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Compounds that conduct electricity either in solution state or in fused state are called electrolytes.

Ex: NaCl, HCl

Compounds that do not conduct electricity either in solution state or in fused state are called non electrolytes.

K_c = \(\frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5][H_2O(l)]}\)

As [H₂O] in excess

K_c = \(\frac{CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5]}\)

“when a constraint applied to a system at equilibrium in a reversible reaction, the equilibrium shifts so as to nullify the constraint”. [Constraint is change in temperature or pressure, addition of reactant or product]

It states that a “change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of change”.

OR

It can also be stated as “if a constraint (change in temperature, pressure, concentration) is applied to a system at equilibrium, the system itself adjust to nullify the effect of the applied constraint”.

No, as the rate of evaporation ≠ Rate of condensation.

If Qc = Kc, the reaction mixture is already at equilibrium.

According to this principle, "a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change"

If Qc > Kc, the reaction will proceed in the direction of the reactants (reverse reactions)

(i) K_p = p[CO₂]

(ii) K_c = [Ag+][Br-]

(iii) K_p = p[CH₃COCH₃]

(iv) K_p = \(\frac{(pCO_2)}{(pCO_4)(pCO_2)^2}\)

(v) K_c = \(\frac{[H_3O^+][PO^{2-}_4]}{[HPO^{2-}_3]}\)

(vi) K_c = \(\frac{[Al^{3+}][H_2(g)]^\frac{3}{2}}{[H^+]^3}\)

K_p = \(\frac{(pH_2)^4}{(pH_2O)^4}\)

If Q_c > K_c , then the reaction would proceed in backward direction.

(i) At equilibrium the rate of forward reaction becomes equal to rate of backward reaction. 

(ii) A chemical equilibrium established only if none of the products are allowed to escape out. 

(iii) Chemical equilibrium can be attained from either direction. 

(iv) Catalyst does not alter the state of equilibrium. 

Chemical equilibrium is dynamic:- Because the velocities of forward and backward reactions are equal also the concentration of reactant and product unchanged at this equilibrium state, (but may not be equal). The equilibrium can be made to shift in either direction by changing the concentration. Pressure, Volume etc., Hence is dynamic in nature.

Two important characteristics of chemical equilibrium are –

(i) Chemical equilibrium is dynamic in nature.

(ii) Chemical equilibrium can be achieved from either of the directions i.e. starting the reaction either by taking the reactants or products.

“ Acid is a substance, which has a tendency to donate a proton to other substance”. “Base is a substance which has a tendency to accept a proton from other substance”.

OR

“Acid is a proton donor and base is proton acceptor”

pH = -log₁₀[H⁺] 

= -log₁₀[0.005 × 2] 

= -log₁₀[0.01] 

= -log₁₀[1 × 10^-2] pH 

= 2 – log₁₀ 

= 2 – 0 

= 2

pH = -log₁₀[H⁺] = -log₁₀[0.01] = -log₁₀[10^-2] = 2

On dilution, i.e., at a very low concentration, weak electrolyte has high degree of ionization.

• Chemical equilibrium is dynamic. 

• Equilibrium reaction will not come to an end. 

• Equilibrium is attained only in a closed system. 

• Catalyst does not alter equilibrium state.

 NH₃(g) = Lewis base, CH₃COOH(aq)= Arrhenius acid , CO₂(g), BF₃ ,Ag+ = Lewis acid and HCO₃ ^-1 is a Bronsted acid.

Extent up to which an acid/base/salt ionize to form ions.

K_p = K_c ( As change in number of moles of reactant and product is Zero).

The correct answer is (iv) 2

The balanced chemical equation corresponding to this expression is

4NO(g) + 6H₂(g) ⇌ 4NH₃(g) + 5O₂(g)

Let the concentration of H₂(g) and I₂(g) at equilibrium be x mol L^-1.

Then for the reaction

H₂(g) + I₂(g) ⇌ 2HI(g)

x           x           x

Equilibrium conc. (mol/L)

K = \(\frac{(0.5\, mol/L)^2}{(x\, mol/L)\times (x\, mol/L)}\) = \(\frac{0.25}{x^2}\)

But K = 54.8

Therefore, \(\frac{0.25}{x^2}\) = 54.8

or, x = \(\sqrt{(0.25/54.8)}\) = 0.068

thus, at equilibrium [H₂(g)] = 0.068 mol/L

and [I₂(g)] = 0.068 mol/L