The thermodynamics equilibrium constant Kp for the homogeneous gaseous reaction is 10^-3. The standard Gibb's free energy change ∆G° for the reaction at 27°C. (Use R= 2 cal K ^−1 mol ^−1) is :

ΔG^ o at 25^°C for the reaction 1/2N2(g) + 3/2H2(g) ⇋ 2NH3(g) is -11.392 kj mole^-1. Hence thermodynamics equilibrium constant is : he equilibrium constant for reaction is 10. (Use R=8.3 JK ^−1 mol ^−1)

Answer key : 10^2

The dissociation equilibrium of a gas AB2 can be represented as : 2AB2(g) ⇋ 2AB(g) + B2(g) The degree of dissociation is "x" and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure P is.

(A) (2KP/P)^1/2  

(B) (KP/P) 

(C) (2KP/P) 

(D) (2KP/P)^1/3

The equilibrium constant for reaction is 10. What will be the value of ΔG o (R=8.314JK −1 mol −1 & T=300K)

Relation between Kp and Kc:

When concentration are expressed in terms of mole fractions.

We know that partial pressure (P) = Mole fraction × Total pressure (Pt)

In Homogeneous gaseous equilibrium, M(g) + 3N(g)--> 4P(g). Initial concentration of M is equal to that of N. If equilibrium concentrations of M and P are equal then what will be the value of KC?

For the equilibrium, PCl 5​ (g)⇌PCl 3​ (g)+Cl 2​ (g) at 298K, K=1.8×10 −7 .Calculate ΔG 0 for the reaction (R=8.314JK −1 mol −1 ).

Equilibrium constant for the reaction given below is 2.0 × 10^-7 at 300 K. calculate standard Gibb's free energy change for the reaction PCl5 (g) --=--PCl3 (g) + Cl2 (g) also calculate standard entropy change if ∆H° = 28.40 Kj mole-1.

Gibb's free energy (∆G) and Equilibrium constant (Kc):

The part f energy which is converted into usefull work called Gibb's free energy or Gibb's function.

 

Energy (H) = Useful work (G) + Non useful (TS)

We van not calculate absolute value of 'G' so we calculate change in Gibb's free energy. 

             ∆G = ∆H -∆TS

       ∆G = ∆H - (∆TS + T∆S). ......(1)

Standard Gibb's energy (∆G°) change at  standard condition is 1 bar  and 298 K

             ∆G° = ∆H° -∆TS°  .......(2) 

Relation between ∆G° and Equilibrium constant (K):

(2) For the equilibrium, PCl 5​ (g)⇌PCl 3​ (g)+Cl 2​ (g) at 298K, K=1.8×10 −7 .Calculate ΔG 0 for the reaction (R=8.314JK −1 mol −1 ).

(2) Equilibrium constant for the reaction given below is 2.0 × 10^-7 at 300 K. calculate standard Gibb's free energy change for the reaction PCl5 (g) --=--PCl3 (g) + Cl2 (g) also calculate standard entropy change if ∆H° = 28.40 Kj mole-1.

Factor's Affecting Equilibrium Constant

(1) Le-Chetelier'e principle

(i) Effect of change in  temperature

(ii) Effect of change in  concentration 

(iii) Effect of change of volume of container at equilibrium

(iv) Effect of change in pressure

(v) Effect of Catalyst at equilibrium

(vi) Effect of Addition of Inert gas

(a) At constant volume

(b) At constant pressure 

          [End of Chapter = ∆∆∆]

Application of Equilibrium Constant (K):

(1) Predicting extent of reactions

(2) Predicting of stability of reactants and products

(3) Predicting direction of reactions at any instant of reaction (Qc)

(4) Predicting concentration of reactants and products at equilibrium.

(5) Degree of dissociation and Kp and Kc in term of DOD.

(6) Degree of dissociation and Vapour density.

Relation between Kp and Kc and Law of Mass of action: Equilibrium constants:

LAW OF MASS ACTION: Law of mass action is applicable for only reversible chemical reactions and it is an imperial law.

The law state that “ At a fixed temperature the rate of a chemical reaction is directly proportional to the product of reactive mass of reactants raised to the their respective Stoichiometric coefficients ” The law of mass action is by Guldberg and Waage.

(1) Equilibrium Constants in term of concentration ( K_C):

At the constant temperature, let us consider the following reversible reaction

According to law of mass action - The rate of forward reaction

The rate of reverse reaction-

Where K_f and K_b  is the rate constant of the forward reaction and backward reaction respectively

We know at equilibrium, the two rates of forward as well as backward are equal. ie

Rate of reaction = Rate of forward reaction – Rate of backward reaction = 0

Kc=Kf/Kb

Unit of Kc= (Conc)ng

(2) Equilibrium Constants in term of Pressure( Kp):

Consider the general chemical  reaction taking place at constant temperature.

From law of mass action- rate forward reaction is directly proportional  to product of active mass of reactants and rate backward reaction is directly proportional  to product of active mass of products.

For an ideal gas PV=nRT

Where 

P= Pressure in atm

V=Volume in liters

n=Number of gaseous moles

R=Gas constant  

  =  0.0821 L atm/mol/K or 1/12 L atm /mole/K

T=Temperature in kelvin

                                      = total number of moles of gaseous products -total number of moles of gaseous reactants

Illustrated examples:

(1) The chemical equilibrium constant Kc for the decomposition of PCl5 is 0.0625 mole L^-1 at 300^C. What will be the value of Kp is ?

(2) The value of Kc for the reaction (formation of ammonia by haber process) is 0.50 at 400º C.What will be the value of Kp at 400ºC when concentration are expressed in mole litre.1 and pressure in atmosphere ?

(3)For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) The Kp is 4.3 × 10^-4 at 375°C. Calculate Kc for the reaction. ?

(4) Methanol (CH3OH) is manufactured industrially by the reaction CO(g) + 2H2(g) ⇌ CH3OH(g) The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is the value of Kp at this temperature ?

(5) At 700 K, the equilibrium constant Kp, for the reaction, dissociation of Sulphur trioxide to Sulphur dioxide is 1.8 × 10^3 kPa What is the numerical value of Kc for this reaction at the same temperature

The chemical equilibrium constant Kc for the decomposition of PCl5 is 0.0625 mole L^-1 at 300^C. What will be the value of Kp is ?

 

The value of Kc for the reaction (formation of ammonia by haber process) is 0.50 at 400º C.What will be the value of Kp at 400ºC when concentration are expressed in mole litre.1 and pressure in atmosphere ?

SOLUTION: 

For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) The Kp is 4.3 × 10^-4 at 375°C. Calculate Kc for the reaction. ?

SOLUTION:

Methanol (CH3OH) is manufactured industrially by the reaction CO(g) + 2H2(g) ⇌ CH3OH(g) The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is the value of Kp at this temperature ?

 SOLUTION: Give Data ,Kc = 10.5 ,T= 220oC = (220 + 273)K = 493 K

At 700 K, the equilibrium constant Kp, for the reaction, dissociation of Sulphur trioxide to Sulphur dioxide is 1.8 × 10^3 kPa What is the numerical value of Kc for this reaction at the same temperature

SOLUTION:

Relation Between vapour density and degree of dissociation : Theory and Numericals:

 

(1) Ammonium carbamate when heated to 200 °C gives a mixture of NH3​ and CO2​ vapour with a density of 13. What is the degree of dissociation of ammonium carbamate?

(2) The observed molecular weight of N 2​O4​ is 80 g mol^−1 at 350 K. The percentage of dissociation of N2​O4​ (g) at 350 K is ?

(3) If PCl5 is 80% dissociated at 523 K, what is the vapour density of the equilibrium mixture at 523 K ?

(4) The vapour density of N2O4 at a certain temperature is 30. What is the percentage dissociation of N2O4 at this temperature?

(5) The vapour density of a mixture containing NO 2​ and N 2​ O 4​ is 38.3 at 27 °C. Calculate the moles of NO2​ in 100 mole mixture.

Ammonium carbamate when heated to 200 °C gives a mixture of NH3​ and CO2​ vapour with a density of 13. What is the degree of dissociation of ammonium carbamate?

The observed molecular weight of N 2​O4​ is 80 g mol^−1 at 350 K. The percentage of dissociation of N2​O4​ (g) at 350 K is ?