**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.**

CHEMICAL EQUILIBRIUM:-

CHEMICAL EQUILIBRIUM:-

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 from

reactants 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 and

backward 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) }+ 4H_{2}O_{ (g) } Fe_{3}O_{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.N_{2} O_{4 (g)} dissociates to give NO_{2 (g)}

K _{c }=

K _{p = }

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)}

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 10^{3}, 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 10^{3}, 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?

i) 3Fe_{ (g)} + 4H_{2}O_{ (g)} Fe_{3}O_{4} + 4H_{2 (g)}

**Solution**

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} =10^{17}

H_{2 (g) }+ Br_{2 (g) } 2HBr _{(g)} K_{C}=10^{9}

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?

b) Which of these reactions would you regard as complete conversion? Why?

**Answers**

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.

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

_{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)

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) }2H_{2 }O _{(g)}

From; K _{p} = K _{c} (RT)^{ Δn}

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

K _{p} = K_{c} (RT)^{-1}

ii) Derive the relationship between K_{ p} and k _{c} for the particular reaction.

2H_{2 (g) +} O_{2 (g) }2H _{2}O _{(g)}

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 _{p} = K _{C}.(RT)^{-2}

b) If K_{c }= 0.105 mol^{-2}dm^{6} at 472^{}C. Calculate K _{p}

[R=8.31 dm^{3} 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:-

CH_{3}CH_{2}OH + CH_{3}COOH CH _{3}COOCH_{2} CH_{3 + }H_{2}O

**Example:**

1. An equilibrium system for the reaction between H_{2} and I_{2}, to form HI at 670K in 5l flask contains 0.4 moles of H_{2}, 0.4 moles of I_{2} and 2.4 moles of HI. Calculate the equilibrium constant 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} I_{2} are placed into a container at 450^{}C. After equilibrium was reached the HI concentration was found to be 1.87 x 10^{. }Calculate the equilibrium constant.

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

a=1.0 x 10^{-3}

a-x = 1.0 x 10^{-3 }– 9.35x 10^{-4}

= 6.5 x 10^{-5}moldm^{-3}

b = 2.0 x 10^{-3}

b -x=2.010^{-3}– 9.3510^{-4}

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

Equilibriu

Concentration

K _{c} = 3.95

K _{c} ≈ 4 shown

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

X^{2} = 60 – 32x + 4x^{2}

3x^{2} – 32x + 60 = 0

X = 8.24 or x = 2.43

Logically; x = 2.43

At equilibrium CH_{3}COOH 3 – 2.43 = 0.57 moles

The value of K_{ c} at 552^{}C is 0.137. If 5moles of CO _{2}, 5moles of H_{2, }1 mole of CO and 1 mole of H_{2} O are initially present, what is the actual concentration of CO_{2}, H_{2,} CO and H_{2}O at equilibrium?

CO_{2 (g)} + H_{2 (g)} CO_{2 (g)} + H_{2} O_{ (g)}

At start; 5 1 1 1

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

3.425 – 1.37x + 0.137x^{2} = X^{2}

3.425 – 1.37x + 0.137 x^{2}–x^{2} =0

3.425 -1.37x – 0.863x^{2} =0

**COMBINING EQUILIBRIUM REACTIONS**

##### 1.** Reversing an equilibrium reaction**

Consider the reaction

##### 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)}

3. Adding the equilibrium.

The equilibrium constant for the reaction

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

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

K _{c (ii)} = 2.1 x 10^{5} (At 25 ^{}c)

Calculate the equilibrium constant for the reaction

iii) 2HCl_{ (g) }+ I_{2 (g) } 2ICl_{ (g) }+ H_{2 (g)}

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

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

= (2.1 x 10^{5}) 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

i) 2H_{2 (g)} +O_{2 (g) } 2H_{2}O_{ (g) }K _{p (i)} = 26.4 atm^{-1}

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)} + H_{2}O_{ (g)}

K _{p (iii)} =?

**Question 2**

.Determine the Equilibrium constant for reaction;

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 10^{30}

=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:-

i) COO_{ (s) }+ H_{2 (g) } CO_{(s) }+ H_{2}O _{(g)} K_{1} = 67

ii) COO_{(s) }+ CO_{ (g) }CO_{(s)} + CO_{2 (g) }K_{2} = 490

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

iii) CO_{2 (g)} +H_{2 (g) }CO_{2 (g)} + H_{2}O_{ (g) }K_{3 }=?

And commercially important water gas reaction

iv) CO_{ (g) }+ H_{2}O_{(g) } CO_{2(g)}+ H_{2(g) }K_{4=?}

Reverse (ii)

(ii)CO_{2 (g)} + CO_{(s) }CO_{ (g) +} COO_{ (g)} K_{2}=2.0408 x 10^{-3}

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

K_{3} = K _{1 }x K_{2}

= 67 x 2.0408 x 10^{-3}

K_{3 }= 0.1367

To find K_{4}

i) COO_{ (s)} +H _{2} CO_{ (s)} + H_{2}O_{ (g)} K_{1}= 67

ii) COO_{(s) }+ CO_{ (g)} CO_{ (s)} + CO_{2 (g)} K_{2} = 490

iii) CO_{2 (g)} + H_{2 (g) }CO_{ (g)} + H_{2}O_{ (g) }K_{3} = 0.1367

iv) CO_{ (g) }+ H_{2}O_{ (g)} CO_{2 (g)} + H_{2 (g) }K_{4 }=?

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

= 7.315

4. The heterogeneous equilibrium

i) Fe_{(s)} + H_{2}O FeO_{(s)} + H_{2 (g) }

ii) Fe_{(s)} + CO_{2 (g) }FeO_{(s)} + CO_{ (g)}

Have been studied at 800 ^{}c and 1000 ^{}c.Also the rate () is constant = 1.81 at 800^{}c and 2.48 at 1000 ^{}c.

i) Why are the ratios constant?

ii) Calculate equilibrium constant at two temperatures of the reaction

iii) H_{2}O_{ (g)} + CO_{ (g) }H_{2 (g) + }CO_{2 (g) }

Answer:

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.

ii) At 800 ^{}C

K _{p1} =2

K _{p2} =1.81

K _{p3} =?

(i)_{ }Fe_{(s)} + H_{2}O_{(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 Kp_{2}=K_{p3}

K_{p3} = K_{p1} x K_{p2}

=2 x 0.55

K_{p3} = 1.1

At 1000 ^{}c

K_{p1} = 1.49

K _{p2} =2.48

i) Fe_{(s) }+ H_{2}O_{ (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 ^{}c 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.

3- 7x + 4x^{2 }= x^{2}

3- 7X + 4X^{2 }– x^{2 }=0

3x^{2 }+ 3 – 7x = 0

3x^{2} -7x + 3 = 0

a b c

50 (2 – 3x + x^{2}) = 4x^{2}

100 – 150x + 50x^{2 }= 4x^{2}

100 – 150x +5 0x^{2 }– 4x^{2 }= 0

100 – 150x + 46x^{2 }= 0

x= 2.33 x = 0.934

x cannot be 2.33

x = 0.934

**At equilibrium:**

Number of moles of H_{2 }= 2- 0.934

=1.066moles

Number of moles of I_{2}= 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 10^{5}. Find the equilibrium pressure of all gases if 10 atm of HBr is introduced into a second container at 1024K.

H_{2 +} Br_{2} 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).

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 250^{}C, K _{c} = 4.0 x 10^{-2}. If the concentration of Cl_{2 }and PCl_{3} 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**. 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 N_{2 }is 0.5 atm and that of 0_{2} 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 PCl_{5}was placed in 1L vessel at 210K.It was found to be 52.6% dissociated into PCl_{3} and Cl_{2}. 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

PCl_{5 }= 4.74 x 10^{-3} moles

PCl_{3 }= 5.26 x 10^{-3} moles

Cl_{2 }= 5.26 x 10^{-3} moles

K _{C} =

=

K _{C} = 5.837 x 10^{-3} moles

2. At 1 atm and 85^{o}C, N_{2}O_{4} is 50%
dissociated, Calculate the equilibrium constant in terms of pressure and
calculate the degree of dissociation of the gas at 10^{0 }c and 55^{}c.

N_{2}O_{4 }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

**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 (PCl_{5}) 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 densi

When = Density of gas mixture of equilibrium

= Density of gas before dissociation

Degree of dissociation

**Examples**1. When PCl_{5} is heated, it gasifies and dissociates into PCl_{3} and Cl_{2}.The density of the gas mixture at 200^{}C is 70.2 Find the degree of dissociation of PCl_{5 }at 200^{}C

**Solution**

Observed density, ρ_{2} =70.2

ρ_{1 }=?

V.D =

= 48.5%

2. At 90^{}C, the V.D of N_{2}O_{4 }is 24.8. Calculate the % dissociation into NO_{2} molecules at this temperature.

N_{2}O_{4 }NO_{2 }+ NO_{2}

= 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.

**Example1:**1.588g of N_{2}O_{4} gives a total pressure of 1atm when partially dissociated at equilibrium in a 500cm^{3} glass vessel at 25^{}C.What is the degree of dissociation at this temperature?

N_{2}O_{4 } 2NO_{2}

M_{1 }= (14x 2) + (16 x4)

= 92gmol^{-1}

M_{2 }=?

From PV= nRT, W here n =

M_{2} =

**FACTORS AFFECTING EQUILIBRIUM REACTION**

These factors are as follows:-

i) Temperature

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**

a) Increasing the temperature, increases the rate of reaction because usually at high temperature the collision factor increases also the number

of molecules having necessary activation energy is large.

b) Effect on the position of equilibrium is explained by using Le-Chateliers principle which states that “when a system at equilibrium is

subjected 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;

âˆ† Hm = change in molar heat

^{ }K = Equilibrium constant

On intergrating the equation above;

Where c = constant

If K_{1} and K_{2} are equilibrium constants corresponding to T_{1} and T_{2 ,}the constant term can be eliminated from the equation above so as to give Vant Hoff`s equation i.e

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