Please Wait... While Loading Full Video#### Class-11 chapter-7 EQUILIBRIUM

### TYPES OF EQUILIBRIUM

`star` HOMOGENEOUS EQUILIBRIA.

`star` Equilibrium Constant in Gaseous Systems.

`star` HETEROGENEOUS EQUILIBRIA.

`star` Equilibrium Constant in Gaseous Systems.

`star` HETEROGENEOUS EQUILIBRIA.

`=>` In a homogeneous system, all the reactants and products are in the same phase. For example, in the gaseous reaction,

`color{red}(N_2(g) + 3H_2(g) ⇌ 2NH_3(g)),`

reactants and products are in the homogeneous phase.

`=>`Similarly, for the reactions,all the reactants and products are in homogeneous solution phase.

`color{red}(CH_3COOC_2H_5 (aq) + H_2O (l) ⇌ CH_3COOH (aq) + C_2H_5OH (aq))` and,` color{red}(Fe^(3+) (aq) + SCN^(–) (aq) ⇌ Fe(SCN)^2+ (aq))`

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`

When in an equilibrium reaction, all the reactants and the products are present in the same phase(i.e. gaseous or liquid), it is called a homogeneous equilibrium.

`color{red}(N_2(g) + 3H_2(g) ⇌ 2NH_3(g)),`

reactants and products are in the homogeneous phase.

`=>`Similarly, for the reactions,all the reactants and products are in homogeneous solution phase.

`color{red}(CH_3COOC_2H_5 (aq) + H_2O (l) ⇌ CH_3COOH (aq) + C_2H_5OH (aq))` and,` color{red}(Fe^(3+) (aq) + SCN^(–) (aq) ⇌ Fe(SCN)^2+ (aq))`

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`

When in an equilibrium reaction, all the reactants and the products are present in the same phase(i.e. gaseous or liquid), it is called a homogeneous equilibrium.

`=>` So far we have expressed equilibrium constant of the reactions in terms of molar concentration of the reactants and products, and used symbol, `color{red}(K_c)` for it.

`=>` For reactions involving gases, however, it is usually more convenient to express the equilibrium constant in terms of partial pressure.

The ideal gas equation is written as,

`color{red}(pV = nRT)`

`=> color{red}(p = n/V RT)`

Here, `color{red}(p)` is the pressure in `color{red}(Pa, n)` is the number of moles of the gas, `color{red}(V)` is the volume in `m^3` and

`color{red}(T)` is the temperature in Kelvin Therefore, `color{red}(n/V)` is concentration expressed in `color{red}(mol//m^3)`

`=>` If concentration `color{red}(c)`, is in `color{red}(mol//L)` or `color{red}(mol//dm^3)`, and `color{red}(p)` is in bar then `color{red}(p = cRT),`

`=>` We can also write `color{red}(p = [gas]RT.)` Here, `color{red}(R= 0.0831)` bar litre/mol K

`=>` At constant temperature, the pressure of the gas is proportional to its concentration i.e., `color{red}(p prop text([gas]))`

`=>` For reaction in equilibrium

`color{red}(H_2 (g) +I_2 (g) ⇌ 2HI (g))`

We can write either `color{red}(K_c = ([ HI (g) ]^2)/([H_2(g) ] [I_2(g)]))`

or `color{red}(K_c = (p_(HI))^2/{(p_(H_2)) (p_(I_2))})` .........(7.12)

Further, since `color{red}(p_(HI) = [HI(g) ] RT)`

`color{red}(p_(I_2) = [I_2(g)]RT)`

`color{red}(p_(H_2) = [H_2(g)]RT)`

Therefore, `color{red}(K_p = (p_(HI))^2/{(p_(H_2))(p_(I_2))} = ([HI(g)]^2 [RT]^2)/{[H_2(g)]RT .[I_2(g)]RT})`

`= color{red}(([HI(g)]^2)/([H_2(g)][I_2(g)]) = K_c)` .......(7.13)

In this example, `color{red}(K_p = K_c)` i.e., both equilibrium constants are equal. However, this is not always the case. For example in reaction

`color{red}(N_2(g) +3H_2(g) ⇌ 2NH_3(g))`

`color{red}(K_p = (p_(NH_3))^2/{(P_(N_2))(p_(H_2))^3})`

` = color{red}(([NH_3(g)]^2 [RT]^2)/([N_2(g)]RT [H_2(g)]^3 (RT)^3})`

` = color{red}({[NH_3(g)]^2 [RT]^(-2)}/([N_2(g)][H_2(g)]^3) = K_c (RT)^(-2))`

or `color{red}(K_p = K_c(RT)^(-2))` ..........(7.14)

`=>` Similarly, for a general reaction

`color{red}(a A + b B ⇌ c C + d D)`

`color{red}(K_p = {(p_C^c)(p_D^d)}/{(p_A^a)(p_B^b)} = {[C]^c [D]^d (RT)^(c+d)}/{[A]^a [B]^b(RT)^(a+b)})`

` = color{red}({[C]^c [D]^d}/{[A]^a [B]^b} (RT)^{(c+d)-(a+b)})`

` = color{red}({ [C]^c[D]^d}/{[A]^a[B]^b} (RT)^(Deltan) = K_c (RT)^(Deltan)) ` .............(7.15)

where `color{red}(Deltan = text{(number of moles of gaseous products) – (number of moles of gaseous reactants)})` in the balanced chemical equation.

`=>` It is necessary that while calculating the value of `color{red}(K_p)`, pressure should be expressed in bar because standard state for pressure is 1 bar.

We know that :

1pascal, `color{red}(Pa=1Nm^(–2))`, and 1bar `= color{red}(10^5 Pa)`

`color{red}(K_p)` values for a few selected reactions at different temperatures are given in Table 7.5

`=>` For reactions involving gases, however, it is usually more convenient to express the equilibrium constant in terms of partial pressure.

The ideal gas equation is written as,

`color{red}(pV = nRT)`

`=> color{red}(p = n/V RT)`

Here, `color{red}(p)` is the pressure in `color{red}(Pa, n)` is the number of moles of the gas, `color{red}(V)` is the volume in `m^3` and

`color{red}(T)` is the temperature in Kelvin Therefore, `color{red}(n/V)` is concentration expressed in `color{red}(mol//m^3)`

`=>` If concentration `color{red}(c)`, is in `color{red}(mol//L)` or `color{red}(mol//dm^3)`, and `color{red}(p)` is in bar then `color{red}(p = cRT),`

`=>` We can also write `color{red}(p = [gas]RT.)` Here, `color{red}(R= 0.0831)` bar litre/mol K

`=>` At constant temperature, the pressure of the gas is proportional to its concentration i.e., `color{red}(p prop text([gas]))`

`=>` For reaction in equilibrium

`color{red}(H_2 (g) +I_2 (g) ⇌ 2HI (g))`

We can write either `color{red}(K_c = ([ HI (g) ]^2)/([H_2(g) ] [I_2(g)]))`

or `color{red}(K_c = (p_(HI))^2/{(p_(H_2)) (p_(I_2))})` .........(7.12)

Further, since `color{red}(p_(HI) = [HI(g) ] RT)`

`color{red}(p_(I_2) = [I_2(g)]RT)`

`color{red}(p_(H_2) = [H_2(g)]RT)`

Therefore, `color{red}(K_p = (p_(HI))^2/{(p_(H_2))(p_(I_2))} = ([HI(g)]^2 [RT]^2)/{[H_2(g)]RT .[I_2(g)]RT})`

`= color{red}(([HI(g)]^2)/([H_2(g)][I_2(g)]) = K_c)` .......(7.13)

In this example, `color{red}(K_p = K_c)` i.e., both equilibrium constants are equal. However, this is not always the case. For example in reaction

`color{red}(N_2(g) +3H_2(g) ⇌ 2NH_3(g))`

`color{red}(K_p = (p_(NH_3))^2/{(P_(N_2))(p_(H_2))^3})`

` = color{red}(([NH_3(g)]^2 [RT]^2)/([N_2(g)]RT [H_2(g)]^3 (RT)^3})`

` = color{red}({[NH_3(g)]^2 [RT]^(-2)}/([N_2(g)][H_2(g)]^3) = K_c (RT)^(-2))`

or `color{red}(K_p = K_c(RT)^(-2))` ..........(7.14)

`=>` Similarly, for a general reaction

`color{red}(a A + b B ⇌ c C + d D)`

`color{red}(K_p = {(p_C^c)(p_D^d)}/{(p_A^a)(p_B^b)} = {[C]^c [D]^d (RT)^(c+d)}/{[A]^a [B]^b(RT)^(a+b)})`

` = color{red}({[C]^c [D]^d}/{[A]^a [B]^b} (RT)^{(c+d)-(a+b)})`

` = color{red}({ [C]^c[D]^d}/{[A]^a[B]^b} (RT)^(Deltan) = K_c (RT)^(Deltan)) ` .............(7.15)

where `color{red}(Deltan = text{(number of moles of gaseous products) – (number of moles of gaseous reactants)})` in the balanced chemical equation.

`=>` It is necessary that while calculating the value of `color{red}(K_p)`, pressure should be expressed in bar because standard state for pressure is 1 bar.

We know that :

1pascal, `color{red}(Pa=1Nm^(–2))`, and 1bar `= color{red}(10^5 Pa)`

`color{red}(K_p)` values for a few selected reactions at different temperatures are given in Table 7.5

Q 3019780610

`PCl_5, PCl_3` and `Cl_2` are at equilibrium at `500 K` and having concentration `1.59M` `PCl_3, 1.59M Cl_2` and `1.41 M PCl_5`.

Calculate Kc for the reaction,

`PCl_5 ⇌ PCl_3 + Cl_2`

Calculate Kc for the reaction,

`PCl_5 ⇌ PCl_3 + Cl_2`

The equilibrium constant `K_c` for the above reaction can be written as,

`K_c = ([PCl_3] [ Cl_2 ])/( [ PCl_5]) = (1.59)^2/(1.41) = 1.49`

Q 3089880717

The value of `K_c = 4.24` at `800K` for the reaction, `CO (g) + H_2O (g) ⇌ CO_2 (g) + H_2 (g)`

Calculate equilibrium concentrations of `CO_2, H_2, CO` and `H_2O` at `800 K`, if only `CO` and `H_2O` are present initially at concentrations of `0.10M` each.

Calculate equilibrium concentrations of `CO_2, H_2, CO` and `H_2O` at `800 K`, if only `CO` and `H_2O` are present initially at concentrations of `0.10M` each.

For the reaction, `tt((CO (g), +, H_2O (g), ⇌, CO_2 (g), +, H_2 (g)) , ("Initial concentration:" , , , , , ,) , ( 0.1M, , 0.1M , , 0 , , 0), ("Let x mole per litre of each of the product be formed." , , , , , , ) , ("At equilibrium: " , , , , , , ) , ((0.1-x)M , , (0.1-x)M , , xM , , xM))`

where x is the amount of `CO_2` and `H_2` at equilibrium.

Hence, equilibrium constant can be written as,

`K_c = x^2/(0.1-x)^2 = 4.24`

`x^2 = 4.24(0.01 + x^2-0.2x)`

`x^2 = 0.0424 + 4.24x^2-0.848x `

`3.24x^2 – 0.848x + 0.0424 = 0`

`a = 3.24, b = – 0.848, c = 0.0424`

(for quadratic equation `ax^2 + bx + c = 0,` )

`x = ((-b pm sqrt(b^2-4ac)))/(2a)`

`x = 0.848±√(0.848)2– 4(3.24)(0.0424)/(3.24×2)`

`x = (0.848 ± 0.4118)/ 6.48`

`x_1 = (0.848 – 0.4118)/6.48 = 0.067`

`x_2 = (0.848 + 0.4118)/6.48 = 0.194`

the value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration.

Hence the equilibrium concentrations are,

`[CO_2] = [H_2] = x = 0.067 M`

`[CO] = [H_2O] = 0.1 – 0.067 = 0.033 M`

Q 3059180914

For the equilibrium, `2NOCl(g) ⇌ 2NO(g) + Cl_2(g)` the value of the equilibrium constant, `K_c` is `3.75 × 10^(–6)` at `1069 K`. Calculate the Kp for the reaction at this temperature?

We know that,

`K_p = K_c(RT)^(Deltan)`

For the above reaction,

`Deltan = (2+1) – 2 = 1`

`K_p = 3.75 ×10^(–6) (0.0831 × 1069)`

`K_p = 0.033`

`=>` Equilibrium in a system having more than one phase is called heterogeneous equilibrium.

`=>` The equilibrium between water vapour and liquid water in a closed container is an example of heterogeneous equilibrium.

`color{red}(H_2O(l) ⇌ H_2O (g))`

In this example, there is a gas phase and a liquid phase. In the same way, equilibrium between a solid and its saturated solution,

`color{red}(Ca(OH)_2(s) +(aq) ⇌ Ca^(2+) (aq) +2OH^(-) (aq))`

is a heterogeneous equilibrium.

`=>` Heterogeneous equilibria often involve pure solids or liquids. For the heterogeneous equilibria involving a pure liquid or a pure solid, as the molar concentration of a pure solid or liquid is constant (i.e., independent of the amount present).

`=>` In other words if a substance `color{red}(‘X’)` is involved, then `color{red}([X(s)])` and `color{red}([X(l)])` are constant, whatever the amount of `color{red}(‘X’)` is taken. Contrary to this, `color{red}([X(g)])` and `color{red}([X(aq)])` will vary as the amount of `color{red}(X)` in a given volume varies. Let us take thermal dissociation of calcium carbonate which is an interesting and important example of heterogeneous chemical equilibrium.

`color{red}(CaCO_3(s) overset(Delta)⇌ CaO (s) +CO_2(g)) ` .........(7.16)

On the basis of the stoichiometric equation, we can write,

`color{red}(K_c = ([CaO(s)] [CO_2(g)])/([CaCO_3(s)]))`

Since `color{red}([CaCO_3(s)])` and `color{red}([CaO(s)])` are both constant, therefore modified equilibrium

constant for the thermal decomposition of calcium carbonate will be

`color{red}(K_c^(') = [CO_2(g) ])` ............(7.17)

or `color{red}(K_p = p_(CO_2))` ............(7.18)

This shows that at a particular temperature, there is a constant concentration or pressure of `color{red}(CO_2)` in equilibrium with `color{red}(CaO(s))` and `color{red}(CaCO_3(s))`. Experimentally it has been found that at 1100 K, the pressure of `color{red}(CO_2)` in equilibrium with `color{red}(CaCO_3(s))` and `color{red}(CaO(s))`, is `color{red}(2.0 ×10^5 Pa).` Therefore, equilibrium constant at 1100K for the above reaction is:

`color{red}(K_p = p_(CO_2) = 2xx10^5 Pa //10^5 Pa = 2.00)`

`=>` Similarly, in the equilibrium between nickel, carbon monoxide and nickel carbonyl (used in the purification of nickel),

`color{red}(Ni(s) +4CO (g) ⇌ Ni(CO)_4 (g))`. the equilibrium constant is written as

`color{red}(K_c = ([Ni (CO)_4])/([CO]^4))`

`=>` It must be remembered that for the existence of heterogeneous equilibrium pure solids or liquids must also be present (however small the amount may be) at equilibrium, but their concentrations or partial pressures do not appear in the expression of the equilibrium constant.

`=>` `color{green}("In the reaction,")`

`color{red}(Ag_2O(s) +2HNO_3(aq) ⇌ 2AgNO_3(aq) +H_2O(l))`

`color{red}(K_c = ([AgNO_3]^2)/([HNO_3]^2))`

`=>` The equilibrium between water vapour and liquid water in a closed container is an example of heterogeneous equilibrium.

`color{red}(H_2O(l) ⇌ H_2O (g))`

In this example, there is a gas phase and a liquid phase. In the same way, equilibrium between a solid and its saturated solution,

`color{red}(Ca(OH)_2(s) +(aq) ⇌ Ca^(2+) (aq) +2OH^(-) (aq))`

is a heterogeneous equilibrium.

`=>` Heterogeneous equilibria often involve pure solids or liquids. For the heterogeneous equilibria involving a pure liquid or a pure solid, as the molar concentration of a pure solid or liquid is constant (i.e., independent of the amount present).

`=>` In other words if a substance `color{red}(‘X’)` is involved, then `color{red}([X(s)])` and `color{red}([X(l)])` are constant, whatever the amount of `color{red}(‘X’)` is taken. Contrary to this, `color{red}([X(g)])` and `color{red}([X(aq)])` will vary as the amount of `color{red}(X)` in a given volume varies. Let us take thermal dissociation of calcium carbonate which is an interesting and important example of heterogeneous chemical equilibrium.

`color{red}(CaCO_3(s) overset(Delta)⇌ CaO (s) +CO_2(g)) ` .........(7.16)

On the basis of the stoichiometric equation, we can write,

`color{red}(K_c = ([CaO(s)] [CO_2(g)])/([CaCO_3(s)]))`

Since `color{red}([CaCO_3(s)])` and `color{red}([CaO(s)])` are both constant, therefore modified equilibrium

constant for the thermal decomposition of calcium carbonate will be

`color{red}(K_c^(') = [CO_2(g) ])` ............(7.17)

or `color{red}(K_p = p_(CO_2))` ............(7.18)

This shows that at a particular temperature, there is a constant concentration or pressure of `color{red}(CO_2)` in equilibrium with `color{red}(CaO(s))` and `color{red}(CaCO_3(s))`. Experimentally it has been found that at 1100 K, the pressure of `color{red}(CO_2)` in equilibrium with `color{red}(CaCO_3(s))` and `color{red}(CaO(s))`, is `color{red}(2.0 ×10^5 Pa).` Therefore, equilibrium constant at 1100K for the above reaction is:

`color{red}(K_p = p_(CO_2) = 2xx10^5 Pa //10^5 Pa = 2.00)`

`=>` Similarly, in the equilibrium between nickel, carbon monoxide and nickel carbonyl (used in the purification of nickel),

`color{red}(Ni(s) +4CO (g) ⇌ Ni(CO)_4 (g))`. the equilibrium constant is written as

`color{red}(K_c = ([Ni (CO)_4])/([CO]^4))`

`=>` It must be remembered that for the existence of heterogeneous equilibrium pure solids or liquids must also be present (however small the amount may be) at equilibrium, but their concentrations or partial pressures do not appear in the expression of the equilibrium constant.

`=>` `color{green}("In the reaction,")`

`color{red}(Ag_2O(s) +2HNO_3(aq) ⇌ 2AgNO_3(aq) +H_2O(l))`

`color{red}(K_c = ([AgNO_3]^2)/([HNO_3]^2))`

Q 3049291113

The value of `K_p` for the reaction, `CO_2 (g) + C (s) ⇌ 2CO (g)` is `3.0 `at `1000 K`. If initially `p_(CO_2)= 0.48` bar and CO

`p = 0` bar and pure graphite is present, calculate the equilibrium partial pressures of `CO` and `CO_2`.

`p = 0` bar and pure graphite is present, calculate the equilibrium partial pressures of `CO` and `CO_2`.

For the reaction,

let ‘x’ be the decrease in pressure of `CO_2`,

then `CO_2(g) +C(g) ⇌ 2CO (g)`

Initial

pressure: `0.48` bar `\ \ \ \ \ \ \ \ \ \ \ 0`

At equilibrium:

`(0.48 – x)` bar `\ \ \ \ \ \ \ \ \ \ 2x` bar

`K_p = (2x)^2/(0.48 – x) = 3`

`4x^2 = 3(0.48 – x)`

`4x^2 = 1.44 – x`

`4x^2 + 3x – 1.44 = 0`

`a = 4, b = 3, c = –1.44`

`x = ((-b pm sqrt(b^2-4ac)))/(2a)`

`= [–3 ± √(3)2– 4(4)(–1.44)]/2 × 4`

`= (–3 ± 5.66)/8`

`= (–3 + 5.66)/ 8` (as value of x cannot be negative hence we neglect that value)

`x = 2.66/8 = 0.33`

The equilibrium partial pressures are,

`p_(CO) = 2x = 2 × 0.33 = 0.66` bar

`p_(CO_2) = 0.48 – x = 0.48 – 0.33 = 0.15` bar