Equilibrium

In the equation K_p = K_c (RT) ^Δn,∆n equals to the difference b/w the number of moles of gaseous products and gaseous reactants where, K_p and K_c are the Equilibrium constants at constant pressure and temperature respectively.

For the reaction NH₄ Cl (s) ⇋ NH₃ (g) + HCl (g)

The no. of moles of reactant is 1 ( 1 molecule of ammonium chloride involved) and the no. of moles of reactant is 2 ( ammonia and HCl )

Then ,∆n = (2 – 1 ) = 1.

• In the viewpoint of thermodynamics, Equilibrium can be explained by the following mathematical equation

∆G =∆G^⊖+ RTlnQ which comes down to ∆G =∆G^⊖+ RTlnK and then to

∆G^⊖= - RTlnK (as at Equilibrium ∆G is zero and K_c is = Q )

And k = e^-∆G⊖/RT (by taking antilog on both sides)

therefore, when ∆G^⊖ > 0, which means positive, it will give a negative K value

( e^G⊖/RT attains a negative value for a positive ∆G^⊖ ) and hence K < 1.

The equilibria involving physical processes can only be achieved when the change occurring is only a type of physical change and not a chemical change

• When gases are being involved in the process then (i) Equilibrium is possible only in a closed system at a given temperature satisfies the equilibria of physical processes because if the system (the vessel) is open then some gases will be escaping and there will be no Equilibrium.

• Again, in the process (ii) All measurable properties of the system remain constant satisfies the property of physical processes too because reaching the Equilibrium at constant temperature and at a closed system the observable properties will always remain constant.

• (iv) The opposing processes occur at the same rate and there is dynamic but stable condition also agrees with the conditions of Equilibrium of physical processes because the rate of the two oppositely directed processes i.e. the forward and backward process are occurring at same rate (reversibility) and therefore the nature of the Equilibrium will be dynamic but in a stable condition.

Reaching the Equilibrium, the rate of the forward and backward process i.e. two oppositely directed processes becomes equal, but that does not necessarily mean that the processes are stopped at either direction, instead there exists a condition which is dynamic in nature but is quite stable.

• For the reaction PCl₅ (g) ⇋ PCl₃ (g) + Cl₂

The reaction Equilibrium constant K_c can be written as :

Hence, =

K_c = 1.8 × 10^-3

• In a mixture of solutions containing iron (III) nitrate and potassium thiocyanate both react to form iron (III) thiocyanate solution which red in colour :

Fe³⁺+SCN^–⇌FeSCN²⁺(Red)

• Now, when oxalic acid is added to the solution, it reacts with the remaining Fe³⁺ ions and forms a stable complex ion [Fe(C₂O₄)₃]^3- thus concentration of free Fe³⁺ decreases, therefore by decreasing the red colour of the given solution.

Fe³⁺ +3 C₂O₄^2-⇌[Fe(C₂O₄)₃]^3- .

• [Co (H₂O)₆]³⁺ (aq) + 4Cl– (aq) ⇋ [CoCl₄] ^2- (aq) + 6H₂O (l)

(pink) (blue)

Upon cooling of the product formed [CoCl₄] ^2- (blue) we are getting the reactant back again [Co (H₂O)₆]³⁺ (pink) which means shift in Equilibrium to the backward direction hence the backward process of the reaction must be exothermic (as heat is evolved when [CoCl₄] ^2- becomes the reactant the opposite case). Therefore the forward reaction is endothermic ∆HH value will be positive and (i) ∆H > 0 for the reaction is the correct answer.

• The formation of hydrogen ions (hydroxonium ions) and hydroxide ions from water is an endothermic process. Using the simpler version of the Equilibrium :

H₂O (l)⇋H⁺ (aq.) + OH^- (aq.)

Then the ionic product of water K_w = [H⁺] [OH^-]

Again, pK_w = pH + pOH (by taking negative log ; -logx=px)

o Hence at 25°C [H⁺] =[OH^-] = 10⁷ (as pH=-log[H+] = 7 )

K_w = [H+] [OH-] = 10^-14

When temperature increases K_w increases at 60°C K_w >10^-14

Therefore, [H⁺] [OH^-] >10^-14

Or [H⁺] > 10^-7

Or pH <7.

o For a weak acid HX (partial ionization) the Equilibrium can be expressed as :

HX (aq.) + H2O (l) ⇋H₃O⁺ (aq.) + X ^– (aq.)

The concentration of this H3O+ can be termed down as [H3O⁺] = √(Ka. C),

where Ka = ionization constant of the acid and C = the initial molar concentration of the acid.

Hence, for the same concentration [H₃O⁺] Ka

But pH=-log[H₃O⁺]

So, larger the value of Ka, larger will be the value of [H₃O⁺]and obviously lower the value of pH.

From the given The K_a values of acetic acid, hypochlorous acid and formic acid are 1.74 × 10^-5, 3.0 × 10^-8 and 1.8 × 10^-4 respectively the order of increasing K_a is :

acetic acid <formic acid <hypochlorous acid

hence the order of pH of 0.1 mol dm^–3 solutions of these acids is

(iv) formic acid > acetic acid > hypochlorous acid as in accordance with the K_a values (larger K_a means lower pH).

For the reaction : H₂ S ⇋ H⁺ + HS^-

the ionization constant -

(i)

^o For the reaction : HS ⇋ H⁺ + S^2-

the ionization constant -

(iii)

From this two above reaction, we can get the following reaction (addition)

H₂S ⇋ 2H⁺ + S^2-

Then Ka₁ and Ka₂ will also be added and Ka₃ will be the result :

From equation (i) and (ii) x (here addition means the multiplication of the Equilibrium constants)

Lewis concept provides us with the concept of acid as the acceptor of electrons.

And this criterion for acid to be an acceptor it has to have 2 properties i. It should have a positive charge or ii. It has to have some electron deficiency to accept new electrons.

For the molecule of BF₃, it is short of an octet of electrons the central atom B has 6 electrons and is electron deficient hence, acts as a good lewis acid to fulfil its valence shell octet in spite of having no protons.

Therefore, it readily accepts lone pair of electrons from NH₃ and reacts with it.

o HCl is a strong acid, whereas NH₄OH is a weak base, their reaction :

HCl + NH₄OH = NH₄Cl + H₂O

Hence all of 0.05 mol dm^-3 of HCl will be neutralized by NH₄OH and after completion of neutralization there will be some amount of NH₄OH left (concentration 0.1 mol dm^-3), so that the reaction mixture results in NH₄OH + NH₄Cl which will form a buffer solution.

Silver chloride (AgCl) reacts to ammonia solvent and forms a soluble complex as NH₃ binds to the Ag⁺ ions to form this complex. Hence AgCl gets dissolved.

Ionization of CH₃COOH occurs as :

CH₃COOH + H₂O⇋H₃O⁺ + CH₃COO^-

Initial concentration: 0.01 0 0

Equilibrium concentration: 0.01 – x x x

ionization constant =

since , x>>0.01 hence, 0.01 – x ~ 0.01 .

x²/0.01 = 1.74 × 10^-5 (given)

X² = 1.74 × 10^-5 x 0.01

X = 4.2 x 10^-4

Or, pH = -log (4.2 x 10^-4 ) = 3.4

Ammonium acetate is a salt which is formed by weak acid CH₃COOH and weak base

NH₄OH

For salts of weak acid and weak base :

pH = 1/2 [pK_w + pK_a – pK_b ]

= 1/2 [14 – log (1.8 × 10^-5) + log (1.8 × 10^-5 )]

= 7

• In the viewpoint of thermodynamics, Equilibrium can be explained by the following mathematical equation :

∆G =∆G^⊖+ RTlnQ which comes down to

∆G = ∆G^⊖ + RTlnK and then to

∆G^⊖= - RTlnK (as at Equilibrium ∆G is zero and Kc is = Q

Now, for the reaction A ⇋ B, at the stage of half completion of the reaction, the concentration of [A] will be equal to [B] i.e. [A] = [B]

Hence, the value of K becomes equal to = =1

The value of ln1 = 0 ,

Hence ∆G^⊖ = - RTlnK = 0

therefore (i) ∆G^⊖ = 0 is the correct answer.

• According to Le Chatelier’s principle: “When the concentration of any of the participants in a reaction (reactant or products) in a reaction at Equilibrium is changed keeping the temperature constant the composition of the Equilibrium mixture changes so as to adjust (minimize) the effect of concentration changes”

For the reaction, N₂ (g) + 3H₂ (g) ⇋ 2NH₃ (g) the increase in total pressure at constant temperature, the Equilibrium concentration will be changed but this is not applicable for the Equilibrium constant K_c as it will not change with pressure or concentration as it depends on temperature and temperature is remaining constant.

• Among the given liquids ether has a minimum boiling point which means ether is the most volatile one amongst the given liquids.

• And as the vapour pressure is the pressure exerted by the molecules of a liquid at liquid-vapour Equilibrium, the more volatility of a liquid indicates that the gaseous molecules will outnumber the liquid molecules which is a clear indication in more vapour pressure . hence amongst the given liquids ether will exert the greatest vapour pressure and water will have the lowest vapour pressure and the increasing order will be in accordance with fact :

(ii) Water < acetone < ether

• For the reaction :

1/2 H₂(g) + 1/2 I₂ (g) ⇋ HI (g) K_c = [HI] = 5 (given)

For the reaction

H₂(g) + I₂ (g) ⇋2HI (g) , ) K_c = [HI]² = (5)² = 25

Hence, for the reaction 2HI (g) ⇋ H₂(g) + I₂ (g) , the value of K_c will be = = 1/25 = 0.04.

• The disturbance or the change in an Equilibrium occurs only when 1.) the partial pressures or 2.) the molar concentrations of the participants changes during the reaction.

• Adding inert gas while keeping the volume constant does not affect the Equilibrium because it cannot change either of the above-mentioned no.1.) or no.2.) major factors as it is not taking part in the reaction.

Hence for each cases of (i) H₂ (g) + I₂ (g) ⇋ 2HI (g)

(ii) PCI₅ (g) ⇋ PCI₃ (g) + CI₂ (g)

(iii) N₂ (g) +3H₂ (g) ⇋ 2NH₃ (g) the Equilibrium will remain unaffected on addition of Argon which is an inert gas.

Hence (iv) is the correct answer.

• For (i)

Generally, the Equilibrium constant for a certain reaction changes with a change in temperature and this change occurs over a dependency on the factor (actually the sign) of ∆H (heat of the reaction or the enthalpy in thermodynamics) which can be expressed by the following relation :

eq:1

where K₁ and K ₂ are the Equilibrium constant at temperatures T₁ and T₂ respectively

For the given reaction N₂O₄ (g)⇋2NO₂ (g), it is informed that the value of K is 50 at 400 K and 1700 at 500 K; i.e. the value of the Equilibrium constant is increasing with an increase in temperature which only possible if the ∆H is positive (eq:1 ).

• For (iii) when NO₂ (g) and N₂ O₄ (g) are mixed at 400 K at partial pressures 20 bar and 2 bar respectively:

At 400 k the reaction quotient, = (20)²/2 = 200. Thus for the given value of k ( 50 at 400 k) Q>k.

Hence Equilibrium will obviously shift backwards to form more of N₂O_4.

• For (iv),

The increase in entropy is shown by the increase in the number of moles of the participant substances.

The option (ii) The reaction is exothermic is incorrect because it is impossible ( ∆H value is positive).

• There is evidence which shows that ice and water which are the solid and liquid phases of the same substance can establish a dynamic Equilibrium between them only when a particular pressure and temperature arrives.

• Therefore, for any pure substance at atmospheric pressure, the temperature at which the solid and liquid phases are at Equilibrium can be called the normal melting point or freezing point of the substance as the two temperatures are the equivalent.

Given equation:-

According to the bronsted-lowry theory, Conjugate acid-base pair is the species which differs by only one proton i.e. when you remove a proton from a parent acid it will become a conjugate base of that acid and that acid will be called a conjugate acid of that base. The two conjugate acid-base pairs in the ionisation of HCl are (HCl-Cl^-) in which HCl is the conjugate acid and Cl^- is the conjugate base similarly the second pair is (H₂O-H₃O⁺) in which H₂O is the conjugate base and H₃O⁺ is the conjugate acid.

The aqueous solution of sugar does not conduct electricity because it does not dissociate into free ions in the solution and remains as molecules in water, not as ions, therefore, it is called a non-electrolyte, while in the case of NaCl it ionizes into sodium ion (Na⁺) and chloride ion (Cl^-) in water which conducts electricity. NaCl has a dissociation rate of 100 i.e. it ionizes completely in water. Conductance depends on the no. of ions present in the solution. More will the no. of ions of NaCl in water more will be the conductivity.

BF₃ is an electron-deficient species as the octet of boron is incomplete (1s² 2s² 2p¹). According to Lewis concept e- deficient species are called lewis acid so BF₃ will act as a lewis acid while NH₃ (N=1s² 2s² 2p³) has a lone pair so it will act as a lewis base and it will donate its lone pair to the empty p-orbital of Boron through a coordinate bond to form an adduct.

pK_b values for Di-methylamine, ammonia, pyridine and urea are respectively 3.29, 4.752, 8.752, and 13.8861.

The value of K_b (base ionisation constant) defines the strength of a base. Weak bases with relatively higher values are strong that bases with lower k_b values.

The decreasing order of bases on the basis of the ionisation constant at Equilibrium will be;-

Di-methylamine > Ammonia > Pyridine > Urea

The strongest base will be Di-methyl amine as its pK_b value is 3.29 and we know that the less the pK_b value strong is the base.

The relative strengths of acids are determined by the extent to which they dissociate in the aqueous solutions. In solutions where the concentration is same, stronger acids tend to ionize to a greater extent; as a result, they yield higher concentrations of hydronium ions than the weaker acids.

Conjugate acids of the following base will be:

OH^- - H₂O CH₃COO^- - CH₃COOH Cl^- - HCl RO^- - ROH

The decreasing order of acidic strength will be;

Cl^- > CH₃COO^- > OH^- >RO^-,

Conjugate base of a strong acid is weak therefore the decreasing order of basic strength will be;

RO^- > OH^- > CH₃COO^- > Cl^-

pH-: It is defined as the negative logarithm to base 10 of the hydrogen ion of that solution.

i. KNO₃ –it is a salt of strong acid (HNO₃)-strong base (KOH), it does not dissociate into H⁺ or OH^-when dissolved into the water so it remains neutral. We know that neutral compounds have a pH of 7.

ii. CH₃COONa –it is a salt of a weak acid (CH₃COOH) and strong base (NaOH) so its aqueous solution will be basic i.e. pH > 7.

iii. NH₄Cl- it is a salt of a strong acid (HCl) and a weak base (NH₄OH). So the aqueous solution will be acidic i.e. pH<7

iv.C₆H₅COONH₄ –it is a salt of a weak acid (benzoic acid) and weak base (NH₄OH) so its aqueous solution will be having pH equal to 7 but slightly less than 7.

The increasing order of pH will be;

CH₃COONa< KNO₃< C₆H₅COONH₄<NH₄Cl

Given equation:

Given, K_C =1 × 10^-4

By law of mass action the Equilibrium constant for the equation will be;

We know that the reaction quotient, Q_c, expresses the relative ratio of products to reactants at a given instant.

[Q_{c =} Reaction quotient]

Here; Q_C >K_C Reaction will proceed in reverse direction.

From the given concentration we can conclude that the solution is very dilute and we know when HCl reacts with water it forms hydronium ion hence, the resulting large concentration of (H+) leads to a decrease in the pH. So, here we cannot neglect the concentration of H₃O⁺ ions produced in the solution. We have to take its concentration into account.

Now total pH will be; [H₃O⁺] = 10^-8 + 10^-7 M 7

Hence the solution will be acidic.

Given pH=5.0, we know that;

As we dilute the solution, the concentration of the solution will reduce by how much times we are diluting it.

So as we are diluting the solution 100 times, the concentration of the acidic solution will decrease by 100 times i.e.

pH = -log[H⁺]

[H⁺]=10^-5 mol/L

Diluting it 100 times

So, the new concentration will be = 10^-7 mol/L

And, pH Value will be = pH = -log [H⁺]

=-log [10^-7] = 7

Hence the pH after diluting solution hundred times will be 7.

Given, solubility of BaSO₄ in water= 8 ×10^-4 g/L

The equation of disassociation of BaSO₄ will be-

Now, we know that,

Now in the presence of 0.01 H₂SO₄ solⁿ

Now,

(S^’ is the solubility of Ba²⁺ in 0.01 HCl)

S <<< 0.01, so it can be neglected

Now, K_sp= (S^’) (0.01)

S^’= 6.4×10^-5

Hence solubility of BaSO₄ in 0.01 mol dm^-3 of H₂SO₄ is 6.4×10^-5

Given, pH =2.85 and C =0.08 mol dm^-3

Now since HOCl is a weak acid its dissociation will be given as-

We know that;

pH = -log [H⁺]

-2.85 = log [H⁺]

[H⁺] = antilog (-2.85)

[H⁺] = 1.41 × 10^-3

We also know that, for a weak mono basic acid-

(On squaring both sides)

K_a = 2.5 × 10^-5

Hence the ionization constant of HOCl will be 2.5 × 10^-5

Given, pH of solution A = 6

[H⁺] of solution A = 10^-6 mol/lit.

pH of solution B = 4

[H⁺] of solution B = 10^-4 mol/lit.

On mixing 1L of each solution we will get total 2L of Solution.

Amount of [H⁺] in 1L: solution A = 10^-6× 1L = 10^-6

: Solution B = 10^-4 × 1L = 10^-4

Total [H⁺] in Solution = 10^-6 + 10^-4

=10^-4 (1+0.01/2)

=10^-4× 1.01/2

=5×10^-5 mol/L

pH = -log[H⁺]= -log[5×10^-5]

=-log(5) + (-5log10)

=-log5 + 5 =4.3

The pH of the Solution formed by mixing will be 4.3

Given, K_sp =2.7×10^-11

The equation of disassociation of Al(OH)₃will be-

We know that,

K_sp= [Al³⁺] [OH^-] ^3-

=(s) × (3s) ³

=27s⁴

s⁴= 10^-12

s= (10^-12)^1/4 =10^-3 mol/L

Now, molar mass of Al(OH)₃ =78

Solubility= molar mass ×s

= 78 ×10^-3

= 7.8 × 10^-2 g/L

NOW, we know that-

pH = 14 - pOH

[OH]= 3s = 3 × 10^-3

pOH= 3-log3

pH = 14 – 3 + log3

= 11.4771

Hence the pH of the solution will be 11.4771 and solubility in g/L will be 7.8×10^-2 g/L.

Given, K_sp of PbCl₂ =3.2 ×10^-8

The equation of disassociation of PbCl₂ will be-

K_sp = [Pb²⁺] [Cl^-] ²

= (x) × (2x) ² = 4x³

4x³ =3.2 × 10^-8

x=2 ×10^-3 mol/L

Solubility = molar mass (PbCl₂) × 2 × 10^-3

=556 × 10^-3 =0.556 g/L

The required volume to get a saturated solution of PbCl₂ is 0.1798 L

Given; NH₃ + BF₃→ H₃N: BF₃

Lewis acid in this reaction is NH₃ (N=1s²2s²2p¹)as it has a lone pair of e^- to donate in its p-orbital and Lewis base Is BF₃ as p-orbital of Boron is empty(B=1s²2s²2p¹) so it will accept lone pair from N and form a dative bond. This is explained by lewis electronic theory. The Hybridisation of N is sp³ and B is sp².

Given; CaCO₃ (s) → CaO₃ (s) + CO₂ (g)

Δ_fH^⊖ [CaO (s)] = -635.1 kJ mol^-1

Δ_fH^⊖ [CO₂ (g)] = -393.5 kJ mol^-1

Δ_fH^⊖ [CaCO₃ (s)] = -1206.9 kJ mol^-1

We know that,

Where is the standard enthalpy change for the formation of one mole of compound from its elements in their most stable states of aggregation.

= Δ_fH^⊖ [CaO(s)] + Δ_fH^⊖ [CO₂(g)]- Δ_fH^⊖ [CaCO₃(s)]

= -635.1 -393.5+1206.9

=178.3 kJ mol^-1

=positive

i.e. the reaction is exothermic.

Hence according to Le Chatelier’s Principle on increasing the temperature the reaction will shift to forward direction.

(i) Liquid ⇋ Vapour – (b) Boiling point

Explanation: when the vapour pressure of a liquid equals the atmospheric pressure surrounding the liquid and the liquid changes into a vapour that temperature is called the boiling point of that substance. This keeps happening until an Equilibrium is attained i.e. the rate of evaporation=rate of condensation

(ii) Solid ⇋ liquid - (d) Melting point

Explanation: For any pure substance at atmospheric pressure, the temperature at which the solid and liquid phases are at Equilibrium is called the melting point of that substance. Here, the systems are in dynamic Equilibrium i.e. there is no change in product or reactant.

(iii) Solid ⇋ Vapour - (c) Sublimation point

Explanation: when a solid start converting into vapour phase without converting into a liquid that process is called the sublimation. This process happens at temperature and pressure below triple point (the point where all the phases co-exist in Equilibrium).

(iv) Solute (s) ⇋ Solute (solution) - (a) saturated solution

Explanation: A solution that contains the maximum concentration of a solute dissolved at a given temperature in the solvent is called a saturated solution. A dynamic Equilibrium exists between the solute molecules in solid-state and the solution.

Given, N₂ (g) + 3H₂ (g) ⇋ 2NH₃ (g)

(i) 2N₂ (g) + 6H₂ (g) ⇋ 4NH₃ (g) - (d) K²_c

Explanation: The Equilibrium constant depends on the stoichiometric coefficient of reactants and products. Here the no of moles is getting double therefore if the initial value of Equilibrium constant was K_c so it will increase two times i.e. K²_c.

(ii) 2NH₃ (g) ⇋ N₂ (g) + 3H₂ (g) - 1/K_c

Explanation: In this equation, the reaction is in reverse order therefore if the initial value of Equilibrium constant was K_c then now it will also get inversed for the given reaction and will become 1/K_c because Equilibrium constant for the reverse reaction is the inverse of the Equilibrium constant for the reaction in the forward direction.

(iii) N₂ (g) + H₂ (g) ⇋ NH₃ (g) - (b) K^1/2_c

Explanation: Here the stoichiometric coefficient are changed by multiplying throughout by 1/2 so if the initial value of Equilibrium constant was K_c then now it will get multiplied by 1/2 throughout for the given reaction and the value of K^’_c will become Kⁿ_c here n=1/2 so K^’_c= K^1/2_c.

(i) ∆G⁰ > 0 - (d) K < 1

We know that,

∆G = ∆G° + RTln K… eq (1)

• ∆G = Gibbs free energy

• ∆G° = Standard Gibbs free energy (1atm pressure and 298k)

• T = temperature (in kelvin)

• R = Gas constant

• K = Equilibrium constant

• At Equilibrium, ∆G = 0

• ∆G° + RTln K = 0

So, RTln K = -∆G°...eq (2)

Fromthe eq (2) we get smaller the magnitude of -∆G°,the higher the rate constant K will be and the faster the reaction.

Explanation: when the value of ∆G° is positive (+) the value of K will become negative hence, reaction will proceed in reverse direction i.e. non spontaneous reaction.

(ii) ∆G⁰ < 0 (a) K > 1

Explanation: when the value of ∆G° is negative (-) the value of K will become positive hence, reaction will proceed in forward direction i.e. spontaneous reaction.

(iii) ∆G⁰ = 0 (b) K = 1

Explanation: we know that, ∆G = ∆G° + RT ln K

At Equilibrium, ∆G = 0

From eq (1), we can write:

∆G° + RTln K = 0

So, RTln K = -∆G°

(∵ Log 1 = 0, so we get, RTln K = 0)

∴ k = 1

Hence value of K will become 1.

According to the bronsted-lowry theory, Conjugate acid-base pair is the species which differs by only one proton i.e. when you remove a proton from a parent acid it will become a conjugate base of that acid and that acid will be called a conjugate acid of that base.

(i) NH₃ - (b) NH⁺⁴

The conjugate acid of a base is formed by the protonation (reception of an H⁺) of the acid. When an ammonium molecule accepts a proton (H⁺)), the ammonium cation (NH₄⁺) is formed, as the nitrogen atom forms a coordinate bond with the hydrogen.

Explanation: The conjugate acid of ammonia is ammonium ion.

(ii) HCO₃^-- (e) H₂CO₃

Explanation: The bicarbonate ion(HCO₃^-) is a negatively charged base HCO3-, when we add a proton (H⁺)i.e. positively charged species on reacting with water we get the neutral conjugate acid, H₂CO_3.

(iii) H₂O - (c) H₃O⁺

Explanation: when water reacts with acid it acts as a base. When the proton (H⁺) will associate with water molecule it will form a hydronium ion ().

(iv) HSO₄^- - (d) H₂SO₄

Explanation: The conjugate acid of hydrogen sulphate ion is H₂SO_4.when hydrogen sulphate ion gains a proton i.e. H⁺ it gets converted into sulphuric acid (H₂SO₄).

• Le chatlier principle: If a system at Equilibrium experiences a change in any factor that determines the Equilibrium then the Equilibrium tends to shifts in a manner as to counteract the imposed change in the Equilibrium

Explanation: According to Le chatlier principle as the time passes accumulation of products occur and depletion of reactant i.e. reaction will proceed in the forward direction.

Explanation: According to Le chatlier principle as time passes accumulation of reactant occurs and depletion of product i.e. the reaction will proceed in the backward direction.

Explanation: According to Le chatlier principle as time passes equal attainment of reactant and product occurs and chemical Equilibrium is obtained. It can be obtained from either direction.

We know that,

∆G = ∆G° + RTln K… eq (1)

• ∆G = Gibbs free energy

• ∆G° = Standard Gibbs free energy (1atm pressure and 298k)

• T = temperature (in kelvin)

• R = Gas constant

• K = Equilibrium constant

• At Equilibrium, ∆G = 0

∆G° + RTln K = 0

So, RTln K = -∆G°...eq (2)

Fromthe eq (2) we get smaller the magnitude of -∆G°,the higher the rate constant K will be and the faster the reaction.

(i) Equilibrium - (c) ∆G = 0

Explanation: When ∆G is equal to zero and K =1, it means Solution that at Equilibrium, the products and the reactants are equally favoured.

(ii) Spontaneous reaction - (d) ∆G < 0, K >1

Explanation: when ΔG<0, then K >1 i.e. reaction will move in the forward direction and can be used to do some useful work.

(iii) Non spontaneous reaction - (a) ∆G > 0, K < 1

Explanation: when ΔG>0, then K <1 i.e. reaction will move in the backward direction and we need external energy if we want to do some useful work.

HA bond strength is a more important factor in determining acidity of an acid than the polar nature of the bond. As the size of halide increases, bond becomes weaker and the acidity increases.

Since order of size of given halides will be F < Cl < Br < I

As the size of halide will increase, the bond between H and halide will weaken and acidity will increase.

Therefore, Increasing order of acidity of hydrogen halides is HF < HCl < HBr < HI.

A buffer is a solution containing either a weak acid and its salt or a weak base and its salt, which is resistant to pH change.

A solution containing a mixture of acetic acid (weak acid) and sodium acetate (salt of weak acid) acts as a buffer solution around pH 4.75 and buffer solution always resist the change in pH on addition of small amount of acid or alkali that’s why a mixture of acetic acid and sodium acetate maintains a constant value of pH.

H₂S ⇋ H⁺ + S^2-

The ionisation of hydrogen sulphide in water is low in the presence of hydrochloric acid. This can be explained on the basis of common ion effect and Le Chatilier principle.

On addition of HCl, concentration of H⁺ will increase as it is common to both HCl and H₂S. So the system accommodates the concentration of H⁺ by combining the H⁺ and S^2- back toH₂S and to do so the Equilibrium will shift in backward direction. Therefore, ionisation of H₂S is low in presence of HCl due to common ion effect.

For any chemical reaction, Equilibrium constant is dependent on temperature. Any reaction will always have the same value of Equilibrium constant if the temperature at which reaction is carried out is kept same.

Therefore, Equilibrium constant at constant temperature is a characteristic property of a reaction i.e. it will always be same if temperature is same.

Equilibrium constant is not independent of temperature.

For e.g. consider an endothermic reaction. The reactant will require heat in order to proceed in forward direction and form products. So, if the temperature is increased, reactants will get heat to carry out reaction and more product will be formed at higher temperature. Thus, Equilibrium constant (which is basically the ratio of concentration of product upon that of reactant) will increase.

Therefore, we can say for endothermic reaction, increased temperature will be favoured and more product will be formed which in turn will increase value of K.

And for exothermic reaction, reverse is true.

So from this it can be said that K is dependent on temperature.

Ammonium carbonate is a salt of weak acid and weak base.

H₂CO₃ + NH₄OH � (NH)₂CO₃

Weak acid weak base

pH = [ + - ]

Where, = -log (K_w)

K_w (ionic product of water)= 10^-14

= -log (10^-14) = 14

On comparing K_a and K_b of H₂CO₃ and NH₄OH,

K_a < K_b

On taking –log on both sides,

-log K_a > -log K_b

pK_a > pK_b

So, pK_a - pK_b > 0

Therefore, pH > [

i.e. pH > 7 which indicates basic solution.

From this it is clearly evident that acidic/basic nature of a solution of a salt of weak acid and weak base depends on K_a and K_b value of the acid and the base forming it.

A buffer is a solution containing either a weak acid and its salt or a weak base and its salt, which is resistant to pH change.

The aqueous solution of ammonium acetate is a salt of weak acid (acetic acid) and weak base (NH₄OH).

Therefore, weak acid (CH₃COOH) and its salt (CH₃COONH₄) constitute a buffer.

At constant temperature and pressure, addition of inert gas does not bring any change in Equilibrium because the addition of inert gas does not change the partial pressure of other gases in the reaction mixture and therefore, it does not affect the dissociation of reactant (PCl₅) also.

Since, helium is an inert gas it will not react with any other gas in reaction mixture and thus, does not remove Cl₂ from field of action.

K_c (Equilibrium constant) is the ratio of concentration of products to that of reactants each raise to their stoichiometric coefficients at Equilibrium.

Whereas, Q_c (reaction quotient) is the ratio of concentration of products to that of reactants each raise to their stoichiometric coefficients at any time during the reaction.

By comparing of values of K and Q we can know that in which direction the reaction will proceed.

(i) If Q < K, it means concentration of products is to be increased so as to reach Equilibrium concentration thus the net reaction proceeds in the forward direction.

(ii) If Q > K, it means reaction concentration of products is to be decreased so as to reach Equilibrium concentration thus the net reaction proceeds backward direction.

(iii) If Q = K, that means Equilibrium is achieved and thus no net reaction occurs.

Given, N₂ (g) + 3H₂ (g) ⇋ 2NH₃ (g), Δ H = -92.38 kJ mol^-1

• Since ΔH is negative that means forward reaction is exothermic.

• Now according to Le Chatelier principle, if the temperature is decreased the reaction will move in forward direction and more yield of product will be obtained.

• But if the temperature is increased the reaction will move in backward direction and yield of product will be less.

• According to Le Chatelier principle, if the pressure is increased the Equilibrium will shift in a direction where the numbers of gas molecules are less. Therefore, on increasing the pressure the Equilibrium will shift in forward direction and yield of product will be increased.

Number of moles of reactants = 1 + 3 = 4

Number of moles of product = 2

• Therefore, low temperature and high pressure are favourable conditions to increase the yield of products.

• Adding an inert gas at constant volume does not result in a shift. This is because the addition of a non-reactive gas does not change the partial pressures of the other gases in the container.

A relationship between the solubility and solubility product for such salt can be derived as follows-

Let the solubility be S.

Initially, assume 1 mole of A_xB_y was present. From which S moles of it was dissolved to give xS and yS moles of A^p+ and B^q- respectively. This can be understood from the following equation.

A^p+_x B^q-_y ⇋ xA^+p + yB^-q

At t=0, 1 0 0

At Equilibrium, 1-S xS yS

Solubility product is the multiplication product of concentration of products each raise to their stoichiometric coefficients.

Hence, Solubility product (K_sp) = [A^+p]^x [B^-q]^y

= (xS)^x(yS)^y

= x^xy^yS^x+y

Relation between ΔG and Q can be given as

ΔG = ΔG° + RT lnQ ...(1)

Where, ΔG° = standard Gibb’s energy

ΔG = Gibb’s energy change

R = gas constant

T = absolute temperature

Q = reaction quotient

This relationship can be further simplified as follows-

We know that at Equilibrium, ΔG° = -RT lnK ...(2)

Where K = Equilibrium constant

Putting (2) in (1),

ΔG = -RT lnK + RT lnQ

ΔG = RT ln(Q/K) ...(3)

As we know, K_c (Equilibrium constant) is the ratio of concentration of products to that of reactants each raise to their stoichiometric coefficients at Equilibrium.

Whereas, Q_c (reaction quotient) is the ratio of concentration of products to that of reactants each raise to their stoichiometric coefficients at any time during the reaction.

(a) If Q< K, it means concentration of products is to be increased so as to reach Equilibrium concentration thus the net reaction proceeds in the forward direction.

When Q= K, that means Equilibrium is achieved and thus no net reaction occurs.

(b) CO(g) + 3H₂(g) ⇋ CH₄ (g) + H₂ O (g)

According to Le Chatiliers principle, on increasing the pressure, the Equilibrium will shift in a direction where the number of gas molecules are less.

In the given reaction,

Number of moles of reactants = 1 + 3 = 4

Number of moles of product = 1 + 1 = 2

And thus the Equilibrium will shift in forward direction as number of moles of product is less.

We know that when Q < K, the Equilibrium shifts in forward direction. Thus, for the following reaction Q < K.