Chemistry 102 Fall 2015
Instructor: Dr. Upali Siriwardane
e-mail: upali@latech.edu
Office: CTH 311
Phone 257-4941
Office Hours: M.W &F, 8:00-9:00 & 11:00-12:00 and Tu ,Th
8:00 - 10:00 am. am. or by appointment
Test Dates
Sept. 29, 2015 (Test 1): Chapter 13
Oct. 22, 2015 (Test 2): Chapter 14 &15
Nove. 17, 2015 (Test 3): Chapter 16 &17
Nvember 19, 2015 (Make-up test) comprehensive:
Chapters 13-17
CHEM 102 Fall 15, LA TECH
14-2-1
Chapter 14. Chemical Equilibrium
14.1
14.2
14.3
14.4
Fetal Hemoglobin and Equilibrium 61 3
The Concept of Dynamic Equilibrium 61 5
The Equilibrium Constant (K) 61 8
Expressing the Equilibrium Constant in Terms of
Pressure 622
14.5 Heterogeneous Equilibria: Reactions Involving Solids
and Liquids 625
14.6 Calculating the Equilibrium Constant from Measured
Equilibrium Concentrations 626
14.7 The Reaction Quotient: Predicting the Direction of
Change 629
14.8 Finding Equilibrium Concentrations 631
14.9 Le Châtelier’s Principle: How a System at Equilibrium
Responds to Disturbances 641
CHEM 102 Fall 15, LA TECH
14-2-2
Law of mass Action
Defines an equilibrium constant (K) for the process
jA+kB
lC+mD
[C]l[D]m
K = ----------------- ; [A], [B] etc are
[A]j[B]k
Equilibrium concentrations
Pure liquid or solid concentrations are not written in
the expression.
CHEM 102 Fall 15, LA TECH
14-2-3
7) What is the difference between initial [A]i and
equilibrium [A]eq concentrations?
N2O4(g)
colorless
Initial
N2O4 NO2
0.00 0.02
0.00 0.03
0.00 0.04
0.02 0.00
CHEM 102 Fall 15, LA TECH
[NO2]2
Keq 
[N2O4 ]
@ Equilibrium
N2O4
NO2
0.0014
0.017
0.0028
0.024
0.0045
0.031
0.0045
0.031
2NO2(g)
Dark brown
Keq
0.21
0.21
0.21
0.21
14-2-4
Equilibrium calculations
We can predict the direction of a reaction by
calculating the reaction quotient.
Reaction quotient, Q
For the reaction: aA + bB
eE + fF
[E]e [F]f
Q =
[A]a [B]b
Q has the same form as Kc with one important difference.
Q can be for any set of concentrations, not just at
equilibrium.
CHEM 102 Fall 15, LA TECH
14-2-5
Equilibrium constant calculations
1) Consider the reaction:
COCl2(g)
CO(g) + Cl2(g)
At equilibrium, [CO] = 4.14 × 10-6 M;
[Cl2] = 4.14 × 10-6 M; and [COCl2] = 0.0627 M.
Calculate the value of the equilibrium constant.
CHEM 102 Fall 15, LA TECH
14-2-6
Terminology
Initial concentration:
concentration (M) of reactants and products
before the equilibrium is reached.
Equilibrium Concentration
Concentration (M) of reactants and products
After the equilibrium is reached.
CHEM 102 Fall 15, LA TECH
14-2-7
Reaction quotient
Any set of concentrations can be given and a Q
calculated. By comparing Q to the Kc value, we
can predict the direction for the reaction.
Q < Kc
Q = Kc
Q > Kc
Net forward reaction will occur.
No change, at equilibrium.
Net reverse reaction will occur.
CHEM 102 Fall 15, LA TECH
14-2-8
Reaction quotient (Q) calculations
2) Consider the reaction system:
C2H5OH (aq) + CH3COOH(aq)
CH3COOC2H5 +H2O (l)
[ethyl acetate]eq
𝐊=
= 𝟎. 𝟗𝟓
[ethanol] eq[acetic acid] eq
𝐐=
[ethyl acetate]in
=?
[ethanol] in[acetic acid] in
The concentrations of both ethanol and acetic acid are 0.45
M and the concentration of ethyl acetate is 1.1 M. Use the
reaction quotient to determine whether the system is at
equilibrium.
CHEM 102 Fall 15, LA TECH
14-2-9
Reaction quotient (Q) calculations
Consider the reaction system:
C2H5OH (aq) + CH3COOH(aq)
CH3COOC2H5 +H2O (l)
The concentrations of both ethanol and acetic acid are 0.45
M and the concentration of ethyl acetate is 1.1 M. Use the
reaction quotient to determine whether the system is at
equilibrium.
[ethyl acetate]in
𝐐=
=?
[ethanol] in[acetic acid] in
CHEM 102 Fall 15, LA TECH
and
Keq=0.95
14-2-10
Determining Equilibrium Constants
ICE Method
1. Derive the equilibrium constant expression for
the balanced chemical equation
2. Construct a Reaction Table with information
(ICE) about reactants and products
3. Include the amounts reacted, x, in the Reaction
Table
4. Calculate the equilibrium constant in terms of x
CHEM 102 Fall 15, LA TECH
14-2-11
Example: An equilibrium is established by
placing 2.00 moles of N2O4(g) in a 5.00 L and
heating the flask to 407 K. It was determined
that at equilibrium the concentration of the
NO2(g) is 0.525 mol/L. What is the value of the
2
[NO
]
2
equilibrium constant?
Kc =
[N2O4]
N2O4(g)
2 NO2(g)
N2O4(g)
[Initial] (mol/L)
[Change]
0.40
-x
0
2x
-1/2 x
x
[Equilibrium]
0.40 - 1/2x
0.40- 0.263= 0.138
CHEM 102 Fall 15, LA TECH
2 NO2(g)
=0+x
0.525
14-2-12
What is the value of the equilibrium
constant?
N2O4(g)
[NO2]2
Kc =
[N2O4]
2 NO2(g )
0.525 = 0 + x
x = 0.525
0.40 - 1/2x
[NO2] = 0.40 - 1/2x
= 0.40 - 1/2(0525)
Kc =
[NO2]2
[N2O4]
CHEM 102 Fall 15, LA TECH
=
(0.525)2
= 2.00
0.138
14-2-13
Equilibrium Calculations
Hydrogen iodide, HI, decomposes according to the
equation
2 HI(g)  H2(g) + I2(g)
When 4.00 mol of HI placed in a 5.00-L vessel at
458ºC, the equilibrium mixture was found to
contain 0.442 mol I2. What is the value of Kc for the
reaction?
CHEM 102 Fall 15, LA TECH
14-2-14
2 HI(g) 
Initial
Change
H2(g) + I2(g)
4.00/5=.80
0
0
-2x
x
x
Equilibrium 0.80-2x
x x=0.442/5
x = 0.0884
Equilibrium concentrations
[HI] = 0.80 - 2x = 0.8 - 2 x 0.0884 = 0.62
[H2] = x = 0.0884
[I2] = x = 0.0884
[H2] [I2]
0.0884 x 0.0884
Kc = ---------------- = ------------------------- = 0.0201
[HI]2
(0.62) 2
CHEM 102 Fall 15, LA TECH
14-2-15
Equilibrium Calculation Example
A sample of COCl2 is allowed to decompose. The
value of Kc for the equilibrium
COCl2 (g)
CO (g) + Cl2 (g)
is 2.2 x 10-10 at 100 oC.
If the initial concentration of COCl2 is 0.095M, what
will be the equilibrium concentrations for each of
the species involved?
CHEM 102 Fall 15, LA TECH
14-2-16
Equilibrium Calculation Example
COCl2 (g) 
Initial conc., M 0.095
Change
-X
in conc. due to reaction
CO (g)
Cl2 (g)
0.000
0.000
+X
+X
Equilibrium M (0.095 -X) X
Concentration,
[ CO ] [ Cl2 ]
Kc =
=
[ COCl2 ]
CHEM 102 Fall 15, LA TECH
X
X2
(0.095 - X)
14-2-17
Equilibrium calculation example
Keq = 2.2 x 10-10 =
X2
(0.095 - X)
Rearrangement gives
X2 + 2.2 x 10-10 X - 2.09 x 10-11 = 0
a X2 + b X - c = 0
This is a quadratic equation. Fortunately,
there is a straightforward equation for their
solution
CHEM 102 Fall 15, LA TECH
14-2-18
Quadratic Equations
An equation of the form
a X2 + b X + c = 0
Can be solved by using the following
x=
-b +
2
b - 4ac
2a
Only the positive root is meaningful in equilibrium
problems.
CHEM 102 Fall 15, LA TECH
14-2-19
Equilibrium Calculation Example
2
X + 2.2 x 10
a
b
2
X =
X=
-b + b
2a
- 2.2 x 10
X = 4.6 x 10
-6
X = -4.6 x 10
-10
-10
X - 2.09 x 10
-11
=0
c
- 4ac
-10 2
-11 1/2
+ [(2.2 x 10
) - (4)(1)(- 2.09 x 10
)]
2
M
-6
M
CHEM 102 Fall 15, LA TECH
14-2-20
Equilibrium Calculation Example
Now that we know X, we can solve for the
concentration of all of the species.
COCl2
CO
Cl2
= 0.095 - X
= X
= X
= 0.095 M
= 4.6 x 10-6 M
= 4.6 x 10-6 M
In this case, the change in the concentration of is
COCl2 negligible.
CHEM 102 Fall 15, LA TECH
14-2-21
Equilibrium calculations using ICE
3) Consider the reaction A
2B,
where the value of Keq is 1.4 × 10-12.
At equilibrium, the concentration of B is 0.45 M.
What is the concentration of A?
ICE Calculation
[A]
[B]
Initial concentration:
Change
Equilibrium concentration:
K = 1.4 x 10-12
CHEM 102 Fall 15, LA TECH
14-2-22
Equilibrium calculations using ICE
4) Calculation of unknown concentration of
reactants or products in an equilibrium mixture At
100o C the equilibrium constant (K) for the
reaction:
H2(g) + I2(g)
2 HI(g) ; K = 1.15 x 102.
If 0.400 moles of H2 and 0.400 moles. If I2 are placed
into a 12.0-liter container and allowed to react at
this temperature, what is the HI concentration
(moles/liter) at equilibrium?
ICE Calculation
[H2]
[I2]
[HI]
Initial concentration:
Change
CHEMEquilibrium
102 Fall 15, LA TECH
concentration:
14-2-23
Equilibrium calculations using ICE
4) H2(g) + I2(g)
2 HI(g) ; K = 1.15 x 102.
If 0.400 moles of H2 and and I2 0.400 moles each are
placed into a 12.0-liter container and allowed to
react at this temperature, what is the HI
concentration (moles/liter) at equilibrium?
ICE Calculation
Initial concentration:
Change
Equilibrium
concentration:
K = 1.15 x
102=
[H2]
0.033
-x
0.033-x
[I2]
0.033
[HI]
0
-x
2x
0.033-x
2x= 0.0281 x 2
=0.056
[HI] 2
𝟐𝒙
𝟐𝒙
=(
)2;𝐊 =
; 10.72=2x/(0.033-x)
𝟎.𝟎𝟑𝟑−𝒙
𝟎.𝟎𝟑𝟑−𝒙
[I2] [H2]
(10.72 x 0.033)-10.72x =2x; (0.357)-10.72x =2x; -12.72x =-.357; x = 0.0281
CHEM 102 Fall 15, LA TECH
14-2-24
Equilibrium calculations using ICE
4) H2(g) + I2(g)
2 HI(g) ; K = 1.15 x 102.
If 0.400 moles of H2 and 0.400 moles. If I2 are placed
into a 12.0-liter container and allowed to react at
this temperature, what is the HI concentration
(moles/liter) at equilibrium?
ICE Calculation
[H2]
[I2]
[HI]
Initial concentration:
Change
Equilibrium
concentration:
K = 1.15 x 102
CHEM 102 Fall 15, LA TECH
14-2-25
What is K (Kc) and Kp
Kc (K) - equilibrium constant calculated based on
[A]-Concentrations.
Kp- equilibrium constant calculated based on partial
pressure
Kp =
CHEM 102 Fall 15, LA TECH
14-2-26
Pressure Equilibrium Constants Kc & Kp
N2 + 3H2
Kc =
Kc =
=
[NH3]2
[N2][H2]3
2NH3
(PNH3/RT)2
=
(PN2/RT) (PH2/RT)3
(PNH3)2 (1/RT)2
(PN2) (1/RT))(PH2)3(1/RT)3)
PNH32
PN2 PH23
CHEM 102 Fall 15, LA TECH
(1/RT)2
(1/RT)(1/RT)3
= Kp
(1/RT) 2
(1/RT)(1/RT)3
14-2-27
Kc vs. Kp
N2 (g) + 3H2 (g)
K c = Kp
2NH3 (g)
(1/RT)2
)-2
=
K
(1/RT
p
3
(1/RT)(1/RT)
In General
Kc = Kp (1/RT)Dn
Kc = Kp (RT)-Dn
where Dn = #moles gaseous products
- # moles gaseous reactants
CHEM 102 Fall 15, LA TECH
14-2-28
What is K (Kc) and Kp
Kc (K) - equilibrium constant calculated based on
[A]-Concentrations.
Kp- equilibrium constant calculated based on partial
pressure (p)
Kc = Kp(RT)-Dn
Kp = Kc(RT) Dn
atm L
R = universal gas constant (0.08206 mol K )
T = Kelvin Temperature,
Dn = (sum of stoichiometric coefficients of gaseous
products) - (sum of the stoichiometric coefficients
of gaseous reactants)
CHEM 102 Fall 15, LA TECH
14-2-29
Partial pressure & Equilibrium
Constants
For the following equilibrium, Kc = 1.10 x
107 at 700. oC. What is the Kp?
2H2 (g) + S2 (g)
Kp
= Kc (RT)Dng
T
= 700 + 273 = 973 K
2H2S (g)
L
R
= 0.08206 atm
mol K
Dng = ( 2 ) - ( 2 + 1) = -1
CHEM 102 Fall 15, LA TECH
14-2-30
Partial pressure & Equilibrium
Constants
Kp = Kc
Dn
(RT) g
[
= 1.10 x 107 (0.08206
= 1.378 x105
CHEM 102 Fall 15, LA TECH
atm L
mol K
) (973 K)
]
-1
14-2-31
Types of Equilibrium
1) Heterogeneous Equilibrium
2) Heterogeneous Equilibrium
3) Acid Dissociation Constant- Ka
4) Base Dissociation Constant- Kb
5) Autoionization Constant- Kw
6) Solubility Product Constant-Ksp
CHEM 102 Fall 15, LA TECH
14-2-32
Heterogeneous Equilibrium
CaCO3(s)
CaO(s) + CO2(g)
[CaO(s)][CO2(g)]
Kc =
[CaCO3(s)]
concentrations of pure solids and liquids
are constant are dropped from expression
Kc = [CO2(g)]
CHEM 102 Fall 15, LA TECH
14-2-33
Solubility Product of Salts in Water
AgCl(s) + H2O (l)
Ag+(aq) + Cl-
Ksp = [Ag+] [Cl-]
Ksp (AgCl) = 1.77 × 10-10
Ksp (BaSO4) = 1.1 x 10-10
CHEM 102 Fall 15, LA TECH
14-2-34
Acid Dissociation Constant
HC2H3O2 (aq) + H2O(l)
K =
H3O+ (aq) + C2H3O2- (aq)
[H3O+][C2H3O2-]
[H2O][HC2H3O2]
Ka = K  [H2O] =
CHEM 102 Fall 15, LA TECH
[H3O+][C2H3O2-]
[HC2H3O2]
14-2-35
Base Dissociation Constant
NH4+ + OH-
NH3 + H2O(l)
K =
[NH4+][OH-]
[H2O][NH3]
Kb = K  [H2O] =
CHEM 102 Fall 15, LA TECH
[NH4+][OH-]
[NH3]
14-2-36
Autoionization of Water
H2O (l) + H2O (l)
K =
H3O+ + OH-
[H3O+][OH-]
[H2O]2
Kw = K [H2O]2 = [H3O+][OH-] = 1.0  10-14
CHEM 102 Fall 15, LA TECH
14-2-37
What is the reaction quotient, Q
(Q) is constant in the equilibrium expression when
initial concentration of reactants and products are
used.
SO2(g)+ NO2(g)  NO(g) +SO3(g)
[NO][SO3]
Q = ---------------[SO2][NO2]
comparing to K and Q provide the net direction to
achieve equilibrium.
CHEM 102 Fall 15, LA TECH
14-2-38
Predicting the Direction of a Reaction
CHEM 102 Fall 15, LA TECH
14-2-39
Q Calculation
Consider the following reaction:
SO2(g) + NO2(g)  NO(g) + SO3(g)
(Kc = 85.0 at 460oC)
Given: 0.040 mole of SO2(g), 0.500 mole of NO2(g),
0.30 mole of NO(g),and 0.020
mole of SO3(g) are mixed in a 5.00 L flask,
Determine:
a) The net the reaction quotient, Q.
b) Direction to achieve equilibrium at 460oC.
CHEM 102 Fall 15, LA TECH
14-2-40
Q Calculation
SO2(g) + NO2(g)  NO(g) + SO3(g) (Kc = 85.0 at 460oC)
[NO][SO3]
Q = ------------[SO2][NO2]
0.040 mole
0.500 mole
0.30 mole
0.020 mole
[SO2] = -------------; [NO2] = ----------- ; [NO] = ------------; [SO3] = ----------5.00 L
5.00L
5.00L
5.00 L
[SO2] = 8 x 10-3mole/L ; [NO2] =0.1mole/L; [NO] = 0.06 mole/L; [SO3] = 4 x 103mole/L
Q=
0.06 (4 x 10-3 )
---------------------8.0 x 10-3 x 0.1
= 0.3
Therefore the equilibrium shift to right
CHEM 102 Fall 15, LA TECH
14-2-41
Le Chatelier’s principle
Any stress placed on an equilibrium system will
cause the system to shift to minimize the effect
of the stress.
You can put stress on a system by adding or
removing something from one side of a
reaction.
N2(g) + 3H2 (g)
2NH3 (g)
What effect will there be if you added more
ammonia? How about more nitrogen?
CHEM 102 Fall 15, LA TECH
14-2-42
Predicting Shifts in Equilibria
Equilibrium concentrations are based on:
• The specific equilibrium
• The starting concentrations
• Other factors such as:
• Temperature
• Pressure
• Reaction specific conditions
Altering conditions will stress a system,
resulting in an equilibrium shift.
CHEM 102 Fall 15, LA TECH
14-2-43
Increase in Concentration
or Partial Pressure
for
N2(g) + 3 H2(g)  2 NH3(g)
an increase in N2 and/or H2 concentration or
pressure, will cause the equilibrium to
shift towards the production of NH3
CHEM 102 Fall 15, LA TECH
14-2-44
Shifts with Temperature
N2O4(g)
colorless
2NO2(g)
Dark brown
N2O4(g)  2 NO2(g) ; D H=? (+or -)
CHEM 102 Fall 15, LA TECH
14-2-45
Predicting Equilibrium Shifts
For the following equilibrium reactions:
H2(g) + CO2(g)  H2O(g) + CO(g) DH = 40 kJ
Predict the equilibrium shift if:
a) The temperature is increased
b) The pressure is decreased
CHEM 102 Fall 15, LA TECH
14-2-46
Shifting of Equilibrium
N2O4(g)  2 NO2(g)
CHEM 102 Fall 15, LA TECH
14-2-47
Changes in pressure
In general, increasing the pressure by decreasing
volume shifts equilibrium towards the side that
has the smaller number of moles of gas.
Unaffected by pressure
H2 (g) + I2 (g)
2HI (g)
Increased pressure, shift to left
N2O4 (g)
CHEM 102 Fall 15, LA TECH
2NO2 (g)
14-2-48
5) How you would increase the products of
following industrially important reactions:
a) CO(g) + H2O (g)
H2 (g) + CO2 (g);DH= −41.2 kJ/mol
b) N2(g) + 3H2(g)
nitrogen
hydrogen
CHEM 102 Fall 15, LA TECH
2NH3(g) H = -92.4 kJ mol-1
ammonia
14-2-49
5) How you would increase the products of
following industrially important reactions:
a) CO(g) + H2O (g)
H2 (g) + CO2 (g);DH= −41.2 kJ/mol
Pressure: no change in Dn,  pressure have no effect: moderate
pressure
Temperature: DH is – (exothermic) heats up: moderate
temperature
Catalysts: Increase the rate at which equilibrium is reached
b) N2(g) + 3H2(g)
2NH3(g) H = -92.4 kJ mol-1
nitrogen
hydrogen
ammonia
Pressure: change in Dn= -2,   pressure shift to more NH3:
higher pressure.
Temperature: DH is – (exothermic) heats up: moderate
pressure
Catalysts: Increase the rate at which equilibrium is reached
CHEM 102 Fall 15, LA TECH
14-2-50
Equilibrium Systems
product-favored if K > 1
exothermic reactions favor products
increasing entropy in system favors
products
at low temperature, product-favored
reactions are usually exothermic
at high temperatures, product-favored
reactions usually have increase in entropy
CHEM 102 Fall 15, LA TECH
14-2-51
Thermodynamics of Equilibrium
a) Enthalpy (DH)
b) Entropy (DS)
c) Free Energy (DG)
(D G is a combined term involving DH, DS and T)
CHEM 102 Fall 15, LA TECH
14-2-52
Probability, Entropy and
Chemical Equilibrium
CHEM 102 Fall 15, LA TECH
14-2-53
Entropy
measure of the disorder in the system
more disorder for gaseous systems than
liquid systems, more than solid systems
Chapter 18. Thermodynamics
DG = DH -TDS
DG = Gibbs Free Energy (- for
spontaneous)
DH = Enthalpy
DS = Entropy
T = Kelvin Temperature
CHEM 102 Fall 15, LA TECH
14-2-54
Equilibrium Reaction Rates
reactions occur faster in gaseous phase
than solids and liquids
reactions rates increase as temperature
increases
reactions rates increase as concentration
increases
rates increase as particle size decreases
rates increase with a catalyst
CHEM 102 Fall 15, LA TECH
14-2-55
Industrial Production of Ammonia
catalysis
N2(g) + 3 H2(g)
 2 NH3(g) ; DH = -
high pressure
and temperature
CHEM 102 Fall 15, LA TECH
14-2-56
Ammonia Synthesis
reaction is slow at room temperature, raising
temperature, increases rate but lowers yield
increasing pressure shifts equilibrium to
products
liquefying ammonia shifts equilibrium to
products
use of catalyst increases rate
CHEM 102 Fall 15, LA TECH
14-2-57
Haber-Bosch Process
CHEM 102 Fall 15, LA TECH
14-2-58
Decrease in Concentration or
Partial Pressure
for
N2(g) + 3 H2(g)  2 NH3(g) ; DH = -
likewise, a decrease in NH3
concentration or pressure will
cause more NH3 to be produced
CHEM 102 Fall 15, LA TECH
14-2-59
Changes in Temperature
for
N2(g) + 3 H2(g)  2 NH3(g) ; DH = -
for an exothermic reaction, an
increase in temperature will cause
the reaction to shift back towards
reactants and vice versa.
CHEM 102 Fall 15, LA TECH
14-2-60
Volume Change
for
N2(g) + 3 H2(g)  2 NH3(g) ; DH = -
an increase in volume, causes the equilibrium to
shift to the left where there are more gaseous
molecules
a decrease in volume, causes the equilibrium to
shift to the right where there are fewer gaseous
molecules
CHEM 102 Fall 15, LA TECH
14-2-61
Shifts with Temperature
N2O4(g)
colorless
2NO2(g)
Dark brown
N2O4(g)  2 NO2(g) ; D H=? (+or -)
CHEM 102 Fall 15, LA TECH
14-2-62
Probability, Entropy and
Chemical Equilibrium
CHEM 102 Fall 15, LA TECH
14-2-63
Entropy
measure of the disorder in the system
more disorder for gaseous systems than
liquid systems, more than solid systems
Chapter 17. Thermodynamics
DG = DH -TDS
DG = Gibbs Free Energy (- for
spontaneous)
DH = Enthalpy
DS = Entropy
T = Kelvin Temperature
CHEM 102 Fall 15, LA TECH
14-2-64
Predicting Equilibrium Shifts
For the following equilibrium reactions:
H2(g) + CO2(g)  H2O(g) + CO(g); DH = 40 kJ
Predict the equilibrium shift if:
a) The temperature is increased
b) The pressure is decreased
CHEM 102 Fall 15, LA TECH
14-2-65
Equilibrium Systems
product-favored if K > 1
exothermic reactions favor products
increasing entropy in system favors
products
at low temperature, product-favored
reactions are usually exothermic
at high temperatures, product-favored
reactions usually have increase in entropy
CHEM 102 Fall 15, LA TECH
14-2-66
Equilibrium Reaction Rates
reactions occur faster in gaseous phase
than solids and liquids
reactions rates increase as temperature
increases
reactions rates increase as concentration
increases
rates increase as particle size decreases
rates increase with a catalyst
CHEM 102 Fall 15, LA TECH
14-2-67