FORM SIX: CHEMISTRY STUDY NOTES-TOPIC 1:  CHEMICAL EQUILIBRIUM

TOPIC 1:  CHEMICAL EQUILIBRIUM

PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM

Chemical reactions which takes place in both directions are called reversible reactions .These reactions do not proceed to completion rather to dynamic equilibrium between products and reactants.

As a result of this, the concentration of reactants and products becomes constant (but not equal)

Reactants    products

At equilibrium, the rate of forward reaction is equal to the rate of backward reaction.

Note: The concentration of all species in the equilibrium remains constant since both opposing reactions proceed at the same rate.

The concentration of reactants decreases with time while those of products increases with time until equilibrium is reached where the

concentrations are constant (not equal).

 

The equilibrium has been attained with high concentration of products than reactants therefore equilibrium lies on product side.

 

 

The equilibrium has been attained with high concentration of reactants than products therefore equilibrium lies on reactants side.

The rate of reaction depends on the concentration of reactants. As the concentration of reactants decreases, the rate of forward reaction decreases

too. The rate of backward reaction increases since the concentration of products increases until equilibrium is attained.

 

Characteristics of chemical equilibrium

1. The equilibrium can be established or attained in a closed system (no part of the reactants or products is allowed to escape out.)
2. The equilibrium can be initiated from either side .The state of equilibrium of a reversible reaction can be approached whether we start fromreactants or products.

Consider the reactions:-

H_2(g) +       2HI_(g)

 

3. Constancy of concentration

When a chemical equilibrium is attained, concentration of various species in the reaction mixture becomes constant.

4. A catalyst cannot change the equilibrium point. When a catalyst is added to a system in equilibrium, it speeds up the rate of both forward andbackward reactions to equal extent.Therefore a catalyst cannot change equilibrium point except that it is reached earlier.

Types of chemical equilibrium

1. Homogeneous equilibrium:

This is equilibrium where reactant and products are in the same physical states i.e. all solids, all liquids or all gases.

E.g.  H_{2 (g)} + I_{2 (g)}   2HI _(g)

N_{2 (g)}   + 3H _2(g)       2NH_{3 (g)}

2. Heterogeneous equilibrium;

This is equilibrium when the reactants and product are in the different physical states

CaCO_3(s)     CaO_(s) + CO_{2 (g)}

3Fe _(S) + 4H₂O _(g)       Fe₃O_4(s) + 4H_{2 (g)}

3. Ionic equilibrium:

This is an equilibrium which involves ions.

E.g.
 

    LAW OF MASS ACTION

The law relates the rates of reactions and the concentration of the reacting substances. The law states that, “the rate of a chemical reaction is directly proportional to the product of the molar concentration of the reactants at  constant temperature, at any given time”.

The molar concentration means number of moles per litre and is also called active mass

 

Consider the following simple reactions:

 

The rate of the reaction, R

R = K   Rate equation

Or        R = K

Where [A] and [B] are the molar concentrations of the reactant A and B respectively and that  K is a constant of proportionality known as rate constant or velocity constant.

If the concentration of each of the reactants involved in the reaction is unity i.e. the concentration of [A] = [B] = 1, thus the rate constant of a

reaction at a given temperature may be defined as a rate of the reaction when the concentration of each of the reactants is unity.

Generally for a reaction

 

Where a and b are stoichiometric coefficients or mole ratio of the reactants, A and B

r = K [A]^a [B]^b or R=K [A]^a [B]^b

 

LAW OF CHEMICAL EQUILIBRIUM AND EQUILIBRIUM CONSTANT

The law of mass action is applied to reversible reaction to derive a mathematical expression for equilibrium constant known as the law of chemical equilibrium.

Consider a simple chemical reaction (reversible)

A+B  C+D

The forward reaction is A+B  C+D

R_{ate of forward (Rf)}  [A] [B]

R_f = K_f [A] [B]
Kf = Rate constant for forward reaction

Similarly for the backward reaction

 

Rate of _backward (Rb) [C] [D]
R_b = K_b [C] [D]
Kb = Rate constant for backward reaction

At equilibrium both rates are equal

R_forward = R_backward

K_f [A] [B] = K_b[C] [D]

=

K_e =

Generally:-

For the reaction

 

The equilibrium constant for this expression is given by

K _e =

K _e is changed to K _c for  equilibrium concentration (equilibrium constant) .

K _c =

 

Where K _c is  equilibrium concentration (equilibrium constant)

[C], [D] etc are molar concentration of species in mol per litre a, b etc are mole ratios or stoichiometric coefficients.

But for gaseous equilibrium we know that PV= nRT

PV = nRT

=

Concentration of a gas is directly proportional to partial pressure therefore the equilibrium constant can be represented in terms of both concentration and partial pressure.

For the above equilibrium;

K _p =

 

Example1.

1.N₂ O_{4 (g)} dissociates to give NO_{2 (g)}

 

K _c =

K _{p =}  

2. 2. H_{2 (g)}   + I_{2 (g)}       2HI _(g)

K _c =

K _{p =}

 

The law of chemical equilibrium states that, “For any system in equilibrium at a given temperature, the ratio of product of concentration of products to the product of the concentration of reactants raised to the power of their mole ratios is constant”.

UNITS OF EQUILIBRIUM CONSTANT

The units of equilibrium constants, K _c and K _p depends on the number of moles of reactants and products involved in the reaction

1. N_{2 (g)}   + O_{2 (g)}  2NO _(g)

K _p =

K _p =    = Unit less

 

2. N_{2 (g)} + 3H_2(g)  2NH_{3 (g)}

K _p =

K _p =

K _p =

K _{c =}

=

K _c =

 

A large value of K _p or K _c means the equilibrium lies on the sides of the product and a small value of K _c and K _p means equilibrium lies on the sides of the reactants thus the equilibrium constant shows to what extent the reactant are converted to the product.

        If K _c is greater than 10³, products pre- dominate over reactants equilibrium therefore the reaction proceeds nearly to completion.

        If K _c is less than 10^-3, reactants dominate over products, the reaction proceeds to very small extent.

        If K _c is in the range of 10^-3 and 10³, appreciable concentrations of both reactants and products are present.

SOLIDS AND PURE LIQUIDS IN EQUILIBRIUM EXPRESSION

The concentration of a solid or pure liquid (but not a solution) is proportional to its density. Therefore, their densities are not affected by any gas pressure hence remain constant; hence their concentration never appear in the equilibrium expression.

Example

1. What is the equilibrium expression for the following reactions?
2.  i)   3Fe _(g) + 4H₂O _(g)     Fe₃O₄ + 4H_{2 (g)}

      Solution

K _c =           _OR     K _p =

 

1.     ii)  PCl_{5 (g)}   PCl_{3 (g)} + Cl _{2 (g)}

K _c =

K _p =

 

iii)  HCl _(g) + Li H_(S)    H_{2 (g)}   +LiCl_(s)

K _c =

K _p =

 

1.    iv)  Cu (OH) _2(s)  Cu²⁺ _(aq) + 2OH ^–_(aq)

K _c = [Cu ²⁺]   [OH ^–] ²

1.    v)   CaCO_3(s) CaO_(s) + CO _2(g)

K _c = [CO₂]

 

1. vi)    [Cu(NH₃)₄]²⁺_(aq)   Cu²⁺_(aq) + 4NH_3(aq)

                    

2. The equilibrium constant for the synthesis of HCl, HBr and HI are given below;

H_{2 (g)} +Cl_{2 (g)}     2HCl _(g) K_C =10¹⁷

H_{2 (g)} + Br_{2 (g)}       2HBr _(g)   K_C=10⁹

H_{2 (g)} +I _2(g)        2HI _(g)    K_C=10

 

3. a) What do the value of K_c tell you about the extent of each reaction?
4. b) Which of these reactions would you regard as complete conversion? Why?

Answers

1. a) For all the 3 reactions, the equilibrium lies on the product side (i.e. RHS) and the extent of formation of HCl is greater than HBr which in turn is greater than HI.
2. b) For the 1^st and 2^nd reaction, it can be regarded as complete conversion because products pre-dominates over reactants at equilibrium.

Characteristics of equilibrium constant

1. Equilibrium constant is applicable only when the concentrations of reactants and products have attained their equilibrium.
2. The value of equilibrium concentration is independent of the original concentration of reactants.
3. The equilibrium constant has a definite value for every reaction at a particular temperature.
4. For a reversible reaction the equilibrium constant for the forwarded reaction is the inverse of the equilibrium constant for the backward reaction.
5. The value of equilibrium constant tells the extent to which reaction proceeds in the forward or reverse direction.
6. Equilibrium constant is independent of the presence of catalyst. This is because the catalyst affects the rate of forward and backward reaction equally.

Relationship between K _c and K _p for a gaseous equilibrium

For any given reaction, K_c or K_p is a function of the reaction itself and temperature.

Consider the following gaseous equilibrium

aA _(g) + bB_(g)      cC _(g) + dD _(g)

K _c =     —————— (1)

K _p =  ————– (2)

From PV= nRT

[X] =  =

I.e. [A] =      , [B] =

Substitute these concentration in terms of partial pressures in equation (1)

K _c =

=  .

= K _p

= K _p

But;   (c+d) = total number of moles in the products.

(a +b) = total number of moles in the reactants.

(c +d) – (a + b) = The difference between moles of products and reactants

So,    (c +d)-(a + b) = Δn

 

Thus, the numerical value of K _p and K _c are equal when there is the same number of moles on products and reactants side numerically.
Example:

1. What is the relationship between K _p and K _c in the following reactions?

2H_{2 (g)} + O_{2 (g)}   2H₂ O _(g)

From; K _p = K _c (RT) ^Δn

= K _c (RT) ^{(2- (1+2))}

K _p = K_c (RT)^-1

1. ii) Derive the relationship between K _p and k _c for the particular reaction.

2H_{2 (g) +} O_{2 (g)}     2H ₂O _(g)

K _c =   …………… (1)

K _p =   ……….. (2)

From  PV = nRT

P = RT

P = [X] RT

= [H ₂] RT ,    = [ ] RT………etc

Substitute the partial pressure in equation (2)

 

K _p =  .

K _p = K _c (RT) ^(2-(1+2))

K _p = K _c (RT)^-1

 

2. a) Derive the relationship between K _p and K _c for ammonic synthesis.

 

N_{2 (g)} + 3H_{2 (g)}  2NH_{3 (g)}

K _c =  ………… (1)

K _P =  …….. (2)

From PV = nRT

P = RT

P = [X] RT

= [ ] RT, = [ ] RT,   = [ ] RT

Substitute the partial pressure in equation (2)

K _p =

K _P =  

K _p = K_C x

   K _p = K _C.(RT)^-2

 

1. b) If K_c = 0.105 mol^-2dm⁶ at 472⁰ Calculate K _p

[R=8.31 dm³ KPa mol^-1 K^-1]

Solution:

K _p = K _c x (RT)^-2

= 0.105 x (8.31x 745)^-2

= 0.105 x (6190.95)^-2

= 2.7395x 10^-9(KPa)^-2

 

DETERMINATION OF EQUILIBRIUM CONSTANT

Consider the reaction:-

CH₃CH₂OH + CH₃COOH    CH ₃COOCH₂ CH_{3 +} H₂O

 

 

Example:

1. An equilibrium system for the reaction between H₂ and I₂, to form HI at 670K in 5l flask contains 0.4 moles of H₂, 0.4 moles of  I₂ and 2.4 moles of HI. Calculate the equilibrium constant K _c.

H_{2 (g)} +I_{2 (g)}    2HI _(g)

K _c =

[HI] =    = 0.48

[H₂] = = 0.08

[I₂] =   = 0.08

K _c =

K _c = 36

 

2. A mixture of 1.0 x10^-3 moldm^-3 H ₂ and 2.0 x 10^-3 moldm^-3 I₂ are placed into a container at 450⁰C. After equilibrium was reached the HI concentration was found to be 1.87 x 10 ^. Calculate the equilibrium constant.

H_{2 (g)} + I_{2 (g)}    2HI _(g)

At   t=0              a        b                   0

Equilibrium:  a-x       b-x               2x

2x = 1.87 x 10^-3

X= 9.35 x 10^-4moldm ^-3

 

a=1.0 x 10^-3

a-x = 1.0 x 10^-3 – 9.35x 10^-4

= 6.5 x 10^-5moldm^-3

 

b = 2.0 x 10^-3

b -x=2.0 10^-3– 9.35 10^-4

=1.065 10^-3moldm^-3

 

K _c =

=

K _c = 50.51

 

3. a) It was found that if 1 mole of acetic acid and half a mole of ethanol react to equilibrium at certain temperature, 0.422 moles of ethyl acetate are produced. Show that the equilibrium constant for this reaction is about 4.

 

Start: –                  1mol               0.5mol                     0                                  0

Equilibrium:       1-0.422          0.5-0.422               0.422                                        x

0.578               0.078                   0.422                         0.422

Equilibrium

Concentration

K _c =

=

K _c = 3.95

K _c ≈ 4   shown

1. b) From the same reaction above, 3moles of acetic acid and 5 moles of ethanol reacted. Find the amounts which will be present equilibrium.

CH₃COOH _(l) + C₂H₅OH _(l)         CH₃COOCH₂CH_{3 (I)} + H₂O _(I)

t= 0                              3                        5                                        0                                0

t = t                            (3-x)             (5-x)                                       x                              x

Equilibrium

Concentration

K _c =    

4 =

4 =

4 =          

X²   = 60 – 32x + 4x²

3x² – 32x + 60 = 0

X = 8.24   or   x = 2.43

Logically;   x = 2.43

At equilibrium CH₃COOH      3 – 2.43 = 0.57 moles

C₂ H₅ OH       5 – 2.43 = 2.57 moles

 

4 .For the reaction;

CO_{2 (g)} + H_{2 (g)}      CO _(g) +   H₂ O _(g)

The value of K _c at 552⁰C is 0.137. If 5moles of CO ₂, 5moles of H_2, 1 mole of CO and 1 mole of H₂ O are initially present, what is the actual concentration of CO₂, H_2, CO and H₂O at equilibrium?

CO_{2 (g)}      +    H_{2 (g)}       CO_{2 (g)} +   H₂ O _(g)

At start;                      5                     1                       1               1

At equilibrium       5-x                   1-x                   1x              1x

K _c   =

0.137 =

0.137 =

3.425 – 1.37x + 0.137x²   = X²

3.425 – 1.37x + 0.137 x²–x² =0

3.425 -1.37x – 0.863x² =0

x =

x =

x =    =     x =1.349

 

COMBINING EQUILIBRIUM REACTIONS

1. Reversing an equilibrium reaction

Consider the reaction

PCl_{5 (g)}  PCl_{3 (g)} +Cl_{2 (g)}………………. (i)

K _{c (i)} =  _{…………………….. (i)}

Reversing the reaction;

PCl_{3 (g)} + Cl_{2 (g)}      PCl_{5 (g)}

K _{c (ii)} =    ………………. (2)

When reaction equation is reversed, the equilibrium constant is reciprocated i.e. from equation (1)   and (2)

K _c(ii) =

 

2. Multiplying an equilibrium reaction by a number

When the stoichiometric coefficient of a balanced equation is multiplied by the same factor, the equilibrium constant for the new equation is old equilibrium constant raised to the power of the multiplied factor.

PCl_{5 (g)}        PCl_{3 (g)} + Cl _(g)

K _{c (i)} =

If the equation (i) is multiplied by ½

PCl_{5 (g)}    PCl_{3 (g)} + Cl_{2 (g)}

 

 

Note; if an equilibrium reaction is multiplied by      then;

K _{c(ii )} =

K _c(ii)   =

 

Example: The K _c for the reaction below is

S _(s) +  O_{2 (g)}    SO_{3 (g);} what is the K _c for SO₃ Equilibriate with S + O₂

2 SO_{3 (g)}      2S_(s) + 30_2(g)

S _(s) +   O_{2 (g)}    SO_{3 (g})………………. (I)

2SO_{3 (g)}         2S_(s) + 30_2(g)…………………. (2)

Equation (2) has been reversed and multiplied by 2

 

When reversed

K _{c (i)} =       = 1.1×10⁶⁵

K _{c (ii)} =

K _{c (ii)} =

K _{c (ii)} = 9.09 x 10^-66

When multiplied

K _{c (ii)} =

 

3. Adding the equilibrium.

The equilibrium constant for the reaction

1. i) 2HCl_(g)   Cl_2(g) +H_2(g)

K _{c (i)} = 4.17x 10^-34   (At 25 ⁰ c)

The equilibrium constant for reaction

1. ii)  I_{2 (g) +} Cl_{2 (g)}   2ICl _(g)

K _{c (ii)} = 2.1 x 10⁵ (At 25 ⁰c)

Calculate the equilibrium constant for the reaction

iii) 2HCl _(g) + I_{2 (g)}      2ICl _(g) + H_{2 (g)}

K _{c (iii)} =?

K _{c (i)} =

K _{c (ii}) =

K _{c (iii)} =

When equation (i) + equation (ii) = equation (iii), Hence

K _{c (iii)} = K _{c (ii)} x K _{c (i)}

= (2.1 x 10⁵) x (4.17 x 10^-34)

K _{c (iii)} = 8.757 x 10^-29 (mol dm^-3)^1/2

Question 1:

The following are reactions which occurs at 3500K

26. i) 2H_{2 (g)} +O_{2 (g)}  2H₂O _(g)    K _{p (i)} = 26.4 atm^-1
27. ii) 2CO _(g) +O_{2 (g)}  2CO_{2 (g)} K _{p (ii)} = 0.376 atm^-1

Calculate  equilibrium constant for reaction

CO _{2 (g)} + H _{2 (g)}       CO _(g) + H₂O _(g)

K _{p (iii)} =?

Question 2

.Determine the Equilibrium constant for reaction;

M_2(g) + O_{2 (g)} + Br_{2 (g)}  NOBr _(g)

Given that:-

i/ 2NO _(g)   N_2(g) + O_2(g)   K _c = 2.4 x

ii/ NO _(g) +  Br_{2 (g)}  NOBr     K _c = 1.4

Answers

26. i) 2H_{2 (g)} + O_{2 (g)}   2H₂ O _(g)                                                                K _{p (i)} =26.4atm^-1
27. ii) 2CO _(g) + O_{2 (g)}    2CO_{2 (g)}                                                                K _{p (ii)} =0.376 atm^-1

iii) CO_{2 (g)} + H_{2 (g)}         CO _(g)   + H₂O _(g)                                        K _{p (iii)} =?

 

K _{p (i)} =

K _{p (ii)} =

K _{p (iii)} =

Reverse (ii), multiply (i) and (ii) by ½

1. i) (2H_{2 (g)} +O_{2 (g)}    H₂O _(g))                                    K _{p (i)} =
2. i) H_{2 (g)} + O ₂  H₂O _(g)                          K _p(i) = 5.138 atm^-1

Reverse (ii)

1. ii) 2CO_{2 (g)}  2CO _(g) +O_{2 (g)}                                   K _{p (ii)} =    

= 2.6595 atm^-1

 

Now multiply by (ii) by ½

(2CO_{2 (g)}    2CO _(g) +0_2(g))                                K _{p (ii)} =

1. ii) CO_{2 (g)} CO _(g) +   O_{2 (g)}                                             =1.63079

 

Add equation (i) + (ii), then

K _{p (iii)} = K _{p (i)} x K _{p (ii)}

= 5.138 x 1.63079

=8.379 atm^-1

 

2. i) 2NO_(g)       N_2(g)     +  O_2(g)                                                             K _c(i) = 2.4 x 10³⁰
3. ii) NO _(g) + Br _(g)      NOBr _(g)                                                           K _{c (ii)} =14

iii)  N_{2 (g)} +    O_{2 (g) +}   Br _(g)         NOBr_(g)                           K _{c (iii)} =?

Reverse ……. (i)

O_{2 (g)}   + N_{2 (g)}    2NO _(g)                                                                       K _{c (i)} =

=4.167 x 10^-31

 

Now multiply (i) by ½

O_{2 (g)}   + N_{2 (g)}   NO _(g)                                                                   K _{c (i)} = ^-31

=6.455 x 10^-16

Equation (i) + (ii) = (iii)

Therefore:

K _{c (iii)} = K _{c (i)} x K _{c (ii)}

= 6.455 x 10^-16 x 1.4

= 9.0369 x 10^-16(mol dm^-3)^1/2 

3. The equilibrium constants for the reactions which have been determined at 878K are as follows:-
4. i) COO _(s) + H_{2 (g)}   CO_(s) + H₂O _(g)              K₁ = 67
5. ii) COO_(s) + CO _(g)   CO_(s) + CO_{2 (g)}                        K₂ = 490

Using these information, calculate K’s (at the same temperature) for;

iii) CO_{2 (g)} +H_{2 (g)}    CO_{2 (g)} + H₂O _(g)            K₃ =?

And commercially important water gas reaction

1. iv) CO _(g) + H₂O_(g)    CO_2(g)+ H_2(g)                                        K_4=?

Reverse (ii)

(ii)CO_{2 (g)} + CO_(s)    CO _{(g) +} COO _(g)          K₂=2.0408 x 10^-3

Equation (i) + (ii) = (iii)

K₃ = K ₁ x K₂

= 67 x 2.0408 x 10^-3

K₃ = 0.1367

To find K₄

1. i) COO _(s) +H ₂   CO _(s) + H₂O _(g)                          K₁= 67
2. ii) COO_(s) + CO _(g) CO _(s) + CO_{2 (g)}                         K₂ = 490

iii) CO_{2 (g)} + H_{2 (g)}            CO _(g) + H₂O _(g)                                  K₃ = 0.1367

1. iv) CO _(g) + H₂O _(g)    CO_{2 (g)} + H_{2 (g)}                                       K₄ =?

When equation (iii) is reversed, it is equal to equation (iv)

K₄ = 

=

= 7.315

 

4. The heterogeneous equilibrium
5. i) Fe_(s) + H₂O     FeO_(s) + H_{2 (g)}                     
6. ii) Fe_(s) + CO_{2 (g)}  FeO_(s) + CO _(g)

Have been studied at 800 ⁰c and 1000 ⁰c.Also the rate ( ) is constant = 1.81 at 800⁰c and 2.48 at 1000 ⁰c.

1. i) Why are the ratios constant?
2. ii) Calculate equilibrium constant at two temperatures of the reaction

iii)  H₂O _(g)    + CO _(g)              H_{2 (g)    +}     CO_{2 (g)}      

Answer:

1. i) The ratios are constant because for any system in equilibrium at a given temperature, the ratio of products of concentration of products to the product of concentration of reactants raised to the point of their mole ratios is constant.
2. ii) At 800 ⁰C

K _p1 =2

K _p2 =1.81

K _p3 =?

(i)       Fe_(s) + H₂O_(l)    Fe_(s) + H_{2 (g)}                                                 K _{p (i)} =2

Reversed (ii) CO _(g) + FeO_(s)                 Fe_(s) + CO_{2 (g)}              K _{p (ii)} =?

Equation (i) + (ii) = (iii)

K_p1 x Kp₂=K_p3

K_p3 = K_p1 x K_p2

=2 x 0.55

K_p3 = 1.1

 

At 1000 ⁰c

K_p1 = 1.49

K _p2 =2.48

1.    i) Fe_(s) + H₂O _(g)       FeO_(s)   +   H_{2 (g)}                     K _p1 =1.49

Reversed     ii)     CO _(g) +   FeO_(s)         CO_{2 (g)} + Fe_(s)                  K_p2 =

=0.403

Equation (i) + (ii) = (iii)

K_p3 =k_p1 x k_p2

= 1.49 x 0.403

K_p3 = 0.6

 

1: When 1 mole of ethanoic acid is maintained at 25 ⁰c with 1 mole of ethanol, 1/3 of ethanoic acid remain when equilibrium is attained. How much would have remained if  3/4 of 1 mole of ethanol had been used instead of 1mole at the same temperature.

CH₃CH₂OH + CH₃COOH    CH₃COOCH₂CH₃ + H₂O

At start:                1mol              1mol                            0                            0

At time:                 1-x                 1-x                              x                           x

At equilibrium:     1 1- x =

 

 

K _c =

=

=

K_c = 4

CH₃CH₂OH + CH₃COOH               CH₃COOCH₂CH₃ +H₂O

At start;                   1mol                   1mol                          0                           0

At time                     1-x                       1-x                           x                          x

At equilibrium      (1 x 3/4)-x               1-x                           x                            x

 

K_c =

4 =     =

4 =X²

3- 7x + 4x² = x²

3- 7X + 4X² – x² =0

3x² + 3 – 7x = 0

3x² -7x + 3 = 0

a       b     c

x =

x =

=

=

X=            or    x =

= 1.76                    =    0.566

X cannot be 1.76

X = 0.566

– 0.566 = 0.184 moles or 23/125 moles

0.184 moles of ethanol would have remained

 

2. 2. The equilibrium constant (K _c) for the reaction; 2HI _(g)     H_{2 (g)}   +I_{2 (g)} is 0.02 of 400 ⁰ If 2 moles of H₂ and 1 mole of I₂ were mixed together in a 1.0dm³ at 400 ⁰c, how many moles of HI, I₂ and H₂ would be present at equilibrium.

2HI _(g)         H_{2 (g)}    +   I_{2 (g)}                          K _c = 0.02

Reverse the equation above

H_{2 (g)}    +   I_{2 (g)}                2HI _(g)                     K_c =  

At start                   2            1                           0                         

At time                 2-x         1-x                  2x                          =    = 50

At equilibrium

K _c =

50 =

50 =

50 (2 – 3x + x²) = 4x²

100 – 150x + 50x² = 4x²

100 – 150x +5 0x² – 4x² = 0

100 – 150x + 46x² = 0

x=

=

=

x=                   or      x=

x= 2.33                                  x = 0.934

x cannot be 2.33

 x = 0.934

At equilibrium:

Number of moles of H₂ = 2- 0.934

=1.066moles

Number of moles of I₂= 1-0.934

=0.066 moles

Number of moles of HI = 2 x 0.934

=1.868 moles

 

3. The equilibrium constant for the reaction; H_{2 (g)} +Br_{2 (g)}      2HBr _(g) at 1024K is 1.6 x 10⁵. Find the equilibrium pressure of all gases if 10 atm of HBr is introduced into a second container at 1024K.

H_{2 +} Br₂       2HBr

PREDICTION OF DIRECTION AND EXTENT OF CHEMICAL EQUILIBRIUM

At each point in the progress of a reaction, it is possible to formulate the ratio of concentration having the same form as the equilibrium

constant expression .This generalized ratio is called reaction quotient (Q).

For the reaction; then:-

Q_C =

Q _p =

 

Q_C differs from K_C in that the concentration in the expression is not necessarily the equilibrium concentration.

When the values of K_C and Q_C are compared, one can predict the direction of the chemical reaction.

        If Q _c  K _c the system is not at equilibrium, the reactants must further be converted to products to achieve equilibrium therefore net reaction proceeds from left to right.

        If Q _c = K _c the system is at equilibrium

        If Q _c > K _c the system is not an equilibrium, the products must converted to reactants to achieve equilibrium therefore a net reaction proceeds from right to left.

Example:

1.Consider the reaction

 

At 250⁰C, K _c = 4.0 x 10^-2. If the concentration of Cl₂ and PCl₃ are both 0.30 M while that of PCl5 is 3.0M, is the system at equilibrium? If not, in which direction does the reaction proceed?

^-2

Q _c =

=

Q _c = 0.03        Q_C ≠ K _c,        Q _c< K _c

The system is not at equilibrium and the reaction proceeds from left to right

2. 2. At 200k, the K _p for the formation of NO is 4 x 10^-4

N_{2 (g)} + O_{2 (g)}                      2NO _(g)

If at 200k the partial pressure of N₂ is 0.5 atm and that of 0₂ is 0.25 atm, that of NO is 4.2 x 10^-3 atm, decide whether the system is at equilibrium, if not in which direction does the reaction proceed.

K _p = 4.0 x 10^-4

Q _p =

=

Q _p = 1.4112 x 10^-4           K _p ≠Q _p,      K _p > Q _p

The system is not at equilibrium, therefore the reaction proceeds from left to right

EQUILIBRIUM CONSTANT WITH DEGREE OF DISSOCIATION ( )

-It gives to what extent the reactants are converted to products by dissociation.

-This has to be treated similar to moles.

Example1.  0.01 moles of PCl₅was placed in 1L vessel at 210K.It was found to be 52.6% dissociated    into PCl₃ and Cl₂. Calculate the K _c at that temperature.

PCl_{5 (g)}                     PCl_{3 (g)} + Cl_{2 (g)}

At start:                      0.01                                0                0

At equilibrium: 0.01-(5.26 x10^-3)            5.26 x 10^-3     5.26×10^-3

=

= 5.26 x

 

PCl₅ = 4.74 x 10^-3 moles

PCl₃ = 5.26 x 10^-3 moles

Cl₂   = 5.26 x 10^-3 moles

K _C =

 

=

K _C = 5.837 x 10^-3 moles

 

2. At 1 atm and 85^oC, N₂O₄ is 50% dissociated, Calculate the equilibrium constant in terms of pressure and calculate the degree of dissociation of the gas at 10⁰ c and 55⁰c.

N₂O₄                          2NO_{2 (g)}      

Start                          1                                        0

Equilibrium:          1                                      2

But = 50% = 0.5

1-0.5                                   2(0.5)

0.5                                      1

n_T = 0.5 + 1

= 1.5 moles

 ₌       x 1                            =    x 1

= 0.33 moles                            = 0.66moles

K _{p =}      

=

K _p = 1.32

 

N₂O₄              2NO_{2 (g)} 

Start:                 1                     0

Equilibrium:    1-                   2

n_T   = (1- ) + 2

=1+

P_T = 10atm

=       x 10                        =  x 10

K _p =

=

=   x 100 x  x

=

1.32 =

1.32 – 1.32 ² = 40 ²

1.32 = 40 ^{2 +} 1.32 ²

1.32 = 41.32 ²

² =0.0319

= 0.1787

Degree of dissociation = 17.87%

 

DEGREE OF DISSOCIATION BY DENSITY MEASUREMENT

This method is used for the determination of degree of dissociation of gases in which 1 molecule produces 2 or more molecules.

i.e. PCl_{5 (g)}    PCl_{3 (g)} + Cl_{2 (g)}

Thus at constant temperature and pressure, the volume increases. The density at constant pressure decreases.

The degree of dissociation can be calculated from the difference in density between the undissociated gas and that of partially dissociated gas at equilibrium.

If we start with 1 mole of the gas (PCl₅) and the degree of dissociation (  ), Then

PCl_{5 (g)}          PCl_{3 (g)} + Cl_{2 (g)}

Moles at equilibrium:  1-

Total moles:   1 –

= 1+

Note:

The density of an ideal gas at constant temperature and pressure is inversely proportional to the number of moles for a given weight.

Hence the ratio of density

 

=

When = Density of gas mixture of equilibrium

                                    = Density of gas before dissociation

Degree of dissociation

 

Examples1. When PCl₅ is heated, it gasifies and dissociates into PCl₃ and Cl₂.The density of the gas mixture at 200⁰C is 70.2 Find the degree of dissociation of PCl₅ at 200⁰C

Solution

Observed density, ρ₂ =70.2

ρ₁ =?

V.D =

=

=104.25

 

 

=

 

= 48.5%

 

2. At 90⁰C, the V.D of N₂O₄ is 24.8. Calculate the % dissociation into NO₂ molecules at this temperature.

N₂O₄        NO₂ + NO₂

Density at equilibrium,   ρ₂ = 24.8

= V.D =  =   = 46

?

=

=

= 85.48%

DETERMINATION OF DEGREE OF DISSOCIATION BY MOLECULAR MASS

Molecular masses are proportional at constant temperature and pressure to the density of their gases; therefore we can substitute the molecular masses for the density in the degree of dissociation.

 

          

Where:-

= Molecular mass of undissociated gas

= Average Molecular mass of gases at equilibrium

 

Example1:1.588g of N₂O₄  gives a total pressure of 1atm when partially dissociated at equilibrium in a 500cm³ glass vessel at 25⁰C.What is the degree of dissociation at this temperature?

N₂O₄          2NO₂

M₁ = (14x 2) + (16 x4)

= 92gmol^-1

M₂ =?

From PV= nRT, W here n =

M₂ =

=

= 77.7gmol^-1

 

 

=

=    0.184

 

 

FACTORS AFFECTING EQUILIBRIUM REACTION

These factors are as follows:-

1. i) Temperature
2. ii) Concentration

iii)  Pressure

        The first three affects both rates and position of chemical equilibrium (i, ii and iii)

        The other three affects the rate of chemical equilibrium

 1) Temperature

1. a) Increasing the temperature, increases the rate of reaction because usually at high temperature the collision factor increases also the numberof molecules having necessary activation energy is large.
2. b)  Effect on the position of equilibrium is explained by using Le-Chateliers principle which states that “when a system at equilibrium issubjected to a change, processes occur which tend to counteract the change” (If a system in equilibrium is disturbed (change in temperature and pressure) the system adjusts itself so as to oppose the disturbance).

 

Consider the reaction

2SO_{2 (g)} + O_{2 (g)}       2SO_{3 (g)} + Heat (negative)

 

If temperature is increased in the system, the equilibrium moves in a direction where there is a absorption of heat and if the temperature is decreased  in the system, the equilibrium moves in a direction where there is release of heat.

Effect of temperature, on the position of equilibrium can be explored by Vant Hoff`s law of mobile chemical equilibrium which states that

“For any system in equilibrium high temperature favours endothermic reactions and low temperature favours exothermic reactions.

The way in which equilibrium constant changes with temperature is found both theoretically and experimentally governed by the following

relationship;

=

∆ Hm = change in molar heat

  K = Equilibrium constant

On intergrating the equation above;

 

Where c = constant

 

If K₁ and K₂ are equilibrium constants corresponding to T₁ and T_{2 ,}the constant term can be eliminated from the equation above so as to give Vant Hoff`s equation i.e.

lnK₁ =   …………………. (1)

 

lnK₂ = …………………. (2)

 

Subtracting equation (1) from (2) i.e.

lnK₂ – lnK₁ = ) )

=    ……………………1

But 1 ln = 2.303log

=   ……………………2

 

But  =

=   ……………………3

Where 1 = 2 = 3

Example1: For the reaction;

N₂O₄       2NO₂      âˆ†H = 61.5KJ mol^-1

K_P = 0.113 at 298K

1. i) What is the value of K_P at 0⁰C?
2. ii) At what temperature will K_P =1?

 

Answers:

1. i) From =

K_P2 = 0.113

K_P1 =?

T ₁ = 273K

T ₂ = 298K

∆Hm = 61.5Kjmol^-1

R=8.314

=

= 3211.9676 (3.0729896 x 10^-4)

= 0.987

To remove ‘log’ make 10^0.987

 

 

 

 

 

1. ii) From log =

K_P2 = 0.113                 T₁ =?

K_P1 = 1                         T₂=298K

∆Hm = 61.5KJ mol^-1

R = 8.314

=

-0.9469 = 3211.9676

-2.948 x 10^-4 =

-0.0878T ₁ = 298 – T ₁

-0.0878T ₁ + T₁ =298

0.912T₁ = 298

T₁ = 326.7k

VARIATION OF EQUILIBRIUM CONSTANT WITH TEMPERATURE

According to Vant Hoff`s law of chemical equilibrium

1. i) Exothermic reaction

-Increasing to temperature decreases the K_C value as the equilibrium shifts to the left hand side to decrease the concentration of products and increase the concentration of reactants.

-Decreasing the temperature increase the K_C value

1. ii) Endothermic reaction

-Increasing the temperature increases the K_C value

-Decrease the temperature decreases the K_C value

2)  Concentration

1. a) Effects on rate of reaction

-Increase in concentration of reactants in the same amount of space/volume increases the number of molecules thus increases the chances

of collision between the molecules and hence this increases the rate of reaction.

-Decreases in concentration result to decrease in rate of reaction as it decreases the number of molecules per unit volume, hence less

collision.

1. b) Effect on the position of equilibrium

-This depends on either the concentration of products or reactants have been increased or decreased

-It also depends on Le-Chateliers principle

Example

1. Consider, 2SO_{2 (g)} + O2 _(g)    2SO_{3 (g)}

State what happens when:

1. i) [SO₂] is increased at the same temperature
2. ii) [O₂] is increased at same temperature

iii) [SO₃] is increased at same temperature

According to Le-Chateliers principle the system will adjust itself so as to cancel out the effect by shifting the equilibrium to the right side i.e.

increase the number of products hence decreasing the amount of SO₂

2. Using;
   CH₃COOH _(g) + CH₃CH₂OH _(g)      CH₃COOCH_{2 (g)} +    H₂O _(l)

What happens when;

1. i) H₂O is added
2. ii) CH₃CH₂OH is added

iii) NaOH is added

1. iv) Anhydrous copper (II) sulphate is added
2. v) CH₃COOCH₂ is added

Answers:

1. i) H₂O will react with CH₃COOH and concentration of CH₃COOH will decrease hence forward reaction, equilibrium will lie on the products side.
2. ii) Forward reaction since it will get on converted to products
3. iv) Reverse reaction since H₂O will react with CUSO₄ to form CUSO₄.5H₂O
4. The following reaction occurs in human body

Hb_(s) + O_{2 (g)}             HbO_{2 (g)}

1. i) What happens in the tissues?
2. ii) What happens in the lung?

Answers:

1. i) In the tissue the amount of O₂ is less and hence according to Le-Chateliers principle the system will adjust itself so as to increase the amount of O₂ by favouring the decomposition of HbO₂ (backward reaction). Hence the equilibrium shifts to left hand side.
2. ii) In the lung, the amount of O₂ is a lot more and hence according to Le-Chateliers principle the system will adjust itself so as to decrease the amount of O₂ by favouring the forward reaction hence, the equilibrium shifts to the right hand side.
3. Consider the following equilibrium

Orange                         yellow

What would you expect to see it:

1. i) Dilute NaOH is added to the equilibrium mixture
2. ii) Dilute HCl is added to the equilibrium mixture

Answers:

1. i) When Dilute NaOH is added to the mixture, it will react with H⁺ to form H₂O and Na⁺. This will decrease the concentration of H⁺ ionshence equilibrium will shift to left hand side. The colour will change yellow to orange.
2. ii) Dilute HCl dissociated to form H⁺ and Cl^– therefore when added to the mixture, the concentration of H ⁺ ions increases. According toLe Chateliers principle the equilibrium will adjust itself in such a way that the concentration of H⁺ ions decreases hence backward

   reaction is favoured

3)  Pressure

1. a) Effect on the rate of reaction

Increase in pressure, increases the rate of reaction. This is because increases in pressure decreases the concentration per unit volume

hence increases effective collision thus increase the rate of reaction, vice versa is also true.

1. b) Effect on position of equilibrium

Effect if number of moles of reactants and products are differ

Homogeneous gaseous equilibrium

Consider the reaction

N₂O₄     2NO₂

What will be the effect of equilibrium, when;

1. i) Pressure is increased?
2. ii) Pressure decreased?

Answers:

1. i) When the pressure is increased according to Le Chateliers principles the equilibrium will adjust itself in such a way that the backwardreaction is favoured
2. ii) According to Le Chateliers principle when pressure is decreased the equilibrium will adjust itself in such a way that the forwardreaction is favoured

N.B:

Pressure has no effect on position of equilibrium if the number of moles in reactants side is equal to number of moles in products side.

Heterogeneous equilibrium

Position of equilibrium is affected by changing the partial pressure of the gases only

E.g.

If partial pressure of CO₂ is decreased, the equilibrium shifts to the right hand side to increase the partial pressure of CO₂. Vice versa is true.

4) Catalyst

The catalyst speeds up the rate of both forward and backward reaction to the some extent since it powers the activation energy of the reaction.

Therefore the equilibrium is not altered when catalyst is added. It only changes the rates at which the reaction approaches the equilibrium.

5) Physical state and sunlight

They do not affect position of equilibrium, only on rate of reaction.

Assignment

Read and write on the important industrial applications of chemical equilibrium (specifically on harber process and contact process.)

MANUFACTURE OF AMMONIA (HARBER PROCESS)

Harber’s process for the manufacture of ammonia involve direct combination (synthesis) of nitrogen and hydrogen

 

1mol     3mol                                                    2mol

1vol      3vol                                                      2vol

This reaction is;

1. i) Reversible
2. ii) Exothermic

iii)              Proceeds with change in volume

According to Le Chateliers principle, the favourable conditions for formation of ammonia are:

1. a) Low temperature

The temperature should be kept as low as possible but at very low temperature, the rate of reaction becomes slow. It has been found that the yield of NH₃ is maximum at about 500⁰C which is the optimum temperature of the reaction.

1. b) High pressure

High pressure favours reaction which is accompanied by a decrease in volume. In actual practice, a pressure of 200-900atm is employed in this process.

1. c) Catalyst

To increase the speed of the reaction, a catalyst should be finely divided iron containing molybdenum or alumina is used as a catalyst. Molybdenum or alumina (Al₂O₃) acts as a promoter and increases the efficiency of the catalyst. A mixture of iron oxide and potassium aluminate has been found to work more effectively. A catalyst iron oxide containing Al₂O₅ and K₂O is also used in the process.

A diagram to show the manufacture of ammonia by Born Harber process

 

 

MANUFACTURE OF SULPHURIC ACID BY CONTACT PROCESS

In contact process sulphur dioxide is oxidized by air in the presence of catalyst Vanadium pentaoxide. Sulphur trioxide produced is absorbed in concentration H₂SO₄ to produce oleum (H₂S₂O₇). Oleum is then reacted with calculated amount of H₂O to form H₂SO₄ of the desired concentration.

The chemistry involved in the contact process is described as follows:

i)    Production of So₂

Sulphur dioxide (  is obtained by burning sulphur or iron pyrites

 

 

1. ii) Catalytic oxidation of SO₂ to SO₃

SO₂ is oxidized by air in the presence of a catalyst to give SO₃

                                  

The reaction is exothermic and proceeds with a decrease in volume. Therefore according to Le Chateliers principle, the favourable conditions for the maximum yield of SO₃ are:-

1. a)  Air or oxygen required for oxidation of SO₂ must be in excess
2. b)  The temperature must be low, a temperature between 350 – 450⁰C gives maximum yield of the products.
3. c)  The pressure must be high, 2atm.
4. d)  Platinised asbestos was used as a catalyst previously, but now days it is replaced by much cheaper Vanadium pentaoxide (V₂O₅).A V₂O₅ remains unaffected the impurities while platinised arbestos is poised by the impurities in the gases.
5. e)  The gases used (SO₂ and O₂) must be free of impurities, viz, dust particles, arsenious oxide e.t.c, to prevent catalyst poisoning

iii)    Conversion of So₃ to Oleum

SO₃ is dissolved in concentratedH₂SO₄ to produce oleum or fumingH₂SO₄. 

             

1. iv) Conversion of oleum to sulphuric acid

Oleum is diluted with a calculated amount of H₂O to get H₂SO₄ of desired concentration.

  

The great advantage of the contact process is that it produces a pure acid of any desired concentration and especially the fuming acid which is

of great value in chemical industry.

 

THE CONTACT PROCESS