Complete PPT On Chemical Equilibria

 

Chemical Equilibrium • What is equilibrium? • Expressions for equilibrium constants, K  constants, K c; • Calculating K  Calculating K c using equilibrium concentrations; • Calculating equilibrium concentrations using initial concentration and K  and K c value; • Relationship between K  between K  and and K   K  ; c

 p

• Factors that affect equilibrium; • Le Chat Chatelier elier’s ’s Princ Principle iple

 

What is Equilibrium?

 

This is not Equilibrium?

 

Chemical Equilibrium in Nature: (The formation of stalagmites and Stalactites)

 

Chemical Equilibrium • Consider the following reactions: CaCO3( s) aq)) ..(1)  s) + CO2(aq aq)) + H2O(l )  Ca2+(aq aq)) + 2HCO3-(aq and Ca2+(aq aq))  CaCO3( s)  s) + CO2(aq aq)) + H2O(l ) ..(2) aq)) + 2HCO3-(aq Reaction (2) is the reverse of reaction (1). At equilibrium the two opposing reactions occur at the same rate. Concentrations of chemical species do not change once equilibrium is established.

 

Expression for Equilibrium Constant Consider the following equilibrium system: wA +  x  xB B  yC  y C +  z D y

[D]]  K c = [C] [D w

[A]] [B [A [B]]

 z 

 x

• The numerical value of K  of K c is calculated using the concentrations of reactants and products that exist at equilibrium.

 

Expressions for Equilibrium Constants • Examples:  N2( g   g )   g ) + 3H2( g 

2NH3( g   g );  K c =

PCl5( g   g );   K c =  g ) + Cl2( g   g )   PCl3( g  CH

( g   g )

4

+H

( g   g )

 

CO( g   g ) + 3H 2

2

 K c =

[CO][H 2 ]

3

[CH 4 ][ ][H H 2 O]

( g   g );

[NH 3 ]

2

[N 2 ][ ][H H2]

3

[PCl 3 ][Cl 2 ] [PCl 5 ]

 

Calculating Equilibrium Constant • Example-1: 1.000 mole of H2 gas and 1.000 mole of I2 vapor are introduced into a 5.00-liter sealed flask. The mixture is heated to a certain temperature and the following reaction occurs until equilibrium is established. H2(g) + I2(g)

⇄

2HI(g)

At equilibrium, the mixture is found to contain 1.580 mole of HI. (a) What are the concentrations of H2, I2 and HI at equilibrium? (b) Calculate the equilibrium constant K  constant K c.

 

Calculating Equilibrium Constant for reaction: H2(g) + I2(g) ⇄ 2HI(g) •

————————— ————————————— ———————— ———————— ———————— ———————  ——— 

H2(g)

• •

• • • •

+

I2(g)

2 HI(g)

————————— ————————————— ———————— ———————— ———————— ———————  ———    Initial    

[ ], M: 0.200 Change in [ ], M: -0.158 Equilibrium [ ], M 0.042

0.200 -0.158 0.042

0.000 + 0.316 0.316

————————— ————————————— ———————— ———————— ———————— ———————  ———  2

 K c =

[HI] [H 2 ][ ][II 2 ]

2

= (0.316) 2 = 57 (0.042 )

 

Calculating Equilibrium Constant • Example-2: 0.500 mole of HI is introduced into a 1.00 liter sealed flask and heated to a certain temperature. Under this condition HI decomposes to produce H2 and I2 until an equilibrium is established. An analysis of the equilibrium equilib rium mixture shows that 0.105 mole of HI has decomposed. Calculate the equilibrium concentrations of H2, I2 and HI, and the equilibrium constant K  constant K c for the following reaction:

H2(g) + I2(g)   ⇄ 2HI(g),

 

Calculating Equilibrium Constant from • The re reaaction on:: H2(g) + I2(g)   ⇄ 2HI(g), proceeds fr right to left. •

————————— ————————————— ———————— ———————— ———————— ———————  ——— 

H2(g) +

• •

• •

I2(g) ⇄ 2HI(g)

————————— ————————————— ———————— ———————— ———————— ———————  ———    Initial  

[ ], ], M   M : 0.000 ], M   M : +0.0525 Change in [ ],

0.000 +0.0525

0.500 -0.105

Equil’m [ ], M  ••   ————————— 0.0525 ———————— 0.05———————— 25 0.395 ———  ————————————— ———————— ——————— 

 K c =

(0.395) (0.0525)

2 2

= 56.6

 

Expression and Value of Equilibrium Constant for a Reaction • The expression for K  for K depends depends on the equation; • The value of K  of K applies applies to that equation; it does not depend on how the reaction occurs; • Concentrations used to calculate the value of K  of K  are those measured at equilibrium.

 

Relationships between chemical equations and the expressions of equilibrium constants • The expression of equilibrium constant depends on how the equilibrium equation is written. For example, for the following equilibrium: 2 [HI]  g ) + I2( g   g );  K c  [H ][  g ) ⇄ 2 HI( g  • H2( g  I2] 2 ][I   

• For the reverse reaction:

[H 2 ][ ][II 2 ]  2 1/K c • 2HI( g   g ) ⇄ H2(  g  g ) + I2(  g  g );  K c '   [HI] 

 



 g ) ⇄ ½H2( g   g ) + ½I2( g   g ); • And for the reaction: HI( g   K c "



 

[H 2 ][ I 2 ]  

[HI] 2



1  K c '



 

 K c

 

Expression and Values of  Equilibrium Constant Using Partial Pressures • Consider the following reaction involving gases:  g )  g ) + O2( g   g ) ⇄ 2SO3( g  2SO2( g 

 K  p =

(PSO3 ) 2 (PSO2 ) 2 (PO2 )

 

The Relationship between K  between K c and and K   K  p • Consider the reaction: 2SO2(g) + O2(g) ⇄ 2SO3(g) (PSO3 ) 2

2

[SO]

=

and  K  p

•  K c = [SO2 ]2 [O 2 ]

(PSO2 ) 2 (PO2 )

• Assuming ideal behavior, • where PV = nRT and P = (n/V)RT = [M]RT

• and PSO3 = [SO3]RT; PSO2 = [SO2]RT; PO2 = [O2]RT  K  p

 



[SO3 ]2 ( RT) 2

[SO3 ]2

-1   ( R T ) [SO2 ]2 ( RT) 2 [O 2 ]( RT) [SO2 ]2 [O 2 ]  





 K c ( RT)

-1

 

Relationship between K  between K c and and K   K  p • For reaction: PCl5( g   g )    PCl3( g   g ) + Cl2( g   g );

 K  p

  (PPCl3 )( PCl2 )   [PCl 3 ]( RT) x [Cl 2 ]( RT) [PCl 5 ]( RT) ( PPCl5 )



 





[PCl 3 ][Cl 2 ] [ PCl 5 ]

( RT) 1



  K C  ( RT) 1

 

Relationship between K  between K c and and K   K  p • In general, for reactions involving gases such that,

• aA + bB ⇄ cC + d D where A, B, C, and D are all gases, and a, b, c, and d  d are are their respective coefficients,

• K  p = K c(RT)

Dn

and Dn = (c (c + d ) –   – ((a + b) (In heterogeneous systems, only the coefficients of the gaseous species are counted.)

 

Relationship between K  between K c and and K   K  p • For other reactions: -1

• 1.

 g ) ⇄ N2O4( g   g ); 2NO2( g 

 K  p = K c(RT)

• 2.

 g ) + I2( g   g ) ⇄ 2 HI( g   g ); H2( g 

 K  p =  K c

• 3.

-2

 g ) + 3H2( g   g ) ⇄ 2 NH3( g   g );  K  N2( g   p = K c(RT)

 

Homogeneous & Heterogeneous Equilibria

Homogeneous equilibria:  g ) + H2O( g   g ) ⇄ CO( g   g ) + 3H2( g   g ); CH4( g   g );  g ) + H2( g   g ) + H2O( g   g ) ⇄ CO2( g  CO( g 

Heterogeneous equilibria:  s) ⇄ CaO(  s) s) + CO2( g   g ); CaCO3( s) aq)) + H2O(l ) ⇄ H3O+(aq aq)) + F-(aq aq)); HF(aq 2+

 s) ⇄ Pb PbCl2( s)

(aq aq))

-

aq)); + 2 Cl (aq

 

Equilibrium Constant Expressions for 

Heterogeneous System • Examples: CaCO3(  s)  g ); s) ⇄ CaO(  s) s) + CO2( g   K c = [CO2]

 

K  p = PCO2;  K  p = K c(RT)

+ • HF(aq aq)); aq)) + F (aq aq)) + H2O(l ) ⇄ H3O (aq

[H 3O ][F- ]   [HF] 

 K a

 

Solubility Eqilibrium

2+

 s) ⇄ Pb PbCl2( s)

aq)) (aq

+

aq)); 2Cl (aq

 K  sp = [Pb2+][Cl-]2 ( K   K  sp is called solubility called solubility product )

 

Combining Equations and Equilibrium Constants • when two or more equations are added to yield a net equation, the equilibrium constant for the net equation, K  equation,  K net  net , is equal to the product of equilibrium constants of individual equations.

• For example, Eqn(1): A + B

C + D;

 K 1

 

[C][D]

  

[A][D] Eqn(2): C + E

B + F;

 K 2

 

[B][F]

  

[C][E]

 

Combining Equations and Equilibrium Constants

•  Net   Net eequation: A + E  K net

 

[D][F]

  

[A][E]

D + F;

= K 1 x  K 2

 Net equation equation, • If Eqn(1) + Eqn(2) =  Net  then  K 1 x K 2 = K net net

 

Equilibrium Exercise #1 A flask is charged with 2.00 atm of nitrogen dioxide and 1.00 atm of dinitrogen tetroxide at 25 oC and allowed to reach equilibrium. When equilibrium is established, the  partial pressure of NO2 has decreased by 1.24 atm. (a) What are the partial pressures of NO2 and N2O4 at equilibrium? (b) Calculate K  Calculate K  p and and K   K c for following reaction at 25 oC.

2 NO2(g)

⇄

 N2O4(g)

(Answer: K  2.80; K c = 68.6) (Answer: K  p = 2.80; K 

 

Equilibrium Exercise #2a • Methanol is produced according to the following equation:

•

CO(g) + 2H2(g)   CH3OH(g)

• In an experiment, 1.000 mol each of CO and H2 were allowed to react in a sealed 10.0-L reaction vessel at 500 K. When the equilibrium was established, the mixture was found to contain 0.0892 mole of CH3OH. What are the equilibrium concentrations of CO, H2 and CH3OH? Calculate the equilibrium constants K  constants K c and and K   K  p for this reaction at 500 K? (R = 0.0821 L.atm/Mol.K) (Answer: [CO] = 0.0911 M  0.0911 M ; [H2] = 0.0822 M  0.0822 M ; [CH3OH] = 0.00892 M; -3

(b) K  (b)  K c = 14.5;  K  p = 8.60 x 10 )

 

Equilibrium Exercise #2b For the reaction: CO(g) + 2H2(g) ⇄ CH3OH(g)  K c = 15.0 at a certain temperature. Is a reaction mixture that contains 0.40 M CO, 0.80 M  0.80 M H H2, and 0.10 M  0.10  M CH CH3OH at equilibrium? If not, in which direction will the net reaction occur? What will be their concentrations when equilibrium is established?

 

Equilibrium Exercise #3 1. The reaction: N2(g) + 3H2(g)  2NH3(g), has equilibrium constant, K  constant, K c = 0.0602 at 500oC. What is the equilibrium constant for the following reaction?

 NH3(g)   ½N2(g) + 3/2H2(g) 2.

For the reaction: 2SO2(g) + O2(g)

2 SO3(g),

 K c = 280 at 1000 K. What is the equilibrium constant for the decomposition of SO3 at 1000 K according to the following equation? (g)  

SO3

(g)

SO2

(g)

+ ½ O2 .

 

Equilibrium Exercise #4 If

N2(g) + ½ O2(g)    N2O(g);  K c(1) = 2.4 x 10-18

and N2(g) + O2(g)   2 NO(g);

 K c(2) = 4.1 x 10-31

What is the equilibrium constant for the reaction?

 N2O(g) + ½ O2(g)   2NO(g)

(Answer: K net = 1.7 x 10-13) (Answer: K 

 

Applications of Equilibrium Constant For any system or reaction: 1. Kno Knowin wing g the eequi quilib libriu rium m co const nstant ant,, we ccan an p pred redict ict whether or not a reaction mixture is at equilibrium, and we can predict the direction of net reaction. •   Qc = K c  equilibrium (no net reaction) •   Qc    K c  a net reverse reaction 2. The value of K   K tells tells us whether a reaction favors the products or the reactants.

 

Equilibrium constant is used to predict the direction of net reaction known K c value, the direction of net reaction • For a reaction of known K  can be predicted by calculating the reaction quotient, Qc. •

Qc is called the reaction quotient , where for a reaction such as:

•

aA + bB bB   cC + d D; D; Qc   

[C]c [ D]d a

 b

[A]] [ B] [A

• Qc has the same expression as K  as K c , but

necessar ily at • Qc is calculated using concentrations that are not necessarily equilibrium.

 

What does the reaction quotient tell us? If Qc = K c,  the reaction is at equilibrium; If Qc  K c,



the reaction is not at equilibrium and there’s a net reaction in the opposite direction.

 

Why is Equilibrium Constant Important?

• Knowing K  Knowing K c and the initial concentrations, we can determine the concentrations of components at equilibrium.

 

Equilibrium Exercise #5 • For the reaction:

CO(g) + 2 H2(g)   CH3OH(g),  K c = 14.5 at 500 K. • Predict whether a mixture that contains 1.50 mol of H2, 1.00 mol of CO, and 0.50 mol of CH3OH in a 10.0-L vessel at 500  K c is at equilibrium. • If not, indicate the direction in which the net reaction will occur to reach equilibrium. (Answer: Qc = 22.2 > K  > K c; net reaction reaction is to the left)

 

Calculating equilibrium concentrations using initial concentrations and value of K  of K c • Consider the reaction: • H2(g) + I2(g)   2 HI(g),

• where where K   K c = 55.6 at 425oC.

• If [H2]0 = [I2]0 = 0.1000 M, and [HI]0 = 0.0 M, what are their concentrations at equilibrium?

 

Using the ICE table to calculate equilibrium concentrations • Equation: •

 

•

  Initial

• •

  Change

•

   

H2(g)

+

I2(g)

 

2 HI(g),

 

[ ], ], M   M 

0.1000

0.1000

[ ], M - x  x - x  x ], M  (0.1000 - x - x)) (0.1000 - x  x)) Equilibrium [ ], M 

0.0000 +2x 2 x

      

[HI]  K c



2

(2 x)   

2

 [H ][ I ]   (0.100 -  x) 2 2 2



55.6

 

Calculation of equilibrium concentrations 2 x (0.100 -  x



55.6  7.46

2 x  0.746 - 7.46 x;   9.46 x  0.746  x 

0.0789 ;  



 

[H 2 ] [I 2 ] 0.0211 M;  [HI]



0.158 M

 

Equilibrium Exercise #6 For the reaction:  g )   N2O4(g);  K  2 NO2( g   p = 1.27 at 353 K.

If the initial pressure of NO2 was 3.92 atm, and initially there was no N2O4, what are the partial  pressures of the gases at equilibrium equilibrium at 353 K? What is the total gas pressure at equilibrium?

(Answer: P NO2 = 1.06 atm; P N2O4 = 1.43 atm; Ptotal = 2.49 atm)

 

Equilibrium Exercise #7 The reaction:  g )   PCl3( g   g ) has  g ) + Cl2( g  PCl5( g  has K   K c = 0.0900. 5 is placed in an empty A 0.1000-mol sample of PCl 1.00-L flask and the above reaction is allowed to come to equilibrium at a certain temperature. How many moles of PCl5, PCl3, and Cl2, respectively, are  present at equilibrium?

(Answer: PCl5 = 0.0400 mol; PCl3 = Cl2 = 0.0600 mol)

 

Le Châtelier’s Principle • The The Le  Le Châtelier's principle principle states that: when factors that influence an equilibrium equ ilibrium are altered, the equilibrium will shift to a new position that tends to minimize those changes. changes.

• Factors that influence equilibrium: Concentration, temperature, and partial pressure (for gaseous)

 

The Effect of Changes in Concentration • Consider the reaction: N2(g) + 3H2(g)   2 NH3(g);  K c

 

[NH] 2



3

[N 2 ][ ][H H2]

• If [N2] and/or [H2] is increased, Qc  K c, and a net reverse reaction

will occur to come to new equilibrium position.

 

 Effects of Pressure Change on Equilibrium • If the volume of a gas mixture is compressed, the overall gas pressure will increase. In which direction the equilibrium will shift in either direction depends on the reaction stoichiometry. • However, there will be no effect to equilibrium if the total gas pressure is increased by adding an inert gas that is not part of the equilibrium system.

 

Reactions that shift right when pressure increases and shift left when pressure decreases Consider the reaction:  g ),  g )   2SO3( g   g ) + O2( g  2SO2( g  1.

Th Thee ttot otal al mol oles es of ga gass d dec ecre reas ases es as re reac acti tion on  proceeds in the forward direction. 2. If p pre ress ssur uree iiss in incr crea ease sed db by yd dec ecre reas asin ing g the the volu volum me

3.

(compression), a forward reaction occurs to reduce r educe the stress. Re Reac acti tion onss ttha hatt rres esul ultt iin n ffew ewer er m mol oles es of ga gass ffav avor or high pressure conditions.

 

Reaction that shifts left when pressure increases,  but shifts right when pressure decreases  g );  g ) + Cl2( g   g )   PCl3( g  Consider the reaction: PCl5( g  1. 2. 3. 4. 5.

Fo Forw rwar ard d re reac acti tion on re resu sult ltss iin nm mor oree g gas as mol olec ecul ules es.. Pr Pres essu sure re iinc ncre reas ases es as rrea eact ctio ion n pro proce ceed edss to towa ward rdss equilibrium. If m mix ixtu ture re iiss co com mpr pres esse sed, d, p pre ress ssur uree in incr crea ease ses, s, aand nd reverse reaction occurs to reduce pressure; If vo volu lume me ex expa pand ndss aand nd pr pres essu sure re dr drop ops, s, fo forw rwar ard d reaction occurs to compensate. This This typ typee of of reac reacti tion onss fav favor orss low low pre press ssur uree con condi diti tion on

 

Reactions not affected by pressure changes Consider the following reactions:  g ) + H2O( g   g )   CO2( g   g ) + H2( g   g ); 1. CO( g   g ));;  g ) + Cl2( g   g )   2HCl( g  2. H2( g 

1. 2. 3.

Re Reac acti tion onss h hav avee ssam amee n num umbe berr o off g gas as mol olec ecul ules es in reactants and products. Redu Reduci cing ng or in incr crea easi sing ng th thee vo volu lume me wi will ll ca caus usee eq equa uall effect on both sides –  sides – no no net reaction will occur. Eq Equi uili libr briu ium m iiss no nott aaff ffec ecte ted d by ch chan ange ge in pr pres essu sure re..

 

The Effect Temperature on Equilibrium • Consider the following exothermic reaction:  N2(g) + 3H2(g)   2NH3(g);

o

D H 

= -92 kJ,

• The forward reaction produces heat => heat is a product. • When heat is added to increase temperature, reverse reaction will take place to absorb the heat; • If heat is removed to reduce temperature, a net forward reaction will occur to produce heat. • Exothermic reactions favor low temperature conditions.

 

The Effect Temperature on Equilibrium Consider the following endothermic reaction: CH4(g) + H2O(g)   CO(g) + 3H2(g), D H o = 205 kJ 1.

Endothermic re reaction ab absorb rbss h heeat  heat is a reactant;

2.

If heat heat is adde added d tto o iinc ncre reasi asing ng the te temp mper erat atur ure, e, it wi will ll ca cause use a net forward reaction.

3.

If heat t is rrem emov oved ed to to re redu duce ce the the temp temper erat atur ure, e, it wil willl ccau ause se a nethea reverse reaction.

4.

Endo Endoth ther ermi micc reac reacti tion onss fav favor or h hig igh h temp temper erat atur uree ccon ondi diti tion on..

 

Equilibrium Exercise #8 Determine whether the following reactions favor high or low pressures?  

2 SO3(g); PCl5(g)   PCl3(g) + Cl2(g);

1. 2.

2SO2(g)

3.

CO(g) + 2H2(g)   CH3OH(g);

4.

N2O4(g)   2 NO2(g);

5.

H2(g) + F2(g)   2 HF(g);

+

O2(g)

 

Equilibrium Exercise #9 •

Determine whether the following reactions favors high or low temperature?

1.

2SO2(g) + O2(g)   2 SO3(g);

D H o

2.

CO(g) + H2O(g)   CO2(g) + H2(g);

D H 

3.

CO(g) + Cl2(g)   COCl2(g);

4.

N2O4(g)   2 NO2(g);

5.

CO(g) + 2H2(g)   CH3OH(g);

 

 

= -180 kJ

o

= -46 kJ

D H 

o

= -108 kJ

D H o

= +57 kJ

o

D H 

= -270 kJ

 

Chemical Equilibria in Industrial Processes Production of Sulfuric Acid, H2SO4; 1. S8(s) + 8 O2(g)  8SO2(g) 2. 2SO2(g) + O2(g) ⇄ 2SO3(g); D H   H = = -198 kJ 3. SO3(g) + H2SO4(l )  H2S2O7(l ) 4. H2S2O7(l ) + H2O(l )  2H2SO4(l ) •

The second reaction is exothermic and has high activation energy;; energy

•

though thermodynamically favored the reaction is very slow at low temperature,. At high temperature reaction goes faster, but the yield would  be very low. An optimum condition is achieved at moderate temperatures

• •

and using catalysts to speed up the reaction. Reaction also favors high pressure.

favors high pressure.  

Chemical Equilibria in Industrial Processes The production of ammonia by the Haber-Bosch the Haber-Bosch  process:  N2(g) + 3H2(g) ⇄ 2NH3(g);

D H   H = =

-92 kJ

• This reaction is exothermic and very slow at low temperature. • Increasing the temperature will increase reaction rate, but will lower the yield. • An optimum condition is achieved at moderate temperature of 250 to 300oC with catalyst added to increase the reaction rate. • Increasing the pressure will favor product formation. • Reaction favors low temperature and high pressure conditions.

 

Chemical Equilibria in Industrial Processes The production of hydrogen gas: • Reaction: CH4( g   g ) + H2O( g   g ) ⇄ CO( g   g ) + 3H2( g   g ); • This reaction is endothermic with D H   H = = 206 kJ • Increasing the reaction temperature will increase both the rate and the yield. • This reaction favors high temperature and low  pressure conditions.