Chemistry TYPES OF EQUILIBRIUM

### Topic to be covered

star HOMOGENEOUS EQUILIBRIA.
star Equilibrium Constant in Gaseous Systems.
star HETEROGENEOUS EQUILIBRIA.

### HOMOGENEOUS 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.

### Equilibrium Constant in Gaseous Systems

=> 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

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

Solution:

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.

Solution:

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?

Solution:

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

### HETEROGENEOUS EQUILIBRIA

=> 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))

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.

Solution:

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