4 mole of an ideal gas having γ = 1.67 are mixed with 2 mole of another ideal gas having γ = 1.4. Find the equivalent value of γ for the mixture.

Answer key:  γ= 1.54

Titration of Weak acid Vs Strong base:

 (II):  Titration of Weak acid Vs Strong base:

Titration of a weak acid by strong base is a bit more complex than both acid and base are strong this is because we have consideration the equilibrium involving the weak acid and its conjugate base.

Titration Table and curve:

S.N	Acid (0.1M)	Base (0.1M )	pH	Nature	Case
1		0  ml	2.85	WA	pH=1/2(Pka –logC)
2	Volume	5 ml	4.22	ABS	pH=(Pka +log [S]/[A]
3	Taken	10 ml	4.7	ABS	pH=Pka Best Buffer
4	20.0 ml	15 ml	5.17	ABS	pH=(Pka +log [S]/[A]
5	Ka for	19 ml	5.98	ABS	pH=(Pka +log [S]/[A]
**6	CH3COOH	20 ml	8.7	SH	pH=7+1/2(Pka+logC)
7	2* 10^-10	21 ml	11.39	SB	NbVb - NaVa=NrVr
8		25 ml	12.04	SB	[OH-]= Nr
9		30 ml	12.3	SB	Strong base
10					Remains present

 

In titration of any weak acid with strong base, The pH at equivalent point will be greater than “7”

TITRATIONS OF DIPROTIC ACID WITH STRONG BASE:

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 :

Calculate the pH of 1.0×10^−3 M sodium phenolate NaOC6​H5​ . Ka​ for C6​H5​OH is 1×10^−10 .

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?

A certain Zero order reaction has rate constant K is 0.025 MS^-1 for disappearance of A . What will be the concentration of A after 15 sec if initial concentration is A is 0.5M.

(A) 2.5 M  (B) 0.5 M (C) 1.5 M  (D) 0.125 M

The rate constant for zero order reaction is 2× 10^-2 Mole L^-1 sec-1. If the concentration of reactant after 25 sec is 0.5M ,the initial concentration must have been ?

(A) 1M (B) 2M (C) 1.5M  (D) 2.5M

Consider a general reaction A--->B follow zero order kinetics.if concentration of A is reduces from 0.1 M to 0.05 M in 10 second then calculate time taken when concentration reduce to 0.025 M.

(A) 2.5 sec  (B) 2 sec (C) 5 sec   (D) 10 sec

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.