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8/7/2019 Reaction Rate Notes Complet
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Rates of Reaction
Introduction to Rate
8/7/2019 Reaction Rate Notes Complet
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Rate of Reaction: the change in “something” of the reactants or
products over time or per unit timemass or moles over time**
concentration over time
color over time
the rate of a reaction can be fast, slow or zero
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Chemical Reactions
are the result of collisions between atoms, ions or molecules
involve bond breaking (releases energy)
NaCl
involve bond making (requires energy)
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Collision Theory
explains why reactions have different rates
1. the reacting particles must collide
2. the particles must have enough energyfor the breaking & making of bonds
3. the colliding particles must collide at thright orientation or geometry
Explains why so
me reaction
s don’t o
ccur
at room temp
Not eno
ugh ener
gy to break & form
bonds
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Good
Orienta
tion
R EAC
TION!!
Wrong O
rientation
NO R E
ACTIO
N!!
Wrong OrientationNO REACTION!!
W r o n g O r i e n t a t i o n
N O R E A C T I O N ! !
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Activation Energy
the min amount of energy that colliding particlesmust have in order to react
“barrier” or “hurdle” reactants mustovercome
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Activation Energy
Progress of Reaction
Energ y
Energy
Reactants
Energy/Enthalpy
Products
Energy of ActivationComplex
Activation
Energy (EA)
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The diagram shows the following reactionA + B →AB
According to this diagram, what is the
activation energy of the reaction?
Progress of reaction
Enthalpy
(kJ/mol)
AB
A + B
-135
035
EA 35 kJ/mol – 0 kJ/mol
EA = 35 kJ/mol
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Enthalpy (kJ)
Progress of the reaction
C + D
A + B
+800
+600
+400
+200
0
-200
-400
-600
-800
-1000
-1200
The graph below shows the potentialenergy changes for: A + B → C + D
What is the activation energy of the
reaction? The enthalpy?
EA
550 – 100 kJ/mol
EA = 450 kJ/mol
-1000 – 100 kJ/mol
∆H = -1100kJ/mol
∆H
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Another Way to Graph EA
Nu
mb
erofPar ti
cles
Energy
EA
Area underthe curverepresents
the # of moleculesthat havereached EA
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Factors Affecting Rates of Reactions
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Factors Affecting Rates:
Various factors will either increase ordecrease a reaction’s rate
We will study: Temperature
Concentration
Particle Size / Surface Area Catalyst
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Temperature:
Increasing the temp, increases the rate
Decreasing temp, decreases the rate
Why?Molecules absorb energy, so…
the reactants reach activation energy &
the number of collisions increase, so reactants step
over the activation energy barrierMore molecules with enough energy to reach E
A
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Nu
mb
erofPar ti
cles
Energy
EA
Room TempAt 50oC
Particlesabsorb
energy, somore
particles(area undercurve) are
able to reachEA
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Concentration:
Increasing concentration, increases the rate
Decreasing concentration, decreases the rate
Why?
Increasing conc increases the number of particles: therefore more frequent collisions
allowing reactants to reach activation energy
Increase Conc
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Particle Size & Surface Area:
Larger surface area (smaller particles) have increased rates
Smaller surface area (larger particles)
have decreased rates Why?
Large SA gives a larger area for collisions,
allowing reactants to reach EA
Increase SA
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Catalysts:
Catalysts increase the rate
Why?Catalysts provide reactants with another reaction
path that has a lower activation energy
Tunnel = Catalyst
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ActivationEnergy (EA)
Progress of Reaction
Energ y
Activation Energy (EA)
with CATALYST
EA is LOWER
with CATALYST
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Nu
mb
erofPar ti
cles
Energy
EA
CATALYST
EA is LOWER
with CATALYSTEA
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Inhibitors
Inhibitors decrease the rate
Why?
Inhibitors provide reactants withanother reaction path that has ahigher activation energy
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EA with
INHIBITOR
Progress of Reaction
Energ y
Activation
Energy (EA)
EA is HIGHER
with INHIBITORS
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Nu
mb
erofPar ti
cles
Energy
EA
INHIBITOR
EA
EA is HIGHER
with INHIBITOR
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Reactants
Products
REACTION PROCEEDS
1--
2--
ENERGY
Curves 1 and 2 on the graph below represent energypathways for the same chemical reaction. The reaction
rate is faster for pathway number 2.
Which factor is responsible for curve 2?
A) Increase in Temp
B) Increase in SA
C) Decrease in Conc
D) Presence of Catalyst
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When you light a fire you use paper and kindling woodrather than trying to light a log with a match.
Use the collision theory to explain why.
Kindling has a greater surface area than a log
Increase SA = more area for collisions
More collisions = faster rateb/c reach EA sooner
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The four graphs below represent the number of moleculesof reactants as a function of their kinetic energy.
Which graph represents the fastest reaction?
Kinetic energy (kJ)
Number of molecules
E E
Kinetic energy (kJ)
Number of molecules
E
Kinetic energy (kJ)
Number of molecules
E
Kinetic energy (kJ)
Number of molecules
A)
B)
C)
D)
EA is the lowest, so
probably reachedfastest!
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The graph shows the energy distribution of thereactants of a reaction at a certain temp
Number of
E
Threshold energy
( )K
E average
molecules
Kinetic Energy
Based on the next graph, explain what has broughtabout an increase in the rate of the chemical reaction.
Number of
Threshold energy
( )
E average
molecules
Kinetic Energy E K
A catalyst has been added.The catalyst lowers the threshold energy so that there
are more effective collisions between molecules
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The energy distribution graph for a givenreaction is shown below
Which graph belowshows the affect of
increasedtemperature?
EA is same, but lessmolec needed to
reach EA
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Pg 424 #18, 20 & 23
Old Exam Question
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The spon reaction of a solid piece of P4 with O2 in air has
an activation energy of 30 kJ/mol and is represented bythe following equation.
P4(s) + 5 O2(g) → P4O10 (s) ∆ H = −700 kJ/mol
Here is a sketch of the kinetic energy distribution curvethat corresponds to this reaction at 20°C and 101 kPa.
Kinetic energy (E k )
Number of moles of
reactant molecules
(n)
a
umber of
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Below is a list of 5 changes in reaction conditions, alongwith 5 graphs. Which graph best corresponds to each
change in conditions listed below?Changes in conditions:
1. The conc of oxygen gas is increased2. An inhibitor is added 3. The temp is lowered
4. A catalyst is added 5. The temp is raised
A B C
D E
E k
n
E k
n
E k
n
E k
n
E k
n
Kinetic energy (E k )
umber of
moles of
reactant molecules
(n)
a
1
234
5
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Rate Laws & Estimating Rate
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Rate Law:
An expression relating the rate of reaction to theconcentration of the reactants
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For: aA + bB → cC +
dD
Rate equation: Rate = k [A]a [B]b
Rate constant:
•determined experimentallyfor each reaction at given
temp
•L/mol•s or s-1
•Large k = fast reaction
Coefficients:
•Coefficientsbecome theexponents
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REMEMBER:Rate in this equation
is dependant on theconcentration of reactants only.
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NO2(g) + O3(g) → NO3(g) + O2(g)
What is the rate equation for the formation of
NO3(g) and O2(g)?Rate = k [NO2]
[O3]If the initial conc of NO2(g) and O 3(g) is
1.0M, what would happen to the rate if weincreased the conc of NO3 to 2.0M?
Rate = k [1][1]
Rate =
Rate = k [2][1]
Rate =
Ratedoubles!
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2NO + O2
→ 2NO2
What is the rate equation for the formation of NO2? Rate = k
[NO]2[O2]If the initial conc of both reactant is 1.0M,what would happen to the rate if weincreased the conc of O2 to 2.0M?
Rate = k
[1]2[1]Rate =1k
Rate = k
[1]2[2]Rate =2k
Rate
doubles!
NO to
2.0M?Rate = k2 Rate =
Rate
quadruples
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Reactions that have reactants:
With oppositely charged ions are very fast
With few or weak bonds are faster than those
with many or strong bonds
Predicting Rates of Reactions:
Ag+ + Cl- AgCl
Mg + HCl MgCl+ H2
C2H5OH + 3O2 2CO2 + 3H2O
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In the same phases are faster that those indifferent phases
Undergoing a two particle collision are fasterthan those requiring many particle collisions
Predicting Rates of Reactions:
H2 (g) + F2 (g) 2HF (g)
Mg (s) + 2H2O (l) 2Mg2+ (aq) + 2OH-
(aq) + H2 (g)
2H2 (g) + O2 (g) 2H2O (g)
4NH3 (g) + 7O2 (g) 4NO2 (g) +
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Rank the following from fastest to slowest.
H2 (g) + F2 (g) 2HF (g)
Ag+
+ Cl-
AgCl
C2H5OH + 3O2 2CO2 + 3H2O
1.
2.3.
2 1 3
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Which would have the greatest rate?
Hg (l) + Br2 (g) HgBr2 (s)
2H2 (g) + O2 (g) 2H2O(g)
4NH3 (g) + 7O2 (g)
4NO2 (g) +6H2O (g)
A.
B.
D.
C
C. Ag+ (aq) + Cl- (aq) AgCl (s)
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Calculating Rate of Reaction
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Calculating Rate:
Recall, rate is the change insomething over time
You can calculate rate if:
1. Given info
2. Given a set of Data
3. Given a Graph
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A bottle of wine contains approximately 72 g of
ethyl alcohol, C2H5OH. When left open, the ethylalcohol changes into acetic acid according to thefollowing equation:
C2H5OH + O2 →CH3COOH + H2O
The ethyl alcohol changed completely into aceticacid after 60 days (d).
What is the average reaction rate, in molesper day (mol/d), of the transformation of thealcohol?
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Calculate the number of moles of alcohol:
moles = 72/ 46moles = 1.57 mol of alc.
Calculate the average rate:
Rate =(d) days of number
(mol) moles of number
Rate = 1.57 / 60
Rate = 0.026 mol/day
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Below is a balanced equation:A + B → C + 2D
A student dissolved 0.15 mol of substanceA in solution B. After a ten-minute periodof reaction, 0.05 mol of A is recovered.
What is the average rate of formation of D in mol/sec?
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Calculate the number of moles of A used in rxn:
Moles used of A = 0.15 – 0.05 mol
moles of A used = 0.1 mol
Calculate the number of moles of D produced:
moles of D produced = 0.2 mol
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Calculate the length of rxn:
10 min = ? sec
time = 600 sec
Calculate the average rate:
Rate =)( sec of number
(mol) D of moles of number
s
Rate = 0.2 / 600
Rate =3.3 x 10-4 mol/sec
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The combustion of a candle is recorded in
the table below.Before
combustionAfter
combustion
Mass of candle
(g)
165.5 162.0
Time (h:min) 1:00 6:00
The equation is:C25 H52 + 38 O2 →25 CO2 + 26 H2O
Calculate the rate of this combustion
reaction in moles/hour
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Calculate the # moles burned:
moles = 3.5/ 352moles = 0.01 mol of candle burned
Calculate the average rate:
Rate =)(hr hours of number
(mol) moles of number
Rate = 0.01 / 5
Rate = 0.002 mol/hr are burned
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G ( ) ( ) C( )
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The graph below shows the # of moles of C
formed by the reaction, as a function of time.
What is the average rate for the formation of
C during the first three seconds?
0.40
0.30
0.20
0.10
1 2 3 4 5
Time (s)
Number of moles of substance C
Given: A(g) + B(g) → C(g)
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Calculate the # C moles formedin first 3 seconds:
moles = 0.25 mol of C formed
Calculate the average rate of C:
Rate =)(s sec of number
(mol) moles of number
Rate = 0.25 / 3
Rate = 0.083 mol/s of C are
formed
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Zinc reacted with HCl, to produce H2 &
MgCl2. The following table shows thevolume of H2 formed during the reaction as
a function of time.Time (s) Volume of H
2(mL)
01050100
150
083245
50
What is the average rate of formation of H2
between the 25th and 75th second?
Vol of H = 39mL 19mL
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1400 40 60 80 100 120 140 Time (s)160
Volume
of H 2
10
20
(mL)
20
10
30
40
50
60
1400 40 60 80 100 120 140 Time (s)160
Volume
of H 2
10
20
(mL)
20
10
30
40
50
60
19mLof H2
39mL
of H2
Vol of H2 = 39mL – 19mL
Vol of H2 = 20 mL
Rate = (s) sec of number
(mL) H of vol 2
Rate = 20 / 50
Rate = 0.40mL/s