14-1CHEM 102, Fall 2011, LA TECH
Instructor: Dr. Upali Siriwardane
e-mail: [email protected]
Office: CTH 311 Phone 257-4941
Office Hours: M,W 8:00-9:00 & 11:00-12:00 am;
Tu, Th, F 8:00 - 10:00am.
Exams: 11:30-12:45 am, CTH 322.
Sept 22, 2011 (Test 1): Chapter 13
Oct 18, 2011 (Test 2): Chapter 14 &15
Nov 15, 2011 (Test 3): Chapter 17 &18
Comprehensive Final Exam: Nov 17, 2011 :
Chapters 13, 14, 15, 16, 17, and 18
Chemistry 102(01) Fall 2011
14-2CHEM 102, Fall 2011, LA TECH
Chapter 14. Chemical Equilibrium 14.1Characteristics of Chemical Equilibrium 14.2The Equilibrium Constant 14.3Determining Equilibrium Constants 14.5The Meaning of Equilibrium Constant14.6Using Equilibrium Constants14.7Shifting a Chemical Equilibrium: Le Chatelier's
Principle 14.8Equilibrium at the Nanoscale14.9Controlling Chemical Reactions: The Haber-Bosch
Process
14-3CHEM 102, Fall 2011, LA TECH
Different types of arrows are used in chemical equations associated with equilibria.
Single arrow
Assumes that the reaction proceeds to completion as written.
Two single-headed arrows
Used to indicate a system in equilibrium.
Two single-headed arrows of different sizes.
May be used to indicate when one side of an equilibrium system is favored.
Chemical equilibrium
14-4CHEM 102, Fall 2011, LA TECH
Chemical EquilibriumBranch of chemistry dealing with reactions
where reactants and products coexist in a dynamic equilibrium
the rates of forward and backward reactions have comparable rates reaction
14-5CHEM 102, Fall 2011, LA TECH
Chemical Equilibrium
Equilibrium region.
A point is finally reached where the forward and reverse reactions occur at the same rate.
H2 + I2 2HI
There is no net change in the concentration of any of the species.
14-6CHEM 102, Fall 2011, LA TECH
Chemical EquilibriumP
art
ial P
ressu
re
Time
HI
I2
H2
Equilibrium
Region
Kinetic
Region
14-8CHEM 102, Fall 2011, LA TECH
Equilibrium
A state where the forward and reverse conditions occur at the same rate.
Dynamic
Equilibrium
I’m in static
equilibrium.
14-9CHEM 102, Fall 2011, LA TECH
This type of plot
shows the energy
changes during
a reaction.
This type of plot
shows the energy
changes during
a reaction.
Forward and Backward Reactions
H
activation
energy
Pote
nti
al
En
erg
y
Reaction coordinate
14-10CHEM 102, Fall 2011, LA TECH
Value of K
rate of forward Reaction k+ K = ------------------------------ = --- rate of backward Reaction k-
K = a (infinity) -> Irreversible reactions
K = 0 -> No reaction
K = between 0 and 1 -> Equilibrium reactions
14-11CHEM 102, Fall 2011, LA TECH
Law of mass ActionDefines an equilibrium constant (K) for the process
j A + k B l C + m D
[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.
14-12CHEM 102, Fall 2011, LA TECH
Equilibrium ExpressionAn equilibrium expression could be written
for any reaction
[HI]2
K = ----------- = 16 L/mol [H2][I2]Keq >> 1 reaction will go mainly to products
Keq ~ 1 reaction will produce roughly equal amounts of product and reactant
Keq << 1 reaction will go mainly to reactants
14-13CHEM 102, Fall 2011, LA TECH
k is constant at a temperature
Initial @ Equilibrium
N2O4 NO2 N2O4 NO2 Keq
0.00 0.02 0.0014 0.017 0.21
0.00 0.03 0.0028 0.024 0.21
0.00 0.04 0.0045 0.031 0.21
0.02 0.00 0.0045 0.031 0.21
N2O4(g)
colorless
2NO2(g)
Dark brown
K eq [ ]
[ ]NON O
2
2 4
2
14-14CHEM 102, Fall 2011, LA TECH
Calculating Stepwise EquilibriumAdd two equations with K1 and K2 to get Keq
Keq = K1 x K2
Subtract one equations with K2 from another with K2 to get Keq
Keq = K1 / K2
Doubling K1 to get Keq
Keq = (K1)2 ;tripling Keq = (K1)3 etc.
Reversing a reaction with K1 get Keq
Keq = (K1)½
14-15CHEM 102, Fall 2011, LA TECH
Stepwise Equilibrium
(1) N2(g) + O2(g) 2NO(g) [NO]
2
Kc1 =
[N2][O2]
(2) 2NO(g) + O2(g) 2NO2(g) [NO2]
2
Kc2 =
[NO]2
[O2]Add to Combine (1.) & (2.)
N2(g) + 2O2(g) 2NO2(g)
[NO]2
[NO2]2
Kc = = Kc1 Kc2
[N2][O2] [NO]2
[O2]
14-16CHEM 102, Fall 2011, LA TECH
Stepwise Equilibrium
Consider the reactions
2NO + O2 <===> 2 NO2 K = a
2 NO2 <===> N2O4 K = b
The value of the equilibrium constant for the reaction
2NO + O2 <===> N2O4 is
a. a + b
b. ab
c. (a/b)2
d. (ab)2
e. ab/2
14-17CHEM 102, Fall 2011, LA TECH
Stepwise Equilibrium
Consider the reactions
2NO + O2 <===> 2 NO2 K = a
2 NO2 <===> N2O4 K = b
The value of the equilibrium constant for the reaction
4NO + 2O2 <===> 2 N2O4 is
a. a + b
b. ab
c. (a/b)2
d. (ab)2
e. ab/2
14-18CHEM 102, Fall 2011, LA TECH
Homogenous equilibrium: Chemical equilibrium where reactants and products are in same phase.
Heterogeneous equilibrium: Chemical Equilibrium where at least one phase of a reactant or product is different from the rest.
Types of Equilibria
14-19CHEM 102, Fall 2011, LA TECH
Homogenous equilibrium: Chemical equilibrium where reactants and products are in same phase.
Heterogeneous equilibrium: Chemical Equilibrium where at least one phase of a reactant or product is different from the rest.
Types of Equilibria
14-20CHEM 102, Fall 2011, LA TECH
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)]
14-21CHEM 102, Fall 2011, LA TECH
Acid Dissociation ConstantHC2H3O2 (aq) + H2O(l) H3O+ (aq) + C2H3O2
- (aq)
[H3O+
][C2H3O2-]
K =
[H2O][HC2H3O2]
[H3O+
][C2H3O2-]
Ka = K [H2O] =
[HC2H3O2]
14-22CHEM 102, Fall 2011, LA TECH
Base Dissociation Constant
NH3 + H2O(l) NH4+ + OH-
[NH4+
][OH-]
K =
[H2O][NH3]
[NH4+
][OH-]
Kb = K [H2O] =
[NH3]
14-23CHEM 102, Fall 2011, LA TECH
Autoionization of Water
H2O (l) + H2O (l) H3O+ + OH-
[H3O+
][OH-]
K =
[H2O]2
Kw = K [H2O]2
= [H3O+
][OH-] = 1.0 10
-14
14-24CHEM 102, Fall 2011, LA TECH
Pressure Equilibrium Constants Kc & Kp
N2 + 3H2 2NH3
[NH3]2
Kc =
[N2][H2]3
=(PNH3/RT)
2
(PN2/RT) (PH2/RT)3
(PNH3)2 (1/RT)
2
Kc =
(PN2) (1/RT))(PH2)3
(1/RT)3
)
PNH32 (1/RT)2
=
PN2 PH23 (1/RT)(1/RT)3
(1/RT) 2
= Kp
(1/RT)(1/RT)3
14-25CHEM 102, Fall 2011, LA TECH
Kc vs. Kp
N2 (g) + 3H2 (g) 2NH3 (g)
In General
Kc = Kp (1/RT)Dn
where Dn = #moles gaseous products
- # moles gaseous reactants
(1/RT)2
Kc = Kp = Kp (1/RT)-2
(1/RT)(1/RT)3
14-26CHEM 102, Fall 2011, LA TECH
What is K (Kc) and Kp
Kc (K) - equilibrium constant calculated based on [A]-Concentrations.
Kp- equilibrium constant calculated based on partial pressure (p)
Kp = K(RT) Dn
R = universal gas constant
T = Kelvin Temperature,
Dn = (sum of stoichiometric coefficients of gaseous products) - (sum of the stoichiometric coefficients of gaseous reactants)
14-27CHEM 102, Fall 2011, LA TECH
For the following equilibrium, Kc = 1.10 x 107
at 700. o
C. What is the Kp?
2H2 (g) + S2 (g) 2H2S (g)
Kp = Kc (RT)Dng
T = 700 + 273 = 973 K
R = 0.08206
Dng = ( 2 ) - ( 2 + 1) = -1
atm L
mol K
Partial pressure & Equilibrium Constants
14-28CHEM 102, Fall 2011, LA TECH
Kp = Kc (RT)Dng
= 1.10 x 107
(0.08206 ) (973 K)
= 1.378 x105
atm L
mol K[ ]-1
Partial pressure & Equilibrium Constants
14-29CHEM 102, Fall 2011, LA TECH
Determining Equilibrium ConstantsICE 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
14-30CHEM 102, Fall 2011, LA TECH
TerminologyInitial concentration:concentration (M) of reactants and products
before the equilibrium is reached.
Equilibrium ConcentrationConcentration (M) of reactants and products
After the equilibrium is reached.
14-31CHEM 102, Fall 2011, LA TECH
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 equilibrium constant?
N2O4(g) 2 NO2(g)
[NO2]2
Kc =
[N2O4]
N2O4(g) 2 NO2(g)
[Initial] (mol/L) 0.40 0
[Change] -x 2x
[Equilibrium]
0.40- 0.243= 0.138 0.525
x-1/2 x
0.40 - 1/2x = 0 + x
14-32CHEM 102, Fall 2011, LA TECH
What is the value of the equilibrium constant?
0.525 = 0 + x [NO2]2
Kc =
[N2O4]
0.40 - 1/2x
x = 0.525 [NO2] = 0.40 - 1/2x
= 0.40 - 1/2(0525)
= 0.138
[NO2]2
(0.525)2
Kc = =
[N2O4] 0.138
= 2.00
NO2(g ) N2O4(g)
14-33CHEM 102, Fall 2011, LA TECH
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?
14-34CHEM 102, Fall 2011, LA TECH
Initial 4.00/5=.80 0 0
Change -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 0.0884Kc = ---------------- = ------------------------- = 0.0201 [HI]2 (0.62) 2
2 HI(g) H2(g) + I2(g)
14-36CHEM 102, Fall 2011, LA TECH
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.
14-37CHEM 102, Fall 2011, LA TECH
We can predict the direction of a reaction by calculating the reaction quotient.
Reaction quotient, Q
For the reaction: aA + bB eE + fF
Q has the same form as Kc with one important difference. Q can be for any set of concentrations, not
just at equilibrium.
Q =[E]
e [F]
f
[A]a [B]
b
Equilibrium calculations
14-38CHEM 102, Fall 2011, LA TECH
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 Net forward reaction will occur.
Q = Kc No change, at equilibrium.
Q > Kc Net reverse reaction will occur.
Reaction quotient
14-40CHEM 102, Fall 2011, LA TECH
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.
Q Calculation
14-41CHEM 102, Fall 2011, LA TECH
Q CalculationSO2(g) + NO2(g) NO(g) + SO3(g) (Kc = 85.0 at 460
oC)
[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 10-3
mole/L ; [NO2] =0.1mole/L; [NO] = 0.06 mole/L; [SO3] = 4 x 10-3
mole/L
0.06 (4 x 10-3
)
Q = ------------------ = 0.3
8.0 x 10-3
x 0.1
Therefore the equilibrium shift to right
14-42CHEM 102, Fall 2011, LA TECH
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?
14-43CHEM 102, Fall 2011, LA TECH
Equilibrium Calculation Example
COCl2 (g) CO (g) Cl2 (g)
Initial conc., M 0.095 0.000 0.000
Change - X + X + X
in conc. due to reaction
Equilibrium M(0.095 -X) X X
Concentration,
Kc = =[ CO ] [ Cl2 ]
[ COCl2 ]
X2
(0.095 - X)
14-44CHEM 102, Fall 2011, LA TECH
Equilibrium calculation example
X2
(0.095 - X)Kc = 2.2 x 10-
10 =
Rearrangement gives
X2 + 2.2 x 10-
10 X - 2.09 x 10
-11 = 0
This is a quadratic equation. Fortunately, there is a
straightforward equation for their solution
14-45CHEM 102, Fall 2011, LA TECH
Quadratic equations
An equation of the form
a X2 + b X + c = 0
Can be solved by using the following
x =
Only the positive root is meaningful in equilibrium problems.
-b + b2 - 4ac
2a
14-46CHEM 102, Fall 2011, LA TECH
Equilibrium Calculation Example
-b + b2 - 4ac
2a
X2 + 2.2 x 10
-10 X - 2.09 x 10
-11 = 0
a b c
X =
X = - 2.2 x 10-10
+ [(2.2 x 10-10
)2 - (4)(1)(- 2.09 x 10
-11)]
1/2
2
X = 4.6 x 10-6
M
X = -4.6 x 10-6
M
14-47CHEM 102, Fall 2011, LA TECH
Equilibrium Calculation Example
Now that we know X, we can solve for the concentration of all of the species.
COCl2 = 0.095 - X = 0.095 M
CO = X = 4.6 x 10-6 M
Cl2 = X = 4.6 x 10-6 M
In this case, the change in the concentration of is COCl2 negligible.
14-48CHEM 102, Fall 2011, LA TECH
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?
14-49CHEM 102, Fall 2011, LA TECH
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.
14-50CHEM 102, Fall 2011, LA TECH
Increase in Concentrationor 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
14-51CHEM 102, Fall 2011, LA TECH
N2O4(g) 2 NO2(g) ; D H=? (+or -)
Shifts with TemperatureN2O4(g)
colorless
2NO2(g)
Dark brown
14-53CHEM 102, Fall 2011, LA TECH
Entropymeasure of the disorder in the systemmore 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
14-54CHEM 102, Fall 2011, LA TECH
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
Predicting Equilibrium Shifts
14-55CHEM 102, Fall 2011, LA TECH
Changes in pressureIn general, increasing the pressure by decreasing
volume shifts equilibrium towards the side that has the smaller number of moles of gas.
H2 (g) + I2 (g) 2HI (g)
N2O4 (g) 2NO2 (g)
Unaffected by pressureUnaffected by pressure
Increased pressure, shift to leftIncreased pressure, shift to left
14-57CHEM 102, Fall 2011, LA TECH
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
14-58CHEM 102, Fall 2011, LA TECH
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
14-59CHEM 102, Fall 2011, LA TECH
Production of Ammonia
N2(g) + 3 H2(g) 2 NH3(g) ; DH = -
catalysis
high pressure
and temperature
14-60CHEM 102, Fall 2011, LA TECH
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
14-62CHEM 102, Fall 2011, LA TECH
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
14-63CHEM 102, Fall 2011, LA TECH
Changes in Temperaturefor 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.
14-64CHEM 102, Fall 2011, LA TECH
Volume Changefor 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
14-65CHEM 102, Fall 2011, LA TECH
At 100o C the equilibrium constant (K) for the reaction:
H2(g) + I2(g) 2HI(g)
is 1.15 x 102
. If 0.400 moles of H2 and 0.400 moles of 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?
Calculating Concentrations at Equilibrium
14-66CHEM 102, Fall 2011, LA TECH
At a certain temperature the value of the equilibrium constant is 3.24 for the
reaction:
H2(g) + CO2(g) H2O(g) + CO(g)
If 0.400 mol H2 and 0.400 mol CO2 are placed in a 1.00 L vessel, what is the
concentration of of CO at equilibrium?
Calculating Concentrations at Equilibrium
14-67CHEM 102, Fall 2011, LA TECH
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?
14-68CHEM 102, Fall 2011, LA TECH
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.
14-69CHEM 102, Fall 2011, LA TECH
Increase in Concentrationor 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
14-70CHEM 102, Fall 2011, LA TECH
N2O4(g) 2 NO2(g) ; D H=? (+or -)
Shifts with TemperatureN2O4(g)
colorless
2NO2(g)
Dark brown
14-72CHEM 102, Fall 2011, LA TECH
Entropymeasure of the disorder in the systemmore 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
14-73CHEM 102, Fall 2011, LA TECH
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
Predicting Equilibrium Shifts
14-74CHEM 102, Fall 2011, LA TECH
Changes in pressureIn general, increasing the pressure by decreasing
volume shifts equilibria towards the side that has the smaller number of moles of gas.
H2 (g) + I2 (g) 2HI (g)
N2O4 (g) 2NO2 (g)
Unaffected by pressureUnaffected by pressure
Increased pressure, shift to leftIncreased pressure, shift to left
14-76CHEM 102, Fall 2011, LA TECH
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
14-77CHEM 102, Fall 2011, LA TECH
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
14-78CHEM 102, Fall 2011, LA TECH
Production of Ammonia
N2(g) + 3 H2(g) 2 NH3(g) ; DH = -catalysis
high pressure
and temperature
14-79CHEM 102, Fall 2011, LA TECH
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
14-80CHEM 102, Fall 2011, LA TECH
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
14-81CHEM 102, Fall 2011, LA TECH
Changes in Temperaturefor 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.