Download Kinetics and Equilibrium

Chemical Kinetics (Reaction
Rates) and Equilibrium
Standards: Reaction Rates
• 8a. rate of reaction is the decrease in
concentration of reactants or the increase in
concentration of products with time.
• 8b.reaction rates depend on such factors as
concentration, temperature, and pressure.
• 8c. the role a catalyst plays in increasing the
reaction rate.
Chemical Reactions and Kinetics
* What is your favorite
dish? What are the main
ingredients and how do
you prepare this dish?
Lets connect to Chemistry
*What is a chemical reaction?
*What does the word “kinetics” remind you
of?
REVIEW: Parts of a chemical reaction
Reaction #1: A + B  AB
Reactants
Product
Reaction #2: 2A + B  A2B
Coefficient
Subscript
REVIEW: Types of a chemical reaction
Direction: Match the chemical reaction with the name of the type of reaction
1. Double
Replacement
A.
P4 + 3 O2  2 P2O3
2. Single
Replacement
B.
2 NO2  2 O2 + N2
3. Combustion
C.
Pb + FeSO4  PbSO4 + Fe
4. Synthesis Reaction
D.
Na3PO4 + 3 KOH  3 NaOH + K3PO4
5. Acid-Base
E.
C6H12 + 9 O2  6 CO2 + 6 H2O +Energy
6. Decomposition
F.
HNO3 + NaOH  H2O + NaNO3
Sketch of what is happening on a
molecular model of a chemical
reaction (specific vs. general)
NaCl (s)  Na+ (aq) + Cl- (aq)
Cl

B

Na
A
Na
A
+
+
Cl
B
Direction: Using the balanced equation below,
create a molecular model of this chemical reaction
(use 3 different colored pencils)
Mg + 2HCl
+
 MgCl2 + H2
+
Collision Theory: reactants forming
products using energy and proper orientation
Rate of reaction-- decrease in [reactants] or the
increase [products] over time.
REAGENTS  PRODUCTS
What are the factors that
determines the rate of
reaction?
Explain how reaction rates depend on such factors as
concentration, temperature, and pressure (use the
pictures below to support your argument)
Low
Pressure
High
Pressure
Catalysts
-- the frequency of collisions between reactants
-- rearrange the orientation of reactants
-- speeds up reaction, not consumed
Ex. biological enzymes, catalase etc.
Energy
Diagram
*Reaction needs to
surpass activation
energy (Ea)
AP Chem Kinetics Activity: Rate Order
Activity 1: The Burning Candle
Activity 2: The Empty Bucket
Activity 3: Penny Decomposition
Goals for each activity:
*Graph data and determine which one
corresponds to zero/1st/2nd rate order
*What are the formulas for each rate order
AP Chem Kinetics Activity: Decomposition
of Crystal Violet with NaOH (Colorimetry)
Goal: Determine the rate order of the decomposition of
crystal violet by colorimetry (if possible spectro-vis)
Direction:
*Groups will collect at least 3 data time points
*Record time and height/ calculate concentration
*Materials: --Crystal Violet 25µM and NaOH 0.2M
Standard
--CV 6mL + 4mL
of water
(concentration
=1.5µM)
CV + NaOH
Decomposing
-- add 10 mL
from class setup
Reaction Mechanism
*Rate determining step-- gives us the rate law (slow step)
hint: look at coefficient (ex. A2 because A + A)
*Intermediate-- it is produced then consumed
*Catalyst – it is consumed first then produced
reaction favors simpler (bi-molecular)
1st rxn: A + A  E
If A + B  C
2nd rxn: E + B  G
3rd rxn: G  C + A
*To check mechanism, if 1st rxn is slow, what is the rate
law, is it similar to the predicted rate law expression? If so,
then the mechanism is correct.
*Draw molecular model for each rxn step.
Reaction Mechanism
Rate = k [H2O2] m
Kinetics Practice Problem
• The reaction 3 A + B  (products) is known to follow the
mechanism:
I. A + B  C + E
II. A + E  G
III.G + A  D + F
a) If the reaction rate is observed to be the first order in A and first
order in B, write the rate law for the reaction.
b) Which step must be the slow step?
c) If step III were the slow step, what would the rate law become?
d) Which letters from the reaction mechanism are used to represent
the products of this reaction?
e) Which letters from the reaction mechanism are used to represent the
unstable intermediate products?
Kinetics Practice Problem
For the reaction A + B  C the following rate data were
obtained at constant temperature:
Exp.
Rate (moles/sec)
[A]0 [B]0
I
2.0 x 10-4
0.10 0.10
II
8.0 x 10-4
0.20 0.20
III
8.0 x 10-4
0.10 0.20
a)Determine the rate order and write the rate law for this reaction.
b) Calculate the rate constant, k. Include units.
c) If each reactant has a molarity of 0.3M, determine the rate
d) Propose a reasonable TWO step mechanism for this reaction and
designate the slow step in the reaction.
I._________________________
II._______________________
Chemical Equilibrium Chp. 14 Concept Map
*Chemical
Equilibrium
*Equilibrium
Constant
*Heterogeneous *Homogeneous
Equilibrium
Equilibrium
*Law of Mass
Action
*Le
Chatelier’s
Principle
*Physical
Equilibrium
*Reaction
Quotient
*Reverse
Reaction
*Forward
Reaction
*K<1 vs. K>1
*Kp vs. Kc
*Multiple
Equilibria
*Predicting
Direction of
Reaction
*How to
calculate Eq.
Concentration
*Factors that
affect Chem Eq.
Standards: Equilibrium
• 9a. Students know how to use Le Chatelier's
principle to predict the effect of changes in
concentration, temperature, and pressure.
• 9b. Students know equilibrium is established
when forward and reverse reaction rates are
equal.
• 9c. Students know how to write and calculate
an equilibrium constant expression for a
reaction.
What can we conclude
from each chart?
How are the two charts
defining what it means
for a reaction to be in
equilibrium?
Which reaction is in equilibrium? Explain why.
H2 + F 2
HCl + NaOH
2HF
H2O + NaCl
Equilibrium --established when forward and
reverse reaction rates are equal and the
concentration of reactants and products
remain constant
Equilibrium Activity
*Each student picks an index card (attach with
paper clip on their shirt or ID badge). Card is
either reactant A or reactant B
A + B+ Heat
A-B
Q: endothermic or exothermic?
For each of the following situations, determine which side of the
reaction is favored. Explain your choice.
Factor that affects
equilibrium
1. Add more products
(change concentration)
2. Add more heat for an
endothermic reaction (hint:
which side of the equation will heat be
part of)
3. Increase the pressure in
the container that contains
this reaction
N2(g)+3H2(g) 2NH3(g)
Which side of equation is favored
(Product or Reactant) AND
which way is favored (Forward
or Reverse Reaction)
Effect on K
value
(Explain)
A + B+ Heat
A-B
Disturbance # 1.
Take away products
A + B+ Heat
A-B
Disturbance # 2.
Cool down this
endothermic reaction
A (g) + B (g) + Heat
A-B (g)
Disturbance # 3.
•Decrease the pressure in
the container that contains
this reaction
Le Chatelier’s Principle
In a reversible chemical reaction, when stress is
applied to a system in an equilibrium, the
reaction will shift in a direction that relieves the
stress and a new equilibrium will be established.
Stress: Changing Concentration
Pressure
Temperature
Graphing Equilibrium:
Ex. The Haber Process is used to produce ammonia.
N2(g) + 3H2(g)
2NH3(g)
Direction: Create 2 graphs, X-axis is time, and Y-axis is
concentration. For Graph #1: plot the forward reaction and
Graph #2: plot the reverse reaction (label your lines with
specific compounds)
How is Kinetics Related to Equilibrium?
• Consider the rxn:
Write the rate law for the forward
and reverse reaction (1st order)
Rate forward =
Rate reverse =
At equilibrium:
A B  C  D
rate  k f [ A][ B]
rate  kr [C][ D]
k f [ A][ B]  rate  kr [C ][ D]
kf
The EQ constant of the reverse
reaction is simply the
inverse/reciprocal of the EQ constant
of the forward reaction.
[C ][ D]

 K eq
k r [ A][ B]
1
[ A][ B]

K eq [C ][ D]
Equilibrium Expression and Calculation
Using the reaction:
Keq=
bR#1 + aR#2  cP
[products]c
[reactant #1 ]b [reactant #2]a
K eq
Keq>>>1
Keq<<<1
Favors this side of
reaction
Products
Reactants
Additional Info:
*In heterogeneous equilibrium, we DON’T include
solids and liquids in calculating Keq. Why?
*If temperature is constant, then partial pressure of a
gas directly related to the concentration (mol/L)
*Kp = Kc (RT)Δn
where R = Univ. gas constant (0.08206 L.atm/K.mol)
Δn = moles of gas products – moles of gas reactants
Practice Problem
1. Which statement correctly describes a chemical
reaction at equilibrium?
(A) The concentrations of the products and
reactants are equal.
(B) The concentrations of the products and
reactants are constant.
(C) The rate of the forward reaction is less than the
rate of the reverse reaction.
(D) The rate of the forward reaction is greater than
the rate of the reverse reaction.
AP Chem FR Practice #1
REMINDERS: Solids and liquids are not counted/ Partial pressure is proportional to # moles
NH4Cl(s)  NH3(g) + HCl(g) H = +42.1 kilocalories
Suppose the substances in the reaction above are at equilibrium
at 600K in volume V and at pressure P. State whether the partial
pressure of NH3(g) will have increased, decreased, or remained
the same when equilibrium is reestablished after each of the
following disturbances of the original system. Some solid NH4Cl
remains in the flask at all times. Justify each answer with a oneor-two sentence explanation.
(a)
A small quantity of NH4Cl is added.
(b)
The temperature of the system is increased.
(c)
The volume of the system is increased.
(d)
A quantity of gaseous HCl is added.
(e)
A quantity of gaseous NH3 is added.
AP Chem FR Practice #1: ANSWER
Answer:
(a)
PNH3 does not change. Since NH4Cl(s) has constant
concentration (a = 1), equilibrium does not shift.
(b)
PNH3 increases. Since the reaction is endothermic, increasing
the temperature shifts the equilibrium to the right and more NH3 is
present.
(c)
PNH3 does not change. As V increases, some solid NH4Cl
decomposes to produce more NH3. But as the volume increases,
PNH3 remains constant due to the additional decomposition.
(d)
PNH3 decreases. Some NH3 reacts with the added HCl to
relieve the stress from the HCl addition.
(e)
PNH3 increases. Some of the added NH3 reacts with HCl to
relieve the stress, but only a part of the added NH3 reacts, so PNH3
increases.
Reaction Quotient
• The reaction quotient, Q is similar to the
equilibrium constant, but describes where the
reaction is at a point in time.
• When Q > Keq:
– not at equilibrium, too many products
– reaction will move to the left, products
form reactants
will
• When Q < Keq:
– not at equilibrium, too many reactants
– reaction will move to the right, reactants will form
products
• When Q = Keq: system at equilibrium
Calculating concentrations at equilibrium
 ICE Method (see pg. 604-606 for steps and example)
Ex. Calculate concentrations at equilibrium if you
have 1 M of each reactant in this reaction:
A+B
Step 1:
2AB
A
B
2C
INITIAL (M)
1
1
0
CHANGE (M)
-x
-x
+2x
EQUILIBRIUM
(M)
1-x
1-x
0+2x
ICE
Calculating concentrations at equilibrium
 ICE Method (see pg. 604-606 for steps and example)
Step 2: Based on your ICE chart, set up your Keq
Step 3: Solve for X  you might have to use the
quadratic equation
Step 4: Evaluate x-values (you will have 2 one
will follow the parameters set in the problem)
Step 5: Plug in x-value to solve for
concentrations at equilibrium
AP Chem: Practice FR #2
CO2(g) + H2(g) H2O(g) + CO(g)
When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the
equation above. In one experiment, the following equilibrium concentrations were
measured.
[H2]
= 0.20 mol/L
[CO2] = 0.30 mol/L
[H2O] = [CO] = 0.55 mol/L
(a) What is the mole fraction of CO(g) in the equilibrium mixture?
(b) Using the equilibrium concentrations given above, calculate the value of Kc, the
equilibrium constant for the reaction.
(c) Determine Kp in terms of Kc for this system.
(d) When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of
the CO(g) is converted back to CO2(g). Calculate the value of Kc at this lower
temperature.
(e) In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in
a 3.0-liter reaction vessel at 2,000 K. Calculate the equilibrium concentration, in
moles per liter, of CO(g) at this temperature.
AP Chem: Practice FR #1 Solution
(a)
(b)
(c)
(d)
(e)
CO = = 0.34
Kc = = 5.04
Since Δn = 0, Kc = Kp
[CO] = 0.55 - 30.0% = 0.55 - 0.165 = 0.385 M
[H2O] = 0.55 - 0.165 = 0.385 M
[H2]
= 0.20 + 0.165 = 0.365 M
[CO2] = 0.30 + 0.165 = 0.465 M
K = (0.385)2/(0.365 X 0.465) = 0.87
let X = Δ[H2] to reach equilibrium
[H2] = 0.50 mol/3.0L - X = 0.167 - X
[CO2] = 0.50 mol/3.0L - X = 0.167 - X
[CO] = +X ; [H2O] = +X
K = X2/(0.167 - X)2 = 5.04 ; X = [CO] = 0.12 M
AP Chem: Practice FR #3
MgF2(s)  Mg2+(aq) + 2 F-(aq)
#2.) In a saturated solution of MgF2 at 18°C, the
concentration of Mg2+ is 1.21X10-3 molar. And F- is
2.42X10-3. The equilibrium is represented by the equation
above.
(a) Write the expression for the solubility-product
constant, Ksp, and calculate its value at 18°C.
(b) Calculate the equilibrium concentration of Mg2+ in
1.000 liter of saturated MgF2 solution at 18°C to which
0.100 mole of solid KF has been added. The KF dissolves
completely. Assume the volume change is negligible.
AP Chem: Practice FR #2 Solution
(a)
(b)
Ksp = [Mg2+][F-]2 = (1.21X10-3)(2.42X10-3)2= 7.09X10-9
X = concentration loss by Mg2+ ion
2X = concentration loss by F- ion
[Mg2+] = (1.21X10-3 - X) M
[F-] = (0.100 + 2.42X10-3 - 2X) M
since X is a small number then (0.100 + 2.42X10-3 - 2X)
 0.100
Ksp = 7.09X10-9 = (1.21X10-3 - X)(0.100)2
X = 1.2092914X10-3
[Mg2+]= 1.21X10-3-1.2092X´10-3 = 7.09X10-7M
AP Chem: Practice FR #4
C(s) + H2O(g)  CO(g) + H2(g)
Hº = +131kJ
#2. A rigid container holds a mixture of graphite pellets (C(s)), H2O(g),
CO(g), and H2(g) at equilibrium. State whether the number of moles of
CO(g) in the container will increase, decrease, or remain the same
after each of the following disturbances is applied to the original
mixture. For each case, assume that all other variables remain constant
except for the given disturbance. Explain each answer with a short
statement.
(a)Additional H2(g) is added to the equilibrium mixture at
constant volume.
(b)The temperature of the equilibrium mixture is increased
at constant volume.
(c)The volume of the container is decreased at constant
temperature.
(d) The graphite pellets are pulverized.
AP Chem: Practice FR #4 Solution
(a)
CO will decrease. An increase of hydrogen gas molecule will
increase the rate of the reverse reaction which consumes CO. A
LeChatelier Principle shift to the left.
(b)
CO will increase. Since the forward reaction is endothermic
(a ΔH > 0) an increase in temperature will cause the forward
reaction to increase its rate and produce more CO. A Le Chatelier
Principle shift to the right.
(c)
CO will decrease. A decrease in volume will result in an
increase in pressure, the equilibrium will shift to the side with fewer
gas molecules to decrease the pressure, , a shift to the left.
(d)
CO will remain the same. Once at equilibrium, the size of the
solid will affect neither the reaction rates nor the equilibrium nor
the concentrations of reactants or products.