Equilibrium Class 11 Chemistry Notes and Questions

Equilibrium Class 11 Chemistry Notes and Questions

CBSE Notes Class 11

Students should go through the Equilibrium Classification Class 11 Chemistry notes provided below. These notes have been designed based on the latest NCERT Book for Class 11 Chemistry. These revision notes cover all important topics in your CBSE books. Students should revise all Class 11 Chemistry Notes as they will help the students to understand all topics given in your books. This will help you to get good marks in the Class 11 Chemistry exams.

CBSE Class 11 Chemistry Chapter 7 Equilibrium Notes

Equilibrium state – When rate of formation of a product in a process is in competition with rate of formation of reactants, the state is then named as “Equilibrium state” .

➣ Equilibrium in physical processes: solid ⇌ liquid ⇌ gas
H2O(s )⇌ H2O(l) ⇌ H2O(vap)

➣ Law of chemical equilibrium: At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.
aA +bB⇌cC + dD
Kc =[C]c [D]d/[A]a [B]b

Chemical equation                Equilibrium constant
aA + b B⇌c C + D                           K
cC + d D⇌a A + b                     B K′c=(1/Kc)
na A + nb B ⇌ncC + ndD           K′″c= (Kcn)
Concentrations or partial pressure of pure solids or liquids do not appear in the expression of the equilibrium constant. In the reaction,
Ag2O(s) + 2HNO3(aq) ⇌2AgNO3(aq) +H2O(l) Kc (Agno3)2/(HN03)2
➣ If Qc >Kc, the reaction will proceed in the direction of reactants (reverse reaction).If Qc <Kc, the reaction will proceed in the direction of the products (forward reaction)
➣ Kp is equilibrium constant in terms of partial pressure of gaseous reactants and products.
➣ Kc is equilibrium constant in terms of molar concentration of gaseous reactants and products.
➣ Kp =Kc (RT)Δn here R is gas constant, T is temperature at which the process is carried out &Δn is no. of moles of gaseous product minus no. of moles of gaseous reactants.
➣ If Kc> 103; Kc is very high i.e. the reaction proceeds nearly to completion.
➣ If Kc<103; Kc is very small i.e. the reaction proceeds rarely.
➣ If Kcis ranging in the range of 103 to 10-3; i.e. reactants and products are just in equilibrium.
➣ ΔG0 = – RT lnK or ΔG0 = – 2.303RT log K
➣ Factors affecting equilibrium constant:- temperature, pressure, catalyst and molar concentration of reactants and products.

➣ Le Chatelier’s principle:- 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 the change.
➣ Arrhenius acids are the substances that ionize in water to form H+.
➣ Arrhenius bases are the substances that ionize in water to form OH.
➣ Lewis acids are lone pair (of e-) accepters while Lewis bases are lone pair donators.
➣ Proton donor are acids while proton accepters are bases(Bronsted-Lowry concept).
➣ The acid-base pair that differs only by one proton is called a conjugate acid base pair. If Brönsted acid is a strong acid then its conjugate base is a weak base and vice versa.
➣ Ionic product of water. Kw = [H+][OH]
➣ pH = -log [H+] ; here[H+] is molar concentration of hydrogen ion.
➣ pH + pOH =14
➣ pKa + pKb =14
➣ Ka x Kb = Kw = ionic product of water=1 x 10-14
➣ Buffer solution :The solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali are called Buffer Solutions.
➣ common ion effect: It can be defined as a shift in equilibrium on adding a substance that provides more of an ionic species already present in the dissociation equilibrium.
➣ Hydrolysis of Salts: process of interaction between water and cations/anions or both of salts is called hydrolysis.
➣ The cations (e.g., Na+, K+,Ca2+, Ba2+, etc.) of strong bases and anions(e.g., Cl, Br, NO3–, ClO4– etc.) of strong acids simply get hydrated but do not hydrolyse, and therefore the solutions of salts formed from strong acids and bases are neutral i.e., their pH is 7.
➣ Salts of weak acid and strong base e.g., CH3COONa are basic in nature.
➣ Salts of strong acid and weak base e.g., NH4Cl, are acidic
➣ Salts of weak acid and weak base, e.g., CH3COONH4. The pH is determined by the formula pH = 7 + ½ (pKa – pKb)
➣ Solubility product- product of the molar concentrations of the ions in a saturated solution, each concentration term raised to the power equal to the no. of ions produced.

Question. Mention the factors that affect equilibrium constant.
Ans. Temperature, pressure, catalyst and molar concentration of reactants and products.

Question. What is ionic products of water?
Ans. Kw = [H+] [OH]

Question. Write conjugate acids of H2O & NH3.
Ans. H3O+& NH4+.

Question. Define Arrhenius acids.
Ans. Arrhenius acids are the substances that ionize in water to form H+.

Question. Define the term degree of ionization.
Ans. Extent up to which an acid/base/salt ionize to form ions.

Question. What are Buffer solutions?
Ans. The solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali are called Buffer Solutions.

Question. Write Kc for the gaseous reaction- N2 + 3H2⇌ 2NH3
Ans. Kc=[NH3]2/[N2] [H2]3

Question. Out of H2O & H3O+ which is stronger acid?
Ans. H3O+.

Question. What is common ion effect?
Ans. Shift in equilibrium on adding a substance that provides more of an ionic species already present in the dissociation equilibrium.

Question. Write relationship between Kp and Kc for the gaseous reaction – N2 + O2 ⇌ 2NO
Ans. Kp = Kc asΔn is zero for the above said reaction.

Question. What is effect of catalyst on equilibrium constant „Kc‟?
Ans. A catalyst does not affect equilibrium constant because it speeds up both forward and backward reactions to the same extent.

Question. State Le Chatelier‟r principle.
Ans. 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 the change.

Question. What is meant by conjugate acid –base pairs? Explain.
Ans. H2O          + HCl⇌H3O+              + Cl
       base acid   conjugate acid        conjugate base

Question. Classify the following bases as strong and weak bases: NaHCO3, NaOH, KOH, Ca(OH)2, Mg(OH)2.
Ans. strong base NaOH, KOH ; weak bases NaHCO3,Ca(OH)2, Mg(OH)2.

Question. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3M. What is its pH ?
Ans. pH = – log[3.8 × 10–3]
= – {log[3.8] + log[10–3]}
= – {(0.58) + (– 3.0)} = – { – 2.42} = 2.42
Therefore, the pH of the soft drink is 2.42and it is acidic.

Question. The species: H2O, HCO3 , HSO4 – and NH3can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and conjugate base.
Ans.

Question. Explain Lewis acids and bases with suitable examples.
Ans. Lewis acids are lone pair (of e-) accepters while Lewis bases are lone pair donators.
AlCl3 is a Lewis acid while NH3 is a Lewis base.

Question. What is difference between alkali and bases? Give examples.
Ans. An alkali is a water soluble base. All the alkalis are bases but all the bases are not alkali. Ex- NaOH is an alkali/base.
Ca(OH)2 is a base but not an alkali.

Question. Explain homogeneous and heterogeneous equilibrium giving examples.
Ans. If all the reactants and products present in an equilibrium mixture are in same phase → homogeneous equilibrium.
If all the reactants and products present in an equilibrium mixture are in different phase→ heterogeneous equilibrium.
N2 (g) + 3H2(g) ⇌2NH3(g) homogeneous equilibrium
CaCO3(s) ⇌ CaO(s) + CO2(g) heterogeneous equilibrium

Question. The pH of some common substances is given bellow. Classify the substances as acidic/basic

Ans. acidic-Human saliva, Lemon juice, milk, vinegar
Basic- Lime water, sea water, milk of magnesia.

Question. Explain general characteristics of acids and bases.
Ans. Most of the acids taste sour .Acids are known to turn blue litmus paper into red and liberate dihydrogen on reacting with some metals. Bases are known to turn red litmus paper blue, taste bitter and feel soapy.

Question. Water is amphoteric in nature. Explain.
Ans. Water can react with acid as well as base
H2O + HCl → H3O+ +Cl water is basic
H2O + NH3 → OH+ NH4 + water is acidic

Question. Describe the effect of :
a) addition of H2
b) addition of CH3OH
c) removal of CO
d) removal of CH3OH
on the equilibrium of the reaction:
2H2(g) + CO (g )⇌ CH3OH (g)
Ans. a) addition of H2        equilibrium will shift on RHS
b) addition of CH3OH          equilibrium will shift on LHS
c) removal of CO                equilibrium will shift on LHS
d) removal of CH3OH          equilibrium will shift on RHS

Question. Classify the following species into Lewis acids and Lewis bases and show how these act as such:
(a) HO (b)F (c) H+ (d) BCl3
Ans. (a) Hydroxyl ion is a Lewis base as it can donate an electron lone pair (:OH ).
(b) Flouride ion acts as a Lewis base as it can donate any one of its four electron lone pairs.
(c) A proton is a Lewis acid as it can accept a lone pair of electrons from bases like hydroxyl ion and fluoride ion.
(d) BCl3 acts as a Lewis acid as it can accept a lone pair of electrons from species like ammonia or amine molecules

Question. For the equilibrium,2NOCl(g) ⇌2NO(g) + Cl2(g)the value of the equilibrium constant, Kcis 3.75 × 10–6 at 1069 K. Calculate the Kp for the reaction at this temperature?
Ans. We know that, Kp= Kc(RT)Δn
For the above reaction,Δn= (2+1) – 2 = 1
Kp= 3.75 ×10–6 (0.0831 × 1069)
Kp= 0.033.

Question. Hydrolysis of sucrose gives, Sucrose + H2O →Glucose + Fructose Equilibrium constant Kc for the reaction is 2 ×1013 at 300K. Calculate ΔG0at 300K.
Ans. ΔG0= – RT lnKc
ΔG0= – 8.314J mol–1K–1J x 300K × ln(2×1013)
ΔG0= – 7.64 ×104 J mol–1

Question. Explain the following :
(i) Common ion effect (ii) solubility products (iii) pH
Ans. (i) Suppression of ionization of weak electrolyte by adding a strong electrolyte having an ion common.
(ii) Product of the molar concentrations of the ions in a saturated Ans. ,each concentration term raised to the power equal to the no. of ions produced.
(iii) Negative logarithm of hydrogen ion concentration.

Question. The values of Ksp of two sparingly soluble salts Ni(OH)2 and AgCN are 2.0 × 10–15 and 6 × 10–17 respectively. Which salt is more soluble? Explain.
Ans. AgCN ⇌ Ag+ + CN
Ksp = [Ag+][CN] = 6 × 10–17
Ni(OH)2 ⇌ Ni2+ + 2OH
Ksp = [Ni2+][OH]2 = 2 × 10–15
Let [Ag+] = S1, then [CN-] = S1
Let [Ni2+] = S2, then [OH] = 2S2
S12 = 6 × 10–17 , S1 = 7.8 × 10–9
(S2)(2S2)2 = 2 × 10–15, S2 = 0.58 × 10–4
Ni(OH)2 is more soluble than AgCN.

Question. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride,PCl5 is 8.3 ×10-3. If decomposition is depicted as, PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) ΔrH0= 124.0 kJ mol–1
a) Write an expression for Kc for the reaction.
b) What is the value of Kc for the reverse reaction at the same temperature?
c) what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased (iii) the temperature is increased ?
Ans. (a) Kc=[PCl3][ Cl2] [ PCl5]
(b) 120.48
(c) (i) equilibrium will shift on RHS
(ii) equilibrium will shift on LHS
(iii) equilibrium will shift on RHS

Question. Dihydrogen gas is obtained from natural gas by partial oxidation with steam asper following endothermic reaction: CH4 (g) + H2O (g) ⇌ CO (g) + 3H2 (g)
(a) Write as expression for Kp for the above reaction.
(b) How will the values of Kp and composition of equilibrium mixture be affected by (i) increasing the pressure(ii) increasing the temperature(iii) using a catalyst?
Ans. (a) Kp = p(CO).p(H2)3/ p(CH4).p(H2O)
(b) (i) value of Kp will not change, equilibrium will shift in backward direction.
(ii) Value of Kp will increase and reaction will proceed in forward direction.
(iii) no effect.

Question. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO2, CN, HClO4, F , OH, CO3 2–, and S2–
Ans. The acid-base pair that differs only by one proton is called a conjugate acid-base pair

Equilibrium Class 11 Chemistry Notes and Questions