determining the Rate constant
How is it done
How will you know if a graph is first order or second order
How do you know if a reaction is zero first or second-order?
If an increase in reactant increases the half life, the reaction has zero-order kinetics. If it has no effect, it has first-order kinetics. If the increase in reactant decreases the half life, the reaction has second-order kinetics.
True or false
Graphical method
Obtaining a straight line:
treat,
ln[A]t = -k ⋅ t. +. ln[A]o
as. y. =. m⋅x + b
and plot a graph
First order
For second order:
Rate = − Δ[A]/delta t= k [A]to the power2
By integration or calculus,it becomes
1/[A]subscript t=kt. +. 1/[A]subscripto
y. = mx. +. b. Line
For determining the rate constant k,
If: ln[A]t = -k ⋅ t + ln[A]o
( y = m⋅x + b )
Then, a graph of ln[A]t versus time has slope m = - k and intercept b = ln[A]o
So the slope =-k
k=-the slope
you get a straight line with a negative slope, then that would be first order. For second order, if you graph the inverse of the concentration A versus time, you get a positive straight line with a positive slope, then you know it’s second order.
In order to determine the rate law for a reaction from a set of data consisting of concentration (or the values of some function of concentration) versus time, make three graphs.
[A] versus t (linear for a zero order reaction)
ln [A] versus t (linear for a 1st order reaction)
1 / [A] versus t (linear for a 2nd order reaction)
The graph that is linear indicates the order of the reaction with respect to A. Then, you can choose the correct rate equation:
For a zero order reaction, rate = k (k = - slope of line) For a 1st order reaction, rate = k[A] (k = - slope of line) For a 2nd order reaction rate = k[A]2 (k = slope of line)
How long will it take for 75% of AB to decompose if the initial
concentration is 1.50 M and k = 0.20 L/mol•s?
For a zero order rate law what’s the equation
What is half life
What is the half life for zero order,first and second order
Points to Remember:
1. Subsequent half lives
get shorter for zeroth order, do not change for first order, and get longer for second order.
2. The values necessary to find k can be read off of the graph.
3. t1/2 provides a third method for determining order.
True ir false
Second-Order Reaction
A chemical decomposes by a second-order reaction (rate=k[AB]2),
AB(g) → A(g) + B(g)
1/[A]subscript t=kt. +. 1/[A]subscripto
[A]subscript t= 1.5 − 0.75(1.5) = 0.375 M
t=(1/[A]subscript t — 1/[A]subscripto)multiplied by 1/k
=(1/0.375–1/1.5)multiplied by 1/0.2
=10seconds
Zeroth-Order Reaction: A → products
What is the implication of the 0th order?
The rate is independent of the concentration.
[A]t = − kt +[A]o
y =mx+b
half-life = t1⁄2 = the time required for the concentration of a reactant to decrease to one half of its initial value ([A]naught➡️ [A]naught/2)
Zero:half life =[A]naught/2k
1st order:half life =0.693/k or In2/k
2nd:Half life=1/k[A]naught
The reaction below is first order in [H2O2]: H2O2(l)”→2H2O(l)+O2(g)
A solution originally at 0.600 M is found to be 0.075 M after 54 min. What is the half-life for this reaction in minutes?
State the relationships between the zero,first and second orders
Half life =ln2/k
Rate =In[A]subscript t /In[A]subscript naught =-kt
Rate=In0.075/0.600=-k(54mins)
k=-1/54 multiplied by In0.075/0.600
= 0.0385 min−1
Half life =In2/0.0385 min−1
=18mins
Rate law- Zero: rate = -k First: rate = -k [A] Second: rate = -k [A]2 Units for k- Zero: mol/L*s First: 1/s Second: L/mol*s Integrated rate law in straight-line form (from reactant A-plot)- Zero: [A]t = -k t + [A]0 First: ln[A]t = -k t + ln[A]0 Second: 1/[A]t = k t + 1/[A]0
Plot for straight line- Zero: [A]t vs. t First:In[A]t vs. t Second: 1/[A]t = t Slope,y intercept - Zero: -k, [A]0 First: -k, In[A]0 Second:k,1/[A]0 Half life - Zero: A]0/2k First: ln 2/k Second: 1/k [A]0
Objective: Understand the RELATIONSHIPS, don’t just memorize equations!
State the empirical methods used to check for rate law
What is the rate law
Useoneoftheempiricalmethods:
– Initial rates - systematically vary concentrations and observe the effect on
rate.
– Integrated rates - solve the calculus (the differential) to give a concentration versus time equation (often uses line plots).
– Half lives - use integrated rate laws solved (for time) specifically at the point where concentration is one-half the original amount.
• Theexperimentalresultsdeterminetheratelaw.
– The rate law is the mathematical expression that describes how the
reaction depends on some key factors (k, reactant concentrations):
Rate = k[A]m[B]n
• Thisishalfofour“scientificmethod”cycle;wehavean empirical law.
– But we need to develop further molecular level insight
What is the Collision model
Empirical observation: Chemical reaction rates often
increase with temperature
The Collision Model (i.e. collision theory)
Reaction rates depend on collisions, which in turn will likely depend on at least 3 factors:
• collision frequency = the number of collisions per second per liter
• collision energy = the fraction of the collisions that are sufficiently forceful
• collision orientation = the fraction of the collisions with correctly oriented molecules
What is activation energy
Consider the Reaction: A + BC → AB + C
• Collision Frequency
higher concentrations ⇒ more frequent collisions
higher temperatures ⇒ more frequent collisions
• Collision Energy
powerful collision ⇒ reaction gentle collision ⇒ no reaction
How much E is enough?
• Collision Orientation
correct alignment ⇒ reaction incorrect alignment ⇒ no reaction
How do the molecules need to orient?
Activation Energy, Ea = the minimum collision energy required for molecules to react
What is an exothetmic and endothermic reactions
Give examples
What is enthalpy
Distribution of Energy
• Typically, only a fraction of the molecules in a sample possess sufficient energy to react.
• The higher the temperature, the higher this fraction.
True or false
An endothermic reaction occurs when energy is absorbed from the surroundings in the form of heat. It occurs as a result of the dissociation of the bonds between the molecules. The energy is then released through the formation of new bonds. energy is absorbed from the surrounding into the reaction
Heat is taken up from the surroundings in such reactions, so the temperature of the system where the reaction is taking place remains cooler. Also, at the end of the reaction, the enthalpy, which is the change in heat energy during the conversion of reactants to products, increases.
Conversely, an exothermic reaction is one in which energy is released from the system into the surroundings.
The energy is released from the system to its environment.
few examples of the endothermic process are photosynthesis, evaporating liquids, melting ice, dry ice, alkanes cracking, thermal decomposition, ammonium chloride in water and much more
A few examples are neutralisation, burning a substance, reactions of fuels, deposition of dry ice, respiration, solution of sulphuric acid into water and much more.
The energy released is caused by the formation of new bonds (products) at a higher level. While the energy required to break up the bonds (reactants) is lower. At the end of the reaction, the enthalpy change decreases as well.
Enthalpy, a property of a thermodynamic system, is the sum of the system’s internal energy and the product of its pressure and volume.
What is the arrhenius equation
https://m.youtube.com/watch?v=dWhMlxuaIkI -how to read an energy profile diagram
The activation energy of a chemical reaction is closely related to its rate. Specifically, the higher the activation energy, the slower the chemical reaction will be.
True or false
https://www.khanacademy.org/science/ap-biology/cellular-energetics/enzyme-structure-and-catalysis/a/activation-energy
The Arrhenius Equation
• If the reaction rate varies with temperature, so must the rate constant, k.
• Arrhenius (1888) proposed the relationship:
k=Ae to the power -Ea/RT
k = rate constant A = frequency factor (related to collision frequency and collision orientation) R = gas constant = 8.314 J/mol⋅K T = temperature in kelvin Higher T ➡️larger k ➡️increased rate Ea=activation energy
Write the linear equation for Arrhenius equation
k=Ae to the power -Ea/RT
Then
In(k)= -Ea/R multiplied by (1/T) +In(A)
y. =. m. x. +. b
m or -Ea/R = the slope
b is the intercept
The decomposition of HI(g),
2HI(g) →H2(g)+I2(g)
has rate constants of 9.51x10-9 L/mol•s at 500 K and
1.1x10-5 L/mol•s at 600 K. What is Ea for this reaction?
In(k1)= -Ea/RT1 + InA
In(k2)= -Ea/RT2 + InA
In(k1) - In(k2)=In(k1/k2)=Ea/R(1/T2 - 1/T1)
Ea=RIn(k1/k2)(1/T2-1/T1) raised to the power -1
= 8.314J/mol•K)ln(1.1x10 to the power -5 / 9.51x10 to the power -9)(1/600 - 1/500) to the power -1
Ea=1.76x10to the power2kJ/mol
Example of a Reaction Mechanism Overall reaction: 2O3(g) →3O2(g) 2-Step Reaction Mechanism: Step 1:O3(g)→O2(g)+O(g) Step 2:O3(g)+O(g)→2O2
What are the rate laws for each of the steps? What is the overall rate law?
What is rate mechanism and elementary step
State the rate law for these reactions A➡️product A+A➡️product A+B➡️product A+A+A➡️product A+B+C➡️product
reaction mechanism = the sequence of elementary steps resulting in an overall chemical reaction
elementary step = a simple reaction with a well- defined rate law that depends on the number of molecules (also, elementary reaction or elementary process)
IF a reaction equation represents an elementary step
then the reaction coefficients are related to the orders of the rate law.
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Through mechanisms, rate laws gain chemical meaning. We do not just have a mathematical rule but more importantly molecular level insight.
Since elementary steps are simple collision processes, their rate laws are determined by their molecularity.
Rate laws: k[A] k[A] to the power 2 k[A][B] k[A]squared[B] k[A][B][C]
How is a mechanism judged as acceptable?
What is the rate detemrinjng step
What is reaction intermediate
Step 1: NO2(g) + F2(g) → NO2F(g) + F(g) (slow)
Step 2: NO2(g) + F(g) → NO2F(g) (fast)
Sum: 2NO2(g) + F2(g) → 2NO2F(g) Slow
Rate = k [NO2] [F2]
This is consistent with experiments. So it is a possible mechanism.
True or false
A mechanism must satisfy two requirements:
(1) The sum of the elementary steps must result in the overall
reaction.
(2) The mechanism must be consistent with the experimentally determined rate law.
(An experiment only provides evidence not proof.)
Rate-determining step:
• All elementary steps do not have the same rate.
• Usually one of the elementary steps is slower than the others and thus limits the overall rate of the reaction- this step is called the rate-determining step (or rate-limiting step).
- Inamultistepmechanism,oneofthestepswillbeslower than all others.
- Theoverallreactioncannotoccurfasterthanthisslowest, rate-determining step
Consider the overall reaction: 2 NO2(g) + F2(g) → 2 NO2F(g)
For which the “accepted” mechanism is:
Step 1: NO2(g) + F2(g) → NO2F(g) + F(g) (slow) Step 2: NO2(g) + F(g) → NO2F(g) (fast) Sum: 2 NO2(g) + F2(g) + F(g) → 2 NO2F(g) + F(g)
Reaction intermediate - a substance that is formed and used up during the overall reaction. Intermediate concentration should not appear in the rate law.
Mechanism with a Slow Initial Step NO2 (g)+CO(g)⎯⎯→NO(g)+CO2 (g) • Rate=k[NO2]2 • Aproposedmechanismforthisreactionis Step 1: NO2 + NO2 ⎯⎯→ NO3 + NO (slow) Step 2: NO3 + CO ⎯⎯→ NO2 + CO2 (fast) • The NO3 intermediate is consumed in the second step. • SinceCOisnotinvolvedintheslow,rate-determining step, it does not appear in the rate law.
For a Mechanism with a Fast Initial Step Reaction: 2 NO(g) + O2(g) → 2 NO2(g) Step 1: NO+O2 !!→NO3 (fast) Step 2: NO3 + NO !!→ 2 NO2 (slow) Rateoverall = Rate2 = k2 [NO3] [NO]
But, k1[NO][O2] = k-1[NO3] or Keq=[NO3] / [NO][O2]
Hence, [NO3] = k1/ksubscript-1[NO][O2]
Rate overall=k2k1/ksusbcript-1 multiplied by [NO]to the power 2[Osubscript2]
= k[NO]2[O2]
One possible mechanism for the decomposition of nitryl chloride, NO2Cl,is:Step 1:NO2Cl(g)#→NO2(g)+Cl(g) fast,equilibrium
Step 2:NO2Cl(g)+Cl(g)#→NO2(g)+Cl2(g)
What is the reaction intermediate? What is the rate law? Does increasing the NO2 concentration increase or decrease the reaction rate?
Solution:
Increasing the NO2 concentration, will decrease the reaction rate.
What is a catalyst
catalyst = a substance that increases the rate of a chemical reaction without undergoing permanent chemical change
Consider: The decomposition of H2O2
Uncatalyzed 2 H2O2 → 2 H2O + O2 (very slow)
Catalyzed 2H2O2 → 2 H2O + O2 (fast)
Q: Why?
Ans: A very different mechanism
Step#1: H2O2 +I- → H2O + IO-
Step#2: H2O2 +IO- → H2O + O2 + I-
Where I is the catalyst in both reactions
Sum: 2H2O2 → 2H2O + O2
Is this 2-step mechanism faster?
Yes, two small energy barriers are easier to surmount than one large energy barrier. Cuz the small energy barriers had catalysts but the bigger one didn’t
