The equilibrium constant K_c for the above reaction can be written as,
K_c = \(\frac { \left[ PCI_{ 3 } \right] \left[ CI_{ 2 } \right] }{ \left[ PCI_{ 5 } \right] } \)
= \(\frac { (1.59)^{ 2 } }{ 1.41 } \) 
= 1.79

For the reaction the reaction quotient Q_c is given by, Q_c = [B] [C] / [A]²
as [A] = [B] = [C] = 3 x 10^-4M
Qc = (3 x 10^-4) (3 x 10^-4) / (3 x 10^-4)² = 1
as Q_c > K_c, so the reaction will proceed in the reverse direction.

(a) The reaction will shift in the forward direction.
(b) Endothermic
(c) (i) lowering the temperature
(ii) increasing pressure.
(iii) increasing concentration of oxygen.

SO₃^2- is Bronsted base because it can accept H⁺ .

Ostwald's dilution law states that the degree of dissociation of weak electrolyte is inversely proportional to square root of its concentration.
\(\alpha =\sqrt { \frac { { K }_{ a } }{ C } } ,\alpha =\sqrt { \frac { K_{ b } }{ C } } \)
Where, K_a and K_b are acid dissociations and base dissociation constants.

The equilibrium constant is not affected by a catalyst.

It is the product of concentration of [H₃O⁺] and [OH^-] at a specific temperature.
K_w =  [H₃O⁺] [OH^-]
= 1.0 x 10^-14 at 298 K

The equilibrium between ions and unionised molecules is called ionic equilibrium.

It states that the rate at which a substance reacts is directly proportional to its molar concentration.

Equilibrium is a state at which rate of forward reaction is equal to the rate of backward reaction.

This is because molar concentration of a pure solid or liquid is independent of the amount present.
Molar concentration = \(\frac { No.of\ moles }{ volume } \times \frac { Mass }{ volume } \times Density\)
Since density of pure liquid or solid is fixed and molar mass is also fixed. Therefore molar concentration are constant.

(i) Heterogeneous equilibrium
(ii) Heterogeneous equilibrium
(iii) Homogeneous equilibrium
(iv) Homogeneous equilibrium
(v) \({ K }_{ c }=\left[ { No }_{ 2 } \right] ^{ 4 }\left[ { O }_{ 2 } \right] \)
(because molar concentrations of pure solids are constant)
(vi) \({ K }_{ c }=\frac { \left[ CH_{ 3 }COOH \right] \left[ { C }_{ 2 }{ H }_{ 2 }OH \right] }{ \left[ { CH }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 } \right] \left[ { H }_{ 2 }O(l) \right] } \)
(vii) \({ K }_{ c }=\frac { 1 }{ \left[ { Fe }^{ 3+ } \right] .\left[ { OH }^{ 1 } \right] ^{ 3 } } \) because [Fe(OH)₃ (s) = 1]
(viii) \({ K }_{ c }=\frac { \left[ { IF }_{ 5 } \right] ^{ 2 } }{ \left[ { F }_{ 2 } \right] ^{ 5 } } \)(because [I₂(s) = 1])

\(p H=p K_a+\log \cdot \frac{[\text { Salt }]}{[\text { Acid }]}, \text { i.e. }, 4.91=4.76 +\log \cdot \frac{[\text { Salt }]}{[\text { Acid }]}\)
or \( \log \cdot \frac{[\text { Salt }]}{[\text { Acid }]}=0.15 \text { or } \frac{[\text { Salt }]}{[\text { Acid }]}=\text { Anti } \log 0.15  =1.41 \)
\( \frac{\text { Moles of Salt }}{\text { Moles of Acid }}=1.41, \text { i.e., } \frac{\frac{0.2}{1000} \times 50}{\frac{0.1}{1000} \times V} \)
\( =1.41 \text { or } \frac{0.01}{1000 \mathrm{~V}}=1.41 \text { or } V=100 / 1.41 \)
\( =70.92 \mathrm{~mL}\)