# Chemical Equilibrium-02- Exercise

STUDY INNOVATIONSEducator en Study Innovations

SECTION-I SUBJECTIVE PROBLEMS Problem 1 : In an experiment, at a total of 10 atmospheres and 400ºC, in the equilibrium mixture 2NH¬3  N2 + 3H2 the ammonia was found to have dissociated to the extent of 96%. Calculate Kp for the reaction Solution : For the reaction; 2NH3  N2 + 3H2 Initial 1 mole 0 0 At equilibrium (1 – x) 1 – x + + = 1 +x Partial pressure Substituting x = 0.96 and P = 10 atmosphere we get, Ans. Problem 2 : At 700 K, CO2 and H2 react to form CO and H2O for this process K is 0.11. If a mixture of 0.45 mole of CO2 and 0.45 mole of H¬2 is heated to 700 K, then (a) Find out the amount of each gas at equilibrium. (b) After equilibrium is reached, another 0.34 mole of CO2 and 0.34 mole of H2 are added to the reaction mixture. Find the composition of the mixture at the new equilibrium state. Solution : (a) The given equilibrium is CO2(g) + H2(g)  CO(g) + H2O(g) At t = 0 0.45 0.45 0 0 At equilibrium (0.45 – x) (0.45 – x) x x x = 0.112  [CO2] = [H2] = 0.45 – x = (0.45 – 0.112) = 0.3379 = 0.34 mole each [CO] = [H2O] = x = 0.112 mole each (b) When 0.34 mole of CO2 and H2 are added in above equilibrium then following case exists. CO2(g) + H2(g)  CO(g) + H2O At t = 0 (0.34) (0.34) + 0.11 0.11 At equilibrium (0.68 – ) (0.68 – ) (0.11 ) + (0.11 + )  = 0.086 [CO2] = [H2] = 0.68 –  = (0.68 – 0.086) = 0.594 mole each [CO] = [H2O] = 0.11 +  = 0.11 + 0.086 = 0.196 mole each Problem 3 : What is the concentration of CO in equilibrium at 25ºC in a sample of a gas originally containing 1.00 mol L–1 of CO2 ? For the dissociation of CO¬2¬ at 25ºC; Kc = 2.96 × 10–92 2CO2(g)  2CO(g) + O2(g) At equilibrium (1 – 2x) (2x) (x) Solution : Applying law of mass action; It can be assured that 1 – 2x = 1.0 as Kc is very small So, 4x3 = 2.96 × 10–92 x = 1.96 × 10–31 mol L–1 Ans. Problem 4 : Under what pressure condition CuSO4 5H2O be efflorescent at 25ºC ? How good a drying agent is CuSO4.3H2O at the same temperature ? Given Kp = 1.086  10–4 atm2 at 25ºC. Vapour pressure of water at 25ºC is 23.8 mm of Hg. Solution : An efflorescent salt is one salt that loses H2O to atmosphere For the reaction;  Given at 25ºC (i.e. 23.8) > 7.92 mm and thus, reaction will proceed in backward direction, i.e. Thus CuSO4 .5H2O will not act as efflorescent but on the contrary CuSO2 .3H2O will absorb moisture from the atmosphere under given conditions. The salt CuSO4 .5H2O will effloresce only on a dry day when the aqueous known or partial pressure of moisture in the air is lesser than 7.72 mm or if relative humidity of air at 25ºC = 7.92/23.8 = 0.333 or 33.3%. Problem 5 : If K1 = 1.8  107 K2 = 5.6  109 Then for What will be the value of equilibrium constant ? Solution : Problem 6 : At 1000K, water vapo

DPP-31 to 32 With Answer por
DPP-31 to 32 With AnswerSTUDY INNOVATIONS
8 vistas6 diapositivas
DPP-73– 75 -PC por
DPP-73– 75 -PCSTUDY INNOVATIONS
13 vistas4 diapositivas
Thermodynamics numericl 2016 por
Thermodynamics numericl 2016nysa tutorial
6.4K vistas45 diapositivas
Energetics-04-SOLVED SUBJECTIVE EXAMPLES por
Energetics-04-SOLVED SUBJECTIVE EXAMPLESSTUDY INNOVATIONS
2 vistas7 diapositivas
Aieee chemistry - 2010 por
Aieee chemistry - 2010Vasista Vinuthan
900 vistas33 diapositivas
DPP-2-Faculty Copy-50 por
DPP-2-Faculty Copy-50STUDY INNOVATIONS
11 vistas52 diapositivas

## Similar a Chemical Equilibrium-02- Exercise

DPP-4-Faculty Copy-93 por
DPP-4-Faculty Copy-93STUDY INNOVATIONS
16 vistas30 diapositivas
DPP-11 to 12 With Answer por
DPP-11 to 12 With Answer STUDY INNOVATIONS
6 vistas4 diapositivas
AIPMT 2014-paper-solution-chemistry por
AIPMT 2014-paper-solution-chemistryALLEN CAREER INSTITUTE
7.7K vistas7 diapositivas
SI #13 Key por
SI #13 Keyjessieo387_1412
24.6K vistas3 diapositivas
Tang 03 enthalpy of formation and combustion por
Tang 03 enthalpy of formation and combustionmrtangextrahelp
9.8K vistas18 diapositivas
DPP-2-Student Copy-70 por
DPP-2-Student Copy-70STUDY INNOVATIONS
19 vistas46 diapositivas

### Similar a Chemical Equilibrium-02- Exercise(20)

Tang 03 enthalpy of formation and combustion por mrtangextrahelp
Tang 03 enthalpy of formation and combustion
mrtangextrahelp9.8K vistas
Electrochemistry-02-Solved Objective Problems por STUDY INNOVATIONS
Electrochemistry-02-Solved Objective Problems
STUDY INNOVATIONS142 vistas
CHEMISTRY-11th (PQRS & J) FINAL TEST Partial Marking.pdf por STUDY INNOVATIONS
CHEMISTRY-11th (PQRS & J) FINAL TEST Partial Marking.pdf

## Más de STUDY INNOVATIONS

Chemistry-40.One Year CRP-Test Paper-6-Solutions por
Chemistry-40.One Year CRP-Test Paper-6-SolutionsSTUDY INNOVATIONS
0 vistas2 diapositivas
Chemistry-38.Test Paper-12 Studying Merg 27 por
Chemistry-38.Test Paper-12 Studying Merg 27STUDY INNOVATIONS
0 vistas11 diapositivas
Chemistry-11.Test Paper-CRP-XII Studying-3-Screening por
Chemistry-11.Test Paper-CRP-XII Studying-3-ScreeningSTUDY INNOVATIONS
0 vistas3 diapositivas
Chemistry-23.Transition Element-Theory por
Chemistry-23.Transition Element-TheorySTUDY INNOVATIONS
0 vistas37 diapositivas
Chemistry-22.Solid State-Theory por
Chemistry-22.Solid State-TheorySTUDY INNOVATIONS
0 vistas28 diapositivas
Chemistry-21.Qualitative Analysis-Theory por
Chemistry-21.Qualitative Analysis-TheorySTUDY INNOVATIONS
0 vistas40 diapositivas

### Más de STUDY INNOVATIONS(20)

Chemistry-40.One Year CRP-Test Paper-6-Solutions por STUDY INNOVATIONS
Chemistry-40.One Year CRP-Test Paper-6-Solutions
Chemistry-11.Test Paper-CRP-XII Studying-3-Screening por STUDY INNOVATIONS
Chemistry-11.Test Paper-CRP-XII Studying-3-Screening
Chemistry-19.Practical Organic Chemistry-Theory por STUDY INNOVATIONS
Chemistry-19.Practical Organic Chemistry-Theory
Chemistry-17.Periodic Chart of the Elements-Theory por STUDY INNOVATIONS
Chemistry-17.Periodic Chart of the Elements-Theory
Chemistry-13.General Organic Chemistry-Theory por STUDY INNOVATIONS
Chemistry-13.General Organic Chemistry-Theory
Chemistry-12.Electrophilic Aromatic Substitution por STUDY INNOVATIONS
Chemistry-12.Electrophilic Aromatic Substitution

## Último

Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37 por
Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37MysoreMuleSoftMeetup
44 vistas17 diapositivas
CUNY IT Picciano.pptx por
CUNY IT Picciano.pptxapicciano
56 vistas17 diapositivas
Thanksgiving!.pdf por
Thanksgiving!.pdfEnglishCEIPdeSigeiro
401 vistas17 diapositivas
Narration lesson plan por
Narration lesson planTARIQ KHAN
64 vistas11 diapositivas
Creative Restart 2023: Atila Martins - Craft: A Necessity, Not a Choice por
Creative Restart 2023: Atila Martins - Craft: A Necessity, Not a ChoiceTaste
38 vistas50 diapositivas
BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (FRIE... por
BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (FRIE...Nguyen Thanh Tu Collection
51 vistas91 diapositivas

### Último(20)

Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37 por MysoreMuleSoftMeetup
Payment Integration using Braintree Connector | MuleSoft Mysore Meetup #37
CUNY IT Picciano.pptx por apicciano
CUNY IT Picciano.pptx
apicciano56 vistas
Narration lesson plan por TARIQ KHAN
Narration lesson plan
TARIQ KHAN64 vistas
Creative Restart 2023: Atila Martins - Craft: A Necessity, Not a Choice por Taste
Creative Restart 2023: Atila Martins - Craft: A Necessity, Not a Choice
Taste38 vistas
BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (FRIE... por Nguyen Thanh Tu Collection
BÀI TẬP BỔ TRỢ TIẾNG ANH 11 THEO ĐƠN VỊ BÀI HỌC - CẢ NĂM - CÓ FILE NGHE (FRIE...
Monthly Information Session for MV Asterix (November) por Esquimalt MFRC
Monthly Information Session for MV Asterix (November)
Esquimalt MFRC91 vistas
11.30.23A Poverty and Inequality in America.pptx por mary850239
11.30.23A Poverty and Inequality in America.pptx
mary85023962 vistas
ANGULARJS.pdf por ArthyR3
ANGULARJS.pdf
ArthyR349 vistas
Guess Papers ADC 1, Karachi University por Khalid Aziz
Guess Papers ADC 1, Karachi University
Khalid Aziz69 vistas
Guidelines & Identification of Early Sepsis DR. NN CHAVAN 02122023.pptx por Niranjan Chavan
Guidelines & Identification of Early Sepsis DR. NN CHAVAN 02122023.pptx
Niranjan Chavan38 vistas
Nelson_RecordStore.pdf por BrynNelson5
Nelson_RecordStore.pdf
BrynNelson544 vistas
Retail Store Scavenger Hunt.pptx por jmurphy154
Retail Store Scavenger Hunt.pptx
jmurphy15447 vistas
Parts of Speech (1).pptx por mhkpreet001
Parts of Speech (1).pptx
mhkpreet00143 vistas

### Chemical Equilibrium-02- Exercise

• 1. Chemistry : Chemical Equilibrium SECTION-I SUBJECTIVE PROBLEMS Problem 1 : In an experiment, at a total of 10 atmospheres and 400ºC, in the equilibrium mixture 2NH3  N2 + 3H2 the ammonia was found to have dissociated to the extent of 96%. Calculate Kp for the reaction Solution : For the reaction; 2NH3 N2 + 3H2 Initial 1 mole 0 0 At equilibrium (1 – x) x 2 3x 2 1 – x + x 2 + 3x 2 = 1 +x Partial pressure 1 x p 1 x   x.p 2(1 x)  3x.p 2(1 x)  Substituting x = 0.96 and P = 10 atmosphere we get, 4 2 2 2 27(0.96) (10) Kp 23,320 16(1 0.96) (1 0.96)     Ans. Problem 2 : At 700 K, CO2 and H2 react to form CO and H2O for this process K is 0.11. If a mixture of 0.45 mole of CO2 and 0.45 mole of H2 is heated to 700 K, then (a) Find out the amount of each gas at equilibrium. (b) After equilibrium is reached, another 0.34 mole of CO2 and 0.34 mole of H2 are added to the reaction mixture. Find the composition of the mixture at the new equilibrium state. Solution : (a) The given equilibrium is CO2(g) + H2(g) CO(g) + H2O(g) At t = 0 0.45 0.45 0 0 At equilibrium (0.45 – x) (0.45 – x) x x 2 2 x Kc (0.45 x)   2 2 x 0.11 (0.45 x)   x = 0.112
• 2. Chemistry : Chemical Equilibrium  [CO2] = [H2] = 0.45 – x = (0.45 – 0.112) = 0.3379 = 0.34 mole each [CO] = [H2O] = x = 0.112 mole each (b) When 0.34 mole of CO2 and H2 are added in above equilibrium then following case exists. CO2(g) + H2(g) CO(g) + H2O At t = 0 (0.34) (0.34) + 0.11 0.11 At equilibrium (0.68 – ) (0.68 – ) (0.11 ) + (0.11 + ) 2 2 2 [CO][H O] Kc [CO ][H ]  2 (0.11 ) 0.11 (0.68 x)       = 0.086 [CO2] = [H2] = 0.68 –  = (0.68 – 0.086) = 0.594 mole each [CO] = [H2O] = 0.11 +  = 0.11 + 0.086 = 0.196 mole each Problem 3 : What is the concentration of CO in equilibrium at 25ºC in a sample of a gas originally containing 1.00 mol L–1 of CO2 ? For the dissociation of CO2 at 25ºC; Kc = 2.96 × 10–92 Solution : Applying law of mass action; 2 2 92 2 c 2 2 2 [O ][CO] x (2x) K 2.96 10 [CO ] (1 2x)        It can be assured that 1 – 2x = 1.0 as Kc is very small So, 4x3 = 2.96 × 10–92 x = 1.96 × 10–31 mol L–1 31 [CO] 2x 2 1.95 10     31 1 3.90 10 molL     Ans. Problem 4 : Under what pressure condition CuSO4 5H2O be efflorescent at 25ºC ? How good a drying agent is CuSO4.3H2O at the same temperature ? 2CO2(g) 2CO(g) + O2(g) At equilibrium (1 – 2x) (2x) (x)
• 3. Chemistry : Chemical Equilibrium Given 4 2 4 2 2 CuSO .5H O(s) CuSO .3H O(s) 2H O(v)  Kp = 1.086  10–4 atm2 at 25ºC. Vapour pressure of water at 25ºC is 23.8 mm of Hg. Solution : An efflorescent salt is one salt that loses H2O to atmosphere For the reaction; 4 2 4 2 2 CuSO .5H O(s) CuSO .3H O(s) 2H O(v)  2 1 2 4 p H O K (P ) 1.086 10    2 1 2 H O P 1.042 10 atm 7.92 mm      Given 2 1 H O P at 25ºC (i.e. 23.8) > 7.92 mm and thus, reaction will proceed in backward direction, i.e. 4 2 2 4 2 CuSO .3H O 2H O(v) CuSO .5H O(s)  Thus CuSO4 .5H2O will not act as efflorescent but on the contrary CuSO2 .3H2O will absorb moisture from the atmosphere under given conditions. The salt CuSO4 .5H2O will effloresce only on a dry day when the aqueous known or partial pressure of moisture in the air is lesser than 7.72 mm or if relative humidity of air at 25ºC = 7.92/23.8 = 0.333 or 33.3%. Problem 5 : If 3 3 2 Ag 2NH Ag(NH )    K1 = 1.8  107 Ag Cl AgCl    K2 = 5.6  109 Then for 3 3 2 AgCl 2NH [Ag(NH ) ] Cl     What will be the value of equilibrium constant ? Solution : 3 2 1 2 3 [Ag(NH ) ] K [Ag ][NH ]    2 [AgCl] K [Ag ][Cl ]    3 2 1 n 2 2 3 [Ag(NH ) ] [Cl ] K K K [AgCl] [NH ]     3 2 1 2 2 3 [Ag(NH ) ] K [Ag ][Cl ] K [AgCl] [Ag ][NH ]        7 2 9 1.8 10 0.32 10 K. 5.6 10        Problem 6 : At 1000K, water vapour at 1 atm has been found to be dinociated into hydrogen and oxygen to the extent of 3 × 10–7 %. Calculate the free energy decrease of the system, assuming ideal behaviour.
• 4. Chemistry : Chemical Equilibrium Solution : 2 2 2 2H O 2H O  2 2 2 2 H O p 2 H O (P ) K (P ) (P )  If the pressure is 1 atm then 2 2 7 7 7 H O 1 P 3.0 10 ; P (3.0 10 ) 1.5 10 2          p 7 2 7 1 K (3.0 10 ) (1.5 10 )      19 20 1 7.4074 10 1.35 10     p p G RT n K 2.303 RTlog K    l 19 2.303 8.314 100 log7.4074 10      = 380447.3 J = 380.4473 kJ Ans. Problem 7 : At 448ºC, the equilibrium constant (Kc) for the reaction 2 2 H (g) I (g) 2HI(g)  is 50.5. Predict the direction in which the reaction will proceed to reach equilibrium at 448ºC, if we start with 2.0 × 10–2 mol of HI, 1.0 × 10–2 mol of H2 and 3.0 × 10–2 mol of I2 in a 2 – O L container. Solution : The initial concentrations are 2 1 2.0 10 [HI] molL 2     = 1.0 × 10–2 mol L–1 2 1 2 1.0 10 [H ] molL 2     = 0.5 × 102 mol L–1 2 1 2 3.0 10 [I ] molL 2     = 1.5 × 10–2 mol L–1 Concentration quotient, 2 2 [HI] Q [H ][I ]  2 3 2 2 (1 10 ) 1.3 10 (.5 10 )(1.5 10 )          Since Q < K; the reaction will proceed in the forward direction to attain equilibrium, so that Q becomes equal to K. Problem 8 : At temperature T, a compound, AB2(g) dissociates according to the reaction 2AB2(g) 2AB(g) + B2(g) with degree of dissociation a, which is small compared with unity. What will the expression of Kp, in term of a and total pressure PT.
• 5. Chemistry : Chemical Equilibrium Solution : For the given equilibrium, the equilibrium constant are 2AB2(g) 2AB(g) + B2(g) Equilibrium Conc. C (1 – a) Ca Ca 2 2 2 2 B AB p 2 AB (P )(P ) K (P )   2 T 2 Ca (Ca) P 2 a [C(1 a)] C 1 2                  3 T p 2 a P K a 2(1 a) a 2           Since a < < 1, 3 T p a P K 2    Ans. Problem 9 : Consider the following reaction 2 2 c H (g) I (g) 2HI(g) K 54.3 at 698 K   If we start with 0.500 mol H2 and 0.500 mol I2(g) in a 5.25 –L vessel at 698 K, how many moles of each gas will be present at equilibrium ? Solution : First, let’s calculate the initial concentration in the 5.25 – L flask. 2 2 0.500 [H ] [I ] 0.0952M [HI] 0 5.25L     Let (–x) represent the changes in concentration of H2 and I2; the change in concentration of H2 and I2; the change in [HI] is +2x because, two moles of HI are formed for every mole of H2 and I2 that react. Thus, The reaction H2(g) + I2(g) 2HI(g) Initial conc. M 0.0952 0.0952 0 Changes, M –x –x +2x Equilibrium M conc. (0.0952 – x) (0.0952 – x) 2x Now we can enter the equilibrium concentration into the Kc expression; 2 2 2 c 2 2 [HI] (2x) (2x) K 54.3 (0.0952 x)(0.0952 x) (0.0952 x) [H ][I ]        The most direct approach is to take the square root of each of the side of the equation and solve for x.
• 6. Chemistry : Chemical Equilibrium 1/ 2 2 1/ 2 2 (2x) (54.3) (0.0952 x)         1/ 2 2x (54.3) 0.0952 x   1/ 2 2x (54.3) (0.0952 x)    2x 7.37 (0.0952 x)    2x = 0.702 – 7.37 x 9.37x = 0.702 x = 0.0749 We can now calculate the equilibrium concentrations [H2] = [I2] = 0.0952 – x = 0.0952 – 0.0749 = 0.0203 M [HI] = 2x = 2  0.0749 = 0.150 M To determine the equilibrium amounts, we multiply the equilibrium concentration by the volume mol H2 = mol I2 – 5.25  0.0203 mol = 0.107 mol. mol HI – 5.25  0.150 mol = 0.788 mol Problem 10 : At 375 K and total pressure of 1 atm sulphuryl chloride is dissociated according to the equation. SO2Cl2 SO2 + Cl2 2 p 2 P K 1     2 2 p P (0.912) 1 K (1 )(1 ) (1 0.912) (1 0.0912)           = 4.944 Net work w = net decrease in free energy = – G But –G = RT ln Kp = 2.303 × 375 × 8.314 × log 4.444 = 4983.6063 J
• 7. Chemistry : Chemical Equilibrium SECTION-II SINGLE CHOICE PROBLEMS Problem 1 : 5 PCl is 50% dissociated into 3 PCl and 2 Cl at 1 atmosphere. It will be 40% dissociated at : (a) 1.75 atm (b) 1.84 atm (c) 2.00 atm (d) 1.25 atm. Solution : 5 3 2 PCl PCl Cl  Let x = degree fo dissociation of 5 PCl at pressure P y = degree of dissociation of 5 PCl at pressure p 2 2 p 2 2 px p y K (1– x ) (1– y )    given, x = 0.50, p = 1 atm y = 0.40, p ?   1 0.25 p 0.16 0.75 0.84     p 1.75atm    (a) Problem 2 : One mole of 2 4(g) N O at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% of 2 4(g) N O decomposes to 2(g) NO . The resultant pressure is (a) 1.2 atm (b) 2.4 atm (c) 2.0 atm (d) 1.0 atm. Solution : 2 4 2 N O 2NO Initially 1 0 At equi. (1–x) 2x Total mole at equilibrium = 1 + x = 1 .2 x 0.2    1 2 2 1 1 2 2 P P P 1 or n T n T 1 300 1.2 600     or 2 P 2.4atm   (b) Problem 3 : For the reversible reaction, 2(g) 2(g) 3(g) N 3H 2NH  at 500°C, the value of p K is 1.44  –5 10 when partial pressure is measured in atmospheres. The corresponding value of c K with concentration in mole –1 litre , is
• 8. Chemistry : Chemical Equilibrium (a) 1.44 –5 –2 10 /(0.082 500)   (b) –5 –2 1.44 10 /(8.314 773)   (c) 1.44 –5 2 10 /(0.082 773)   (d) –5 –2 1.44 10 /(0.082 773)   . Solution : g – n c p K K (RT)   Since g n –2    –5 2 c K 1.44 10 (0.082 773)      (d) Problem 4 : For which of the following reactions, the degree of dissociation cannot be calculated from the vapour density data I (g) 2(g) 2(g) 2HI H I  II 3(g) 2(g) 2(g) 2NH N 3H  III (g) 2(g) 2(g) 2NO N O  IV 5(g) 3(g) 2(g) PCl PCl Cl  (a) I and III (b) II and IV (c) I and II (d) III and IV. Solution : The degree of dissociation cannot be calculated from the vapour density data if the number of moles remains unchanged before and after reaching equilibrium.  (a) Problem 5 : At 27°C NO and 2 Cl gases are introduced in a 10 litre flask such that their initial partial pressures are 20 and 16 atm respectively. The flask already contains 24 g of magnesium. After some time, the amount of magnesium left was 0.2 moles due to the establishment of following two equilibria (g) 2(g) (g) 2NO Cl 2NOCl  –1 2(g) (s) 2(s) p Cl Mg MgCl ; K 0.2 atm   The final pressure of NOCl would be (a) 7.84 atm (b) 18.06 atm (c) 129.6 atm (d) 64.8 atm.
• 9. Chemistry : Chemical Equilibrium Solution : 2 p Cl 1 K 0.2 P    2 Cl P at equilibrium = 5 atm (g) 2(g) (g) 2NO Cl 2NOCl  Initially 20 16 0 At equi. 20–2x 16–x–y 2x 2(g) (s) 2(s) Cl Mg MgCl  Initially 16 1 0 At equi. 16–x–y 1–y y = 0.2 y= 0.8 (since y is the moles while y is pressure in atm) 0.8 0.082 300 y 1.97 10     We know that, 16 – x – y 5  or, x = 9.03; NOCl P 18.06 atm   (b) Problem 6 : An acid reacts with glycerine to form complex and equilibrium is established. If the heat of reaction at constant volume for above reaction is 1200 cal more than at constant pressure and the temperature is 300 K, then which of the following expression is true ? (a) p c K K  (b) c p K K  (c) p c K K  (d) none of these. Solution : Given : E – H 1200 cal    H E nRT      n –2   n p c K K (RT)  p –3 c K 1.648 10 K   p c K K   (a)
• 10. Chemistry : Chemical Equilibrium Problem 7 : Steam at pressure of 1 atm is passed through a furnace at 2000 K wherein the reaction –5 1 2 2 (g) 2(g) 2(g) p H O H O ; K 6.4 10    occurs. The percentage of oxygen in the exit steam would be (a) 0.32 (b) 0.08 (c) 0.04 (d) 0.16 Solution : 1 2 2 (g) 2(g) 2(g) H O H O  Initially 2 atm 0 0 At equi. 2 – x x x/2 1/ 2 –5 x(x / 2) 6.4 10 2 – x   3/ 2 –5 x 6.4 10 2(2 – x)   Dissociation of 2 H O is so small that we can assume 2 H O P 2  atm i.e. 2 – x 2  3/2 –5 x 2 2 6.4 10    x or 3/2 3/2 –6 x (2 16) 10    x = 2 –4 16 10 0.0032     % of 2 O in the exit steam 0.0032 / 2 100 0.08% 2     (b) Problem 8 : Solubility of a substance which dissolves with a decrease in volume and absorption of heat will be favoured by (a) high P and high T (b) low P and low T (c) high P and low T (d) low P and high T. Solution : The substance dissolves with a decrease in volume and absorption of heat. Therefore according to Le Chalelier’s principle the process of dissolution will be favoured by high pressure and high temperature.  (a) Problem 9 : At temperature T, a compound AB2(g) dissociates according to the reaction 2AB2(g) 2AB(g) + B2(g) with degree of dissociation , which is small compared with unity. The expression for Kp, in terms of  and the total pressure, PT is (a) 3 T P 2  (b) 2 T P 3  (c) 3 T P 3  (d) 2 T P 2  . Solution : For the given equilibrium, the equilibrium concentrations are
• 11. Chemistry : Chemical Equilibrium 2AB2(g) 2AB(g) + B2(g) Equilib. conc. c(1 – ) c c 2   2 2 2 2 T B AB P 2 2 AB c (c ) P (P )(P ) 2 K (P ) [c(1– )] [c(1 )] 2          ; 3 T P 2 P K 2(1– ) 1 2             Since,  is small compared to unity, so 1 –  1 and 1 + 1 2  .  3 T P P K 2     (a) Problem 10: 2.0 mol of PCl5 were introduced in a vessel of 5.0 L capacity at a particular temperature. At equilibrium, PCl5 was found to be 35% dissociated into PCl3 and Cl2. The value of Kc for the reaction is (a) 1.89 (b) 0.377 (c) 0.75 (d) 0.075. Solution: Moles of PCl5 dissociated = 2 35 100  = 0.7 Moles of PCl5 left undissociated = 2 – 0.7 = 1.3 mol [PCl5] = 1.3 5 M, [PCl3] = 0.7 5 M, [Cl2] = 0.7 5 M 3 2 5 0.7 0.7 [PCl ][Cl ] 5 5 K 0.75 1.3 [PCl ] 5                    (c)
• 12. Chemistry : Chemical Equilibrium SECTION-III MULTIPLE CHOICE PROBLEMS (with one or more than one answer correct) Problem 1 : For the following reaction 2 2 1 A B (g) 2AB(g), K 2 2 2 1 1 A (g) B (g) AB(g), K 2 2  2 2 3 2AB(g) A (g) B (g), K  2 2 4 1 1 AB(g) A (g) B (g), K 2 2  then which of the relation(s) is/are correct ? (a) 1 2 K K 1   (b) 1 4 K K 1   (c) 3 2 K K 1   (d) None Solution : (a, b, c) Problem 2 : For the reaction 5 3 2 PCl (g) PCl (g) Cl (g)  the forward reaction at constant temperature is favoured by (a) Introducing PCl5 at constant volume (b) Introducing chlorine gas at constant volume (c) Introducing an inert gas at constant pressure (d) Increasing the volume of the container Solution : (a, c, d) Problem 3 : For which of the following statements about the reaction quotient, Q, are correct (a) the reaction quotient, Q, and the equilibrium constant always have the same numerical value (b) Q may be > < = Keq (c) Q (numerical value) varies as reaction proceeds (d) Q = 1 at equilibrium. Solution : (b, c) Problem 4 : When NaNO3 is heated in a closed vessel oxygen is liberated and NaNO2 is left behind. At equilibrium, which are not correct : (a) Addition of NaNO2 favours reverse stem (b) Addition of NaNO3 favours forward stem (c) Increasing temperature favours forward direction (d) Increasing pressure favours reverse stem Solution : (a, b, d) Problem 5 : The value of equilibrium constant of a reversible reaction at a given temperature
• 13. Chemistry : Chemical Equilibrium (a) depends on the initial concentration of reactants (b) depends on the concentration of products at equilibrium (c) gets reversed when the mode of representation of the reaction is reversed (d) changes when the unit of active mass is changed. Solution : (c, d) Problem 6 : Which of the following equilibrium is (are) not affected by pressure ? (a) 2 2 3 2SO (g) O (g) 2SO (g)  (b) 2 2 H (g) I (g) 2HI(g)  (c) 1 2 2 2NO (g) O (g) 2NO (g)  (d) 2 2 N (g) O (g) 2NO(g)  Solution : (b, d) Problem 7 : For the gas phase reaction, 2 4 2 2 6 [C H H C H 32.7 k cal]    carried out in a vessel, the equilibrium concentration of C2H4 can be increased by (a) Increasing temperature (b) Decreasing pressure (c) Removing some H2 (d) Adding some C2H6 Solution : (b, c, d) Problem 8 : Which of the following relation(s) hold good for gaseous and reversible reactions (a) p ( n)g c K (RT) K   (b) p ng x K (P) K   (c) ( n)g c x K P K RT         (d) ( n)g c x K RT K P         Solution : (a, b, c) Problem 9 : The reaction A + B C + D is studied in a one litre vessel at 250ºC. The initial concentration of A was 3n and the initial concentration of B was n. After equilibrium was attained then equilibrium concentration of C was found to be equal to equilibrium concentration of B. What is the concentration of D at equilibrium? (a) n 2 (b) n 3n 2        (c) n n 2        (d) n Solution : (a) Problem 10 : Which of the following will favour the formation of products in the following reaction? (a) An increase in temperature (b) An increase in pressure (c) Addition of B (d) Removal of C Solution : (a, c, d)
• 14. Chemistry : Chemical Equilibrium SECTION-IV COMPREHENSION TYPE PROBLEMS  Write up – I According to Le-Chatellier’s principle, when external factors or conditions (like P, T, C, inert gas) is applied to the chemical equilibrium, it moves in such a direction so that the effect of external factors get minimized. On increasing pressure, the reaction moves on that direction where volume is reduced and vice-versa. On increasing temperature or heat, the equilibrium moves on that direction, in which direction, heat is consumed and vice-versa. The effect of temperature has great significance on equilibrium constants which are released as under with the equilibrium constant 2 1 1 2 K H 1 1 ln K R T T          Where K1 and K2 are equilibrium constants at temperature T1 and T2 K respectively. There is also great effect of addition of inert gas at equilibrium at constant pressure QC constant volume. At every condition, temperature remains constant. Consider the following gaseous reversible reaction in which no molecules of inert gas is added, aA(g) + bB(g) cC(g) + dD(g) If PT is the total pressure exerted by the gas at equilibrium, then n d c a b D C T p a b c d n A B P P P a .b K P P C .d             where n = sum of number of moles of gaseous products – sum of number of moles of gaseous reactants = (c + d) – (a + b). n = Total number of moles of gases in the equilibrium mixture = a + b + c + d + no The effect of addition of inert gas can be understand from equation (1). If there is no change in Kp, then the equilibrium remains undisturbed. So, effect of inert gas at constant volume and constant pressure can be understood from equation (1). On the basis of the write up, answer the following questions. Problem 1 : What will be the effect of inert gas addition in the following gaseous reversible reaction at constant volume and at constant temperature ? (a) Backward reaction will be favoured (b) Forward reaction will be favoured (c) The equilibrium will remain undisturbed (d) Equilibrium constant will decrease Solution : (c)
• 15. Chemistry : Chemical Equilibrium Problem 2 : What will be the effect of addition of inert is at constant pressure and at constant temperature for the following equilibrium reaction. 2 2 3 N 3H NH  (a) yield of NH3 will decrease (b) yield of NH3 will increase (c) No effect of addition of inert gas in the equilibrium (d) The equilibrium constant will decrease Solution : (a) Problem 3 : What will be the effect of addition of catalyst at equilibrium at constant temperature (a) The equilibrium constant will remain constant (b) H of the reaction will remain constant (c) Kf and Kb will increase with equal times. (d) All (a), (b) and (c) Solution : (A) Problem 4 : What will be the effect on the equilibrium constant on increasing temperature, if the reaction neither absorbs heat nor release heat ? (a) Equilibrium constant will remain constant (b) Equilibrium constant will decrease (c) Equilibrium constant will increase (d) cannot be predicted Solution : (a)  Write up – II Ammonia is manufactured by the Haber process represented by the equilibrium: 2(g) 2(g) 3(g) N 3H 2NH  H0 = –184.4 kJ The optimum conditions for the production of ammonia are a pressure of about 200 atm and a temperature of around 700K. Iron oxide containing small amounts of K2O and Al2O3 is used as catalyst so as to increase the rate of attainment of equilibrium. Annual world production of ammonia now exceeds 10 million tonnes. Problem 5 : According to Le-Chateliers principle the favourable condition for good yield of NH3 (a) Low pressure and low temperature (b) High pressure and low temperature (c) Low pressure and high temperature (d) High pressure and high temperature Solution : (b)
• 16. Chemistry : Chemical Equilibrium Problem 6 : In absence of catalyst during Haber’s process of ammonia synthesis (a) yield of NH3 will decrease (b) yield of NH3 will remain the same (c) equilibrium position will get shifted in the backward direction (d) equilibrium will never set up Solution : (b) MATCHING TYPE PROBLEMS Problem 7 : List-I (Entities) List-II (Units) (p) Kc when N = –2 (w)JV–1 mol–1 (q) Kc when N = –2 (x) (mol/L)–2 (r) R (gas constant) (y) bar–2 (s) F (Faraday constant) (z) KPa dm3 mol–1 K–1 P Q R S (a) y x z w (b) x y z w (c) x y w z (d) y x w z Solution : ASSERTION – REASONING TYPE PROBLEMS The following questions consist of two statements, one labelled as ASSERTION (A) and REASON (R). Use the following key to chose the correct appropriate answer. (a) (A) and (R) both are correct and (R) is the correct explanation of (A) (b) (A) and (R) both are correct but (R) is not the correct explanation of (A) (c) (A) is correct but (R) is incorrect. (d) (A) is incorrect but (R) is correct ASSERTION (A) REASONING (R) Problem 8 : At equilibrium, G° at unit activity = –RTlnKc Since G = G° + RT ln Q and at equilibrium, G = 0, and Q becomes Kc at unit activity. Solution : (a)
• 17. Chemistry : Chemical Equilibrium Problem 9. The endothermic reactions are favoured at lower temperature and the exothermic reactions are favoured at higher temperature. When a system in equilibrium is disturbed by changing the temperature, it will tend to adjust itself so as to overcome the effect of change. Solution : (d) Problem 10. The equilibrium constant of the exothermic reaction at high temperature decreases. Since 2 1 1 2 K H 1 1 ln K R T T          and for exothermic reaction, H = –ve and thereby 2 1 K ln ve K   Solution : (a)
• 18. Chemistry : Chemical Equilibrium ASSIGNMENTS SECTION - I SUBJECTIVE QUESTIONS LEVEL-I 1. A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted to CO on addition of graphite. Calculate the value of K if the total pressure at equilibrium is 0.8 atm. 2. Calculate the amount of NH3 and NH4Cl required to prepare a buffer solution of pH 9.0 when total concentration of buffering reagent is 0.6 mol L–1 . pKb for NH3 = 4.7, log 2 =0.30. 3. The solubility product of SrF2 in H2O is 8.0  10–10 . Calculate its solubility in 0.1 M NaF aqueous solution. 4. At 973 K, Kp is 1.50 for the reaction (s) 2(g) (g) C CO 2CO  Suppose the total gas pressure at equilibrium is 1.0 atm. What are the partial pressures of CO and CO2 ? 5. When S in the form of S8 is heated at 900K, the initial pressure of 1 atmosphere falls by 29% at equilibrium. This is because of conversion of some S8 to S2. Find the Kp for reaction. 6. The equilibrium constant for the reaction: H3BO3 + glycerine (H3BO3 + glycerine complex) is 0.90. How much glycerine should be added to one litre of 0.10 M H3BO3 solution, so that 60% of the H3BO3 is converted to boric acid glycerine complex? 7. At 273K and one atm, ‘a’ litre of N2O4 decomposes to NO2 according to equation 2 4(g) 2(g) N O 2NO . To what extent has the decomposition proceeded when the original volume is 25% less than that of existing volume? 8. N2O4 dissociates as 2 4 2 N O 2NO . At 55°C and one atmosphere, % decomposition of N2O4 is 50.3%. At what P and same temperature, the equilibrium mixture will have the ratio of N2O4:NO2 as 1:8? 9. At 627°C and one atmosphere pressure SO3 is partially dissociated into SO2 and O2by 3(g) 2(g) 2(g) 1 SO SO O . 2  The density of the equilibrium mixture is 0.925 g/litre. What is the degree of dissociation? 10. When 20g of CaCO3 were put into 10 litre flask and heated to 800°C 35% of CaCO3 remained unreacted at equilibrium. Calculate Kp for decomposition of CaCO3.
• 19. Chemistry : Chemical Equilibrium LEVEL-II 1. At 440ºC the equilibrium constant (K) for the following reaction is 49.5 H2(g) + I2(g) 2HI(g) If 0.2 mole of H2 and 0.2 mole of I2 are placed in 10 litre vessel and permitted to react at this temperature. What will be the concentration each substance at equilibrium. 2. Nicotinic acid (Ka = 1.4  10–5 ) is represented by the formula HniC. Calculate its percentage dissociation in a solution which contains 0.10 mole of nicotinic acid per 2.0 litre of solution. 3. Calculate the simultaneous solubility of AgSCN and AgBr.Ksp for AgSCN and AgBr are 1.0 × 10–12 and 5.0 × 10–13 respectively. 4. An aqueous solution of aniline of concentration 0.24 M is prepared. What concentration of sodium hydroxide is needed in this solution so that anilium ion concentration remains at 1  10–8 M ? 5. 15g sample of BaO2 is heated to 794°C in a closed evacuated vessel of 5 litre capacity. How many g of peroxide is converted to BaO(s) 2(s) (s) 2(g) p 2BaO 2BaO O , K 0.5atm   6. The density of an equilibrium mixture of N2O4 and NO2 at 1 atm and 348 K is 1.84 dm–3 . Calculate the equilibrium constant, Kp of the reaction, 7. When heated, phosphorus pentachloride dissociates according to the equation 5(g) 3(g) 2(g) PCl PCl Cl  At a certain temperature, 1.5 moles out of 2 moles of PCl5 in a closed 10 litre vessel are decomposed. Calculate the equilibrium constant (Kc) at this temperature. 8. For the equilibrium 2 3(s) 3(s) 3 p LiCl 3NH LiCl NH 2NH , K 9 atm     at 40°C. A 5 litre vessel contains 0.1 mole of LiClNH3. How many mole of NH3 should be added to the flask at this temperature to derive the backward reaction for completion? 9. The dissociation pressure of solid ammonium hydro-sulphide at 27°C is 60 cm. What will be the total pressure when it dissociates at the same temperature in presence of NH3 at a pressure of 45 cm? Assume NH4HS dissociates completely into NH3 and H2S. 10. One mole of N2 and 3 mole of PCl5 are placed in a 100 litre vessel heated to 227°C. The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate degree of dissociation of PCl5 and Kp of the reaction: 5 3 2 PCl PCl Cl 
• 20. Chemistry : Chemical Equilibrium LEVEL-III 1. Ammonia under a pressure of 15 atm at 27°C is heated to 347°C in a closed vessel in the presence of a catalyst. Under these conditions NH3 is partially decomposed according to the equation. 3 2 2 2NH N 3H  The vessel is such that the volume remains effectively constant, whereas pressure increases to 50 atm. Calculate the percentage of NH3 actually decomposed. Pressure of NH3 at 27°C or 300K = 1 5 atm. 2. Sulphide ion in alkaline solution reacts with solid sulphur to form polysulphide ions having formulae 2 2- 2- 2 3 4 S , S , S ,  and so on. The equilibrium constant for the formation of 2 2 S  is 12 and for the formation of 2 3 S  is 130, both form S and S2– . Find the equilibrium constant for the formation of 2 3 S  from 2 2 S  and S. 3. Pure PCl3 is introduced into an evacuated chamber and comes to equilibrium at 250°C and 2 atmosphere. The equilibrium mixture contains 40.7% Cl2 by volume. a) What are the partial pressures of each constituent at equilibrium? b) What are Kp and Kc? c) If the gas mixture is expanded to 0.200 atm at 250°C, calculate i) The % of PCl5 dissociated at this equilibrium ii) The partial pressure of each at equilibrium 4. In thermal dissociation of I2 into atoms at 1000°C and 1 atm the dissociation of I2 is 40%. Find out the total pressure at which dissociation of I2 is reduced to 20% at the same temperature. 5. The equilibrium concentration of the reactants and products for the given equilibrium in a two litre container are shown below: 3(g) 2(g) 5(g) 2M 1M 4M 2PCl Cl PCl  i) If 2 mole of Cl2 are added in the container, find the new equilibrium concentration of each. ii) If the equilibrium mixture reported initially is transferred into 4 litre vessel, what would be the new concentrations at equilibrium. 6. H2 and I2 are mixed at 400°C in a 2.0 litre container and when equilibrium was established . [HI] = 0.49M, [H2] = 0.08M and [I2] = 0.06 M. If now an additional 0.4 mole of HI are added, calculate the new equilibrium concentrations.
• 21. Chemistry : Chemical Equilibrium 7. The degree of dissociation is 0.4 at 400K and 1.0 atm for the gaseous reaction PCl5 PCl3 + Cl2. Assuming ideal behaviour of all the gases, calculate the density of equilibrium mixture at 400K and 1.0 atmosphere. (Atomic mass of P = 31.0 and Cl = 35.5) 8. Two solid compounds A and B dissociate into gaseous products at 20°C as i) (s) (g) 2 (g) A A H S  ii) (s) (g) 2 (g) B B H S  At 20°C pressure over excess solid A is 50 mm and that over excess solid B is 68 mm. Find; i) the dissociation constant of A and B. ii) relative number of mole of A and B in the vapour phase over a mixture of the solids A and B. iii) Show that the total pressure of gas over the solid mixture would be 84.4 mm. 9. Solid NH4I on rapid heating in a closed vessel at 257°C develops a constant pressure of 275 mm Hg owing to partial decomposition of NH4I into NH3 and HI but the pressure gradually increases further (when the excess solid residue remains in the vessel) owing to the dissociation of HI. Calculate the final pressure developed at equilibrium. Kp for HI dissociation is 0.015 at 257°C. 10. The moisture content of a gas is often expressed in terms of the dew point. The dew point is the temperature to which the gas must be cooled before the gas becomes saturated with water vapour. At this temperature, water or ice (depending on the temperature) will be deposited on a solid surface. Dew point of H2O is –43°C at which vapour pressure of ice formed is 0.07 mm. Assuming that the CaCl2 owes its desiccating properties to the formation of CaCl22H2O, calculate: i) Kp at that temperature of the reaction ii) G   2 2 (g) 2 2 (s) CaCl 2H O CaCl 2H O  
• 22. Chemistry : Chemical Equilibrium SECTION - II SINGLE CHOICE QUESTIONS 1. c K for the reaction (g) (g) (g) A B 2C  is 3.0 at 400 K. In an experiment a mol of A is mixed with 3 mol of B in a 1-L vessel. At equilibrium 3 mol of C is formed. The value of a will be (a) 4.5 mol (b) 9.5 mol (c) 2.5 mol (d) 3.5 mol. 2. 2 XY dissociates as 2(g) (g) (g) XY XY Y  . When the initial pressure of 2 XY is 600 mm of Hg, the total pressure developed is 800 mm of Hg. p K for the reaction is (a) 200 (b) 50 (c) 100 (d) 150. 3. In a gas-phase reaction 2A B 3C 2D   it was found that when 1.00 mole of A, 2.00 mole of B, and 1.00 mole of D were mixed and allowed to come to equilibrium at 25°C, the resulting mixture contained 0.90 mole of C at a total pressure of 1.00 bar. The value of p K for the reaction is (a) 6.86 (b) 4.86 (c) 68.6 (d) 10.86. 4. Which of the following graphs represents an exothermic reaction? (a) 1/T lnKp (b) 1/T lnKp (c) 1/T lnKp (d) 1/T lnKp 5. 2 4 2 c N O 2NO , K 4  . This reversible reaction is studied graphically as shown in figure. Select the correct statements out of I, II and III. I. Reaction quotient has maximum value at point A Time A D E F G C B Concentration II. Reaction proceeds left to right at a point when 2 4 2 [N O ] [NO ] 0.1M   III. C K Q  when point D or F is reached : (a) I, II (b) II, III (c) I, III (d) I, II, III.
• 23. Chemistry : Chemical Equilibrium 6. For the following equilibrium reaction 2 4(g) 2(g) N O 2NO 2 NO is 50% of the total volume at a given temperature. Hence, vapour density of the equilibrium mixture is : (a) 30.5 (b) 25.0 (c) 23.0 (d) 20.0. 7. Before equilibrium is set-up for the chemical reaction 2 4 2 N O 2NO , vapour density d of the gaseous mixture was measured. If D is the theoretical value of vapour density, variation of x with D/d is given by the graph. What is value of D/d at point A ? D d x A (a) 0 (b) 0.5 (c) 1 (d) 1.5. 8. 10 g of solid 3 CaCO is taken in a vessel of volume 1.0 litre at a temperature of 1000 K when the 3 CaCO decomposed to give solid CaO and gaseous 2 CO . If the p K for 3(s) (s) 2(g) CaCO CaO CO  is 8.21 atm, the pressure of 2 CO observed in the vessel is (a) 7.4 atm (b) 0.74 atm (c) 8.21 atm (d) 0.821 atm. 9. In an aqueous solution of volume 500 ml, when the reaction of 2 2Ag Cu Cu 2Ag     reached equilibrium the 2 [Cu ]  was x M. When 500 ml of water is further added, at the equilibrium 2 [Cu ]  will be (a) 2 x M (b) x M (c) between x M and x/2 M (d) less than x/2 M. 10. In the system (s) (g) (g) AB A B  doubling the quantity of (s) AB would : (a) increase the amount of A to double its value (b) increase the amount of B to double its value (c) increase the amounts of both A and B to double their values (d) cause no change in the amounts of A and B. 11. At a certain temperature p K for the equilibrium (g) 2(g) 2CO CO C  (graphite) is 0.1. Calculate the ratio of partial pressures of CO and 2 CO under equilibrium taking the total pressure to be one atm. (a) 9.1 (b) 9.8 (c) 10.9 (d) 11.1.
• 24. Chemistry : Chemical Equilibrium 12. For A + B  C + D, the equilibrium constant is 1 K and for C + D  A + B, the equilibrium constant is 2 K . The correct relation between 1 K and 2 K is : (a) 1 2 K K 1   (b) 1 2 K (K –1) 0   (c) 1 2 K 1 K  (d) all of these. 13. For the dissociation of 5 PCl into 3 PCl and 2 Cl in gaseous phase reaction, d is the observed vapour density and D the theoretical vapour density with x as degree of dissociation. Variation of D d with x is given by following graph : (a) D d x (b) D d x (c) D d x (d) D d x 14. The equilibrium 3(g) 2(g) 2(g) 2SO 2SO O  was established in a vessel of volume one litre at 600°C taking SO3 initially in the vessel. If one mole of O2 was formed under equilibrium conditions how much of SO3 was taken initially. c K for 2 2 3 2SO O 2SO  is 4.5 at 600°C (a) 4.5 moles (b) 3.91 moles (c) 4.62 moles (d) 6.24 moles. 15. The vapour density of PCl5 is 104.25 but when heated to 230C, its vapour density is reduced to 62. The degree of dissociation of PCl5 at this temperature will be (a) 6.8% (b) 68% (c) 46% (d) 64%. 16. If XY2 dissociates, XY2(g) XY(g) + Y(g). Initial pressure of XY2 is 600mm Hg. The total pressure at equilibrium is 800 mm Hg. Assuming volume of system to remain constant, the value of Kp is: (a) 50 (b) 100 (c) 200 (d) 400 17. For an equilibrium reaction involving gases, the forward reaction is Ist order while the reverse reaction is IInd order. The units of KP for the forward equilibrium is (a) atm (b) atm2 (c) atm–1 (d) atm–2
• 25. Chemistry : Chemical Equilibrium 18. The reaction C2H4(g) + H2(g) C2H6(g); H = –32.7 kcal is carried out in a vessel. The equilibrium concentration of C2H4 can be increased by (a) increasing temperature (b) decreasing the pressure (c) removing some hydrogen (d) all the above 19. For the reaction 1 2 2(g) 2(g) 2 5(g) 2NO O N O  If the equilibrium constant is p K , then the equilibrium constant for the reaction. 2 5(g) 2(g) 2(g) 2N O 4NO O  would be (a) 2 P K (b) P 2/ K (c) 2 P 1/ K (d) P 1/ K 20. For an equilibrium reaction (g) (g) (g) (g) A B C D , H ve      an increase in temperature would cause (a) an increase in the value of eq K (b) a decrease in the value of eq K (c) no change in the value of eq K (d) a change in eq K which cannot be qualitatively predicted. SECTION - III MULTIPLE CHOICE QUESTIONS (with one or more than one answer correct) 1. The dissociation of ammonium carbamate may be represented by the reaction 2 4(s) 3(g) 2(g) NH COONH 2NH CO  H° for the forward reaction is negative. The equilibrium will shift from right to left if there is (a) a decrease in pressure (b) an increase in temperature (c) an increase in the concentration of ammonia (d) an increase in the concentration of carbondioxide
• 26. Chemistry : Chemical Equilibrium 2. Catalyst 2(g) 2(g) 3 500 C N 3H 2NH heat    In the above reaction, the direction of equilibrium will be shifted to the right by (a) increasing the concentration of nitrogen (b) compressing the reaction mixture (c) removing the catalyst (d) decreasing the concentration of ammonia 3. The equilibrium of which of the following reactions will not be disturbed by the addition of an inert gas at constant volume (a) 2(g) 2(g) (g) H I 2HI  (b) 2 4(g) 2(g) N O 2NO (c) (g) 2(g) 3 (g) CO 2H CH OH  (d) (s) 2 (g) (g) 2(g) C H O CO H   4. Which of the following factors will increase solubility of NH3(g) in H2O 3(g) 2 (aq) 4 (aq) NH H O NH OH  (a) increase in pressure (b) addition of water (c) liquefaction of NH3 (d) decrease in pressure 5. For the gas phase reaction C2H4(g) + H2(g) C2H6(g); H = –32.7 kcal carried out in a vessel, the equilibrium concentration of C2H4 can be increased by (a) increasing temperature (b) decreasing pressure (c) removing some H2 (d) adding some C2H6 6. When NaNO3 is heated in a closed vessel, O2 is liberated and NaNO2 is left behind. At equilibrium. (a) addition of NaNO2 favours reverse reaction (b) addition of NaNO3 favours forward reaction (c) increasing temperature favours forward reaction (d) increasing pressure favours forward reaction 7. For which of the following statements about the reaction quotient, Q, are correct? (a) the reaction quotient, Q, and the equilibrium constant always have the same numerical value (b) Q may be >< = Keq (c) Q (numerical value) varies as reaction proceeds (d Q = 1 at equilibrium 8. Which of the following factors will affect solubility of CaO in H2O? (a) pressure (b) temperature (c) addition of water (d) volume
• 27. Chemistry : Chemical Equilibrium 9. For the following equilibrium 4 (s) 3(g) 2 (g) NH HS NH H S  partial pressure of NH3 will increase: (a) if NH3 is added after equilibrium is established (b) if H2S is added after equilibrium is established (c) temperature is increased (d) volume of the flask is decreased 10. Which are true statements for the following equilibrium? 2 (l) 2 (g) H O H O (a) increase in pressure will result in the formation of more liquid water (b) increase in pressure will increase boiling point (c) decrease in pressure will vaporize H2O(l) to a greater extent (d) increase in pressure will liquefy steam. SECTION - IV COMPREHENSION TYPE QUESTIONS  WRITE-UP I A chemical reaction in the laboratory is carried out under the condition of constant temperature and pressure. The condition of spontaneity in terms of enthalpy and entropy are, respectively, H  0 and S  0 whereas condition of equilibrium are, respectively, H = 0 and S = 0. An endothermic reaction is driven by increase in entropy i.e. increase in disorderness. The two criteria combined together, the condition of reversibility and irreversibility are as follows: GP, T = H – TS  0, where equal to sign refers to reversibility and less than sign refers to irreversibility. Reversibility is the condition of equilibrium whereas irreversibility is the condition of spontaneity. A reversible reaction is characterised by equilibrium constant (K), the magnitude of which measures the position of equilibrium i.e. how far a chemical reaction will go to completion before attainment of equilibrium. The position of equilibrium constant of a reaction with temperature is given by Van’t Hoff equation of thermodynamics which is as follows: 2 dln K H dT RT   This equation can be integrated assuming H to be independent of temperature. Within the short interval of temperature. For a gaseous equilibria k can be expressed Kc and Kp related as g n p c K K (RT)  
• 28. Chemistry : Chemical Equilibrium Where the symbols have their usual meanings. ng of a reaction can be found out from the unit of equilibrium constant, though it is not customary to write unit of equilibrium constant. Another way of finding ng of a reaction is to use the equation H = E + ng RT Where the terms have their usual meanings. Knowing K and reaction quotient it is possible to calculate free energy change of a reaction using the equation: G = G0 + RTlnQ At equilibrium Q = K and GP, T = 0 So, G0 = –RTlnK where G0 is the free energy change of the reaction in the standard state. The standard state of a substance is defined as the state of unit activity at 25°C. In the case of solution of a substance the activity is taken to be molar concentration while for a gas it is pressure in atm. The standard free energy of an element is taken to be zero. 1. At 27°C the heat of reaction at constant pressure is 600 cals more than that at constant volume. The ratio of Kp to Kc of the reaction is: (a) 24.63 (b) (24.63)2 (c) 0.6 (d) 0.36 2. For the reaction: 4 (s) 3(g) 2 (g) NH HS NH H S  Kp and Kc are interrelated as (a) Kp = KcRT (b) Kp = c 2 K (RT) (c) Kp = Kc(RT)2 (d) Kp = Kc/RT 3. The Kp of a reaction is 10 atm–2 at a temperature T on Kelvin scale. Hence (a) Kp = Kc (b) Kp  Kc (c) Kp  Kc (d) Can’t be predicted  WRITE-UP II Life at high attitudes and hemoglobin production: In the human body, countless chemical equilibria must be maintained to ensure physiological well being. Transport of oxygen by blood depends on the reversible combination of oxygen with haemoglobin. In blood, haemoglobin, oxygen and oxyhaemoglobin are in equilibrium. 2 2 haemoglobin (Hb) + O Oxyhaemoglobin [Hb(O )] The equilibrium constant is, 2 c 2 [Hb(O )] K [Hb][O ] 
• 29. Chemistry : Chemical Equilibrium The pH of blood stream is maintained by a proper balance of H2CO3 and NaHCO3 concentrations. According to Henderson’s equation a [salt] pH log K log [acid]    An important component of blood is the buffer combination of 2 4 H PO ion and the 4 HPO ion. Consider blood with a pH of 7.44. Given 1 3 a K 6.9 10   , 2 8 a K 6.2 10   and 3 13 a K 4.8 10   . 4. What volume of 5M NaHCO3 solution should be mixed with a 10 mL sample of blood which is 2M in H2CO3, in order to maintain a pH of 7.47? Ka for H2CO3 in blood is 7.8 × 10–7 . (a) 78.32 mL (b) 88.32 mL (c) 68.32 mL (d) None 5. What is the ratio of 2 4 2 4 [H PO ] [HPO ]   ? (a) 0.59 (b) 0.69 (c) 0.79 (d) None 6. What will be the pH, when 25% of the 2 4 HPO  ions are converted to 2 2 4 H PO  ion? (a) 7.16 (b) 8.16 (c) 9.16 (d) None MATCHING TYPE QUESTIONS 7. Match the following Column-I Column-I (p) Homogeneous equilibrium (w) Le-Chatelier (q) Law of mass action (x) CaCO3 CaO + CO2 (r) Active mass (y) PCl5 PCl3 + Cl2 (s) Dynamic equilibrium (z) Molar concentration Codes (p) (q) (r) (s) (a) (w) (z) (x) (y) (b) (z) (w) (x) (y) (c) (z) (w) (y) (x) (d) (w) (z) (y) (x)
• 30. Chemistry : Chemical Equilibrium 8. Match the following Column-I Column-I (p) 2(g) 2(g) 3(g) A 3B 2AB  (w) 1 2 p c K K (RT)  (q) 2(g) 2(g) (g) A B 2AB  (x) 0 p c K K (RT)  (r) (s) 2(g) 3(g) A 1.5B AB  (y) 1 2 p c K K (RT)   (s) 3(g) (g) 2(g) AB AB B  (z) 2 p c K K (RT)  Codes (p) (q) (r) (s) (a) (z) (w) (y) (x) (b) (z) (x) (y) (w) (c) (x) (z) (y) (w) (d) (z) (x) (w) (y) MISCELLANEOUS TYPE QUESTIONS The following questions consist of two statements, one labelled as ASSERTION (A) and REASON (R). Use the following key to chose the correct appropriate answer. (a) (A) and (R) both are correct and (R) is the correct explanation of (A) (b) (A) and (R) both are correct but (R) is not the correct explanation of (A) (c) (A) is correct but (R) is incorrect. (d) (A) is incorrect but (R) is correct ASSERTION (A) REASONING (R) 9. Even at constant temperature, on changing the stoichiometric coefficients of the reaction, the equilibrium constant changes. Because the equilibrium depends upon the ratio of concentration of product and reaction in which the stoichiometric coefficients becomes, thereby the equilibrium constant depends upon the stoichiometric coefficients. 10. For the reaction (g) (g) (g) (g) A B C D   at the given temperature, there will be no effect by addition of inert gas either at constant pressure or at constant volume. For the reaction where n = 0, there is effect of inert gas either at constant volume or at constant pressure because Kp becomes purely a number. 11. for the physical equilibrium ice water on increasing temperature and increasing pressure more water will form. Since forward reaction is endothermic in nature and volume of water is lesser than that of the volume of ice.
• 31. Chemistry : Chemical Equilibrium 12. The equilibrium remains undisturbed by the addition of inert gas at constant volume. Since Kp remains constant at constant volume and at constant temperature by the addition of inert gas at equilibrium. 13. The catalyst does not alter the equilibrium constant at constant temperature Because for the catalysed reaction and uncatalysed reaction H remains same and equilibrium constant depends on H and temperature. 14. The addition of inert gas for the reaction N2(g) + 3H2 3(g) 2NH at constant pressure decreases the rate of formation of NH3. Because the addition of inert gas at constant pressure favours the backward reaction. 15. The reaction : 2NO(g)+ O2 2NO2 is favoured in the forward direction with increase of pressure. The reaction is exothermic. SECTION - V QUESTIONS ASKED IN IIT-JEE OBJECTIVE A. Fill in the Blanks 1. For a given reversible reaction at a fixed temperature, equilibrium constant Kp and Kc are related by ………….. 2. A ten–fold increase in pressure on the reaction, N2(g) + 3H2(g) 2NH3(g) at equilibrium, results in …………..in Kp. 3. For a gaseous reaction 2BA, the equilibrium constant Kp is ………….. to/than Kc. B. True / False 1. When a liquid and its vapour are at equilibrium and the pressure is suddenly decreased, cooling occurs. 2. If equilibrium constant for the reaction, 2 2 A B 2AB  , is K, then for the backward reaction 2 2 1 1 AB A B 2 2  the equilibrium constant is 1 K . 3. The rate of an exothermic reaction increases with increasing temperature.
• 32. Chemistry : Chemical Equilibrium C. Only one option is correct (Objective Questions) 1. Pure ammonia is placed in a vessel at a temperature where its dissociation constant () is appreciable. At equilibrium : (a) Kp does not change significantly with pressures (b)  does not change with pressure (c) concentration of NH3 does not change with pressure. (d) concentration of hydrogen is less than that of nitrogen 2. A certain weak acid has a dissociation constant of 1.0 × 10–4 . The equilibrium constant for its reaction with strong base is : (a) 1.0 × 10–4 (b) 1.0 × 10–10 (c) 1.0 × 1010 (d) 1.0 × 1014 3. For the chemical reaction 3 3X(g) Y(g) X Y(g)  the amount of X3Y at equilibrium is affected by : (a) temperature and pressure (b) temperature only (c) pressure only (d) temperature pressure and catalyst 4. For the reversible reaction, 2 2 3 N (g) 3H (g) 2NH (g)  at 500°C, the value of Kp is 1.44 × 10–5 when partial pressure is measured in atmosphere. The corresponding value of Kc with concentration in mol/L is : (a) 5 2 1.44 10 (0.082 500)     (b) 5 2 1.44 10 (8.314 773)     (c) 5 2 1.44 10 (0.082 773)    (d) 5 2 1.44 10 (0.082 773)     5. At constant temperature, the equilibrium constant (Kp) for the decomposition reaction, 2 4 2 N O 2NO , is expressed by 2 p 2 (4x P) K (1 x )   , where P = Pressure, x= extent of decomposition. Which one of the following statement is true : (a) Kp increases with increases of P (b) Kp increases with increases of x (c) Kp increases with decrease of x (d) Kp remains constant with change in P and x
• 33. Chemistry : Chemical Equilibrium 6. Consider the following equilibrium in a closed container 2 4 2 N O (g) 2NO (g) At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements hold true regarding the equilibrium constant (Kp) and degree of dissociation () ? (a) Neither Kp nor  changes (b) Both Kp and  changes (c) Kp changes but  does not change (d) Kp does not change but  changes 7. Haber-Bosch process for the formation of NH3 at 298K is 2(g) 2(g) 3(g) p N 3H 2NH ; H 46.0kJl K      41 The correct statement is (a) On adding N2, the equilibrium will spontaneously shift to the right and S will be positive (b) Even though at 298 K, the yield is more, still industrial manufacture of NH3 is carried out at 400 K (Kp = 41). This is because at 298 K, the catalyst increases the rate by factor of two for both forward and backward reaction. But at 400K, the iron catalyst increases the rate of forward reaction by a factor of two and that of backward reaction by a factor of 1.7 (c) The condition for equilibrium is 2 2 3 N H NH G 3G 2G   where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent. (d) Catalyst will not affect the rate of forward or backward reaction. 3 3 3 f Ag NH [Ag(NH )] ; K 6.8 10       3 3 3 3 2 f [Ag(NH )] NH [Ag(NH ) ] ; K 1.6 10       Over all formation constant (a) 1.08 × 10-6 (b) 6.8 × 10–3 (c) 1.08 × 10–5 (d) 8.4 × 10–3 D. More than one options are correct (Objective Questions) 1. For the gas phase reaction. 2 4 2 2 6 C H H C H  (H = –32.7 kcal) Carried out in a vessel, the equilibrium concentration of C2H4 can be increased by : (a) increasing the temperature (b) decreasing the pressure (c) removing some H2 (d) adding some C2H6
• 34. Chemistry : Chemical Equilibrium 2. When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At equilibrium. (a) addition of NaNO2 favours reverse reaction (b) addition of NaNO3 favours forward reaction (c) removing temperature favours forward reaction (d) increasing pressure favours reverse reaction 3. The equilibrium SO2Cl2(g) SO2(g) + Cl2(g) is attained at 25°C in a closed container and an inert gas, helium is introduced. Which of the following statements are correct? (a) Concentration of SO2, Cl2 and SO2Cl2 change (b) More chlorine is formed (c) Concentration of SO2 is reduced (d) All are incorrect 4. For the reaction; 5 3 2 PCl (g) PCl (g) Cl (g)  The forward reaction at constant temperature is favoured by : (a) introducing an inert gas at constant volume (b) introducing chlorine gas at constant volume (c) introducing an inert gas at constant pressure (d) increasing the volume of the container (e) introducing PCl5 at constant volume SUBJECTIVE 1. At temperature T, a compound AB2(g) dissociates according to the reaction: 2AB2(g) 2AB(g) + B2(g) with a degree of dissociation ‘x’ which is small compared to unity. Deduce the expression for ‘x’ in terms of the equilibrium constant Kp and the total pressure P. 2. The KC for A2(g) + B2(g) 2AB(g) at 100°C is 49. If one litre flask containing one mole of A2 is connected with a two litre flask containing 2 mole of B2, how many mole of AB will be formed at 100°C? 3. The Kp values for the reaction; H2 + I2 2HI, at 460°C is 49. If the initial pressure of H2 and I2 is 0.5 atm respectively, determine the partial pressure of each gas at equilibrium. 4. One mole of H2, two mole of I2 and three mole of HI are injected in one litre flask. What will be the concentration of H2, I2 and HI at equilibrium at 500°C. KC for reaction; 2 2 H I 2HI is 45.9  5. KC for CO(g) + H2O(g) CO2(g) + H2(g) at 986°C is 0.63. A mixture of 1 mole of H2O(g) and 3 mole CO(g) is allowed to react to come to an equilibrium. The equilibrium pressure is 2.0 atm.
• 35. Chemistry : Chemical Equilibrium i) How many mole of H2 are present at equilibrium? i) Calculate partial pressure of each gas at equilibrium 6. A sample of air consisting of N2 and O2 was heated to 2500K until the equilibrium 2(g) 2(g) (g) N O 2NO  was established with an equilibrium constant KC = 2.1 × 10–3 . At equilibrium, the moles % of NO were 1.8. Estimate the initial composition of air in mole fraction of N2 and O2. 7. At 700K, CO2 and H2 react to form CO and H2O. For this reaction KC is 0.11. If a mixture of 0.45 mole of CO2 and 0.45 mole of H2 is heated to 700K. a) Find out amount of each gas at equilibrium. b) When equilibrium has been reached, another 0.34 mole of CO2 and 0.34 mole of H2 are added to the reaction mixture. Find composition of mixture at new equilibrium 8. The Kp for the reaction N2O4 2NO2 is 640 mm at 773K. Calculate the percentage dissociation of N2O4 at equilibrium pressure of 160 mm. At what pressure, the dissociation will be 50%? 9. An equilibrium mixture at 300K contains N2O4 and NO2 at 0.28 and 1.1 atmosphere respectively. If the volume of container is doubled, calculate the new equilibrium pressureof two gases. 10. At same temperature and under a pressure of 4 atm, PCl5 is 10% dissociated. Calculate the pressure at which PCl5 will be 20% dissociated, temperature remaining same. 11. The degree of dissociation is 0.4 at 400K and 1.0 atm for the gaseous reaction PCl5 PCl3 + Cl2. Assuming ideal behaviour of all the gases, calculate the density of equilibrium mixture at 400K and 1.0 atmosphere. (Atomic mass of P = 31.0 and Cl = 35.5) 12. Calculate the value of log Kp for the reaction 2(g) 2(g) 3(g) N 3H 2NH  at 25°C. The standard enthalpy of formation of NH3(g) is – 46kJ and standard entropies of N2(g), H2(g) and NH3(g) are 191, 130, 192 JK–1 mol–1 respectively (R = 8.3 JK–1 and mol–1 ). 13. Calculate the percent dissociation of H2S(g) if 0.1 mole of H2S is kept in 0.4 litre vessel at 1000K. For the reaction 2H2S(g) 2H2(g) + S2(g) the value of KC is 1.0 × 10–6 . 14. When 3.06g of solid NH4HS is introduced into a two litre evacuated flask at 27°C, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide. i) Calculate the KC and Kp for the reaction at 27°C, ii) What would happen to the equilibrium when more solid NH4HS is introduced into the flask? 15. A vessel at 1000K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on addition of graphite. Calcualte the value of K, if total pressure at equilirum is 0.8 atm. 16. At 817°C, Kp for the reaction between CO2(g) and excess hot graphite(s) is 10 atm.
• 36. Chemistry : Chemical Equilibrium a) What are the equilibrium concentration of the gases at 817°C and a total pressure of 5 atm? b) At what total pressure, the gas contains 5% CO2 by volume? 17. For the reaction, CO(g) + 2H2(g) CH3OH(g); H2 is introduced into a five litre flask at 327°C, containing 0.2 mole of CO(g) and a catalyst till the pressure is 4.92 atmosphere. At this point 0.1 mole of CH3OH(g) is formed. Calculate the KC and Kp. 18. When 1-pentyne (A) is treated with 4N alcoholic KOH at 175°C, it is converted slowly into an equilibrium mixture of 1.3% 1-pentyne (A), 95.2% 2-pentyne (B)_ and 3.5%, of 1- 2pentadiene (C). The equilibrium was maintained at 175°C. Calculate G0 for the following equilibria.: 0 1 B A : G ?   0 2 B C: G ?   From the calculated value 0 1 G  and 0 2 G  indicate the order of stability of (A), (B) and (C). 19. In the following equilibrium 2 4 2 N O (g) 2NO (g) When 5 moles of each are taken, the temperature is kept at 298 K the total pressure was found to be 20 bar. Given that o f 2 4) o f 2 G (N O 100kJ G NO 50kJ     (i) Find G of the reaction (ii) The direction of the reaction in which the equilibrium shifts
• 37. Chemistry : Chemical Equilibrium ANSWERS EXERCISE – 1 1. B 2. N2O4 3. False 4. 0.333 5. 4NO(g) + 6H2O(g)  4NH3(g) + 5O2(g) 6. n = 1 7. While writing chemical equilibrium equation, molar concentration or partial pressure of the reactants and products are taken as unity thus, a) 2 2 2 CO c p 2 CO [CO] p K ; K [CO ] p   b) 2 2 c 2 [Hg ] K [Hg ]    c) 2 2 3 3 H 2 c p 3 3 2 H O p [H ] K , K [H O] p   d) 2 2 2 2 c 2 2 p H O K [H ] [O ], K p p   8. 1.8 × 10–5 EXERCISE – 2 1. a) 0; b) 1.64.06M–1 2. right to left 3. a) forward; b) backward; c) no EXERCISE – 3 1. a) Kp = KcRT; b) Kp = KcRT; c) Kp = Kc 2. 3.02 × 10–5 3. 1.067 4. Kp = 0.0313, Kc = 1.28 × 10–3 EXERCISE – 4 1. Shifts in backward direction 2. [H+ ]1 = 4.2 × 10–3 , [H+ ]2 = 1.8 × 10–5 . There is decrease in [H+ ] 3. a) decreases; b) decreases; c) increases 4. a) no effect; b) forward; c) backward EXERCISE – 5 1. a) 80; b) 0.15; c) 26.08 2. a) 41.4; b) 82.8; c) 11.1% 3. a) 2.57 atm; b) 4.25% EXERCISE – 6 1. 500K 2. 16.06kJ 3. a) – 2.42kJ; b) K = 0.982 SECTION - I
• 38. Chemistry : Chemical Equilibrium SUBJECTIVE QUESTIONS Level-I 1. Kc = 3.1420 2 0.167 atm 3 57.92 4. CO = 0.686 atm, CO2 = 0.314 atm 5. 2.55 (atm)3 6. 1.67 M 7. 33% 8. 0.19 atm 9. 34.08g 10. 1.145 atm Level-II 1. 2 2 2 H O CO CO H P P 7.34 atm, P 12.66 atm, P 2.66 atm     2. Kc = 4.012 4. 0.3136 5. 9.633g 6. 5.206 atm 7. 0.45 8. 0.7837 mole 9. 75 cm 10. 33.13%, Kp = 0.20 Level-III 1. 61.3% 2. 10.83 3. a) 2 3 5 Cl PCl PCl P P 0.814, P 0.372atm    b) 1.78 atm, 0.04 mol litre–1 c) i) 94.8%, ii) 0.0053, 0.097 atm 4. 4.57 atm 5. i) PCl3 = 1.5M, Cl2 = 1.5M ii) PCl3 = 1.225M, Cl2 = 0.725 M 6. H2 = 0.1516M, I2 0 0.1316M, HI = 0.5468M 7. 4.53 g/litre 8. (i) A = 625 (mm)2 , B = 1156 (mm)2 (ii) 0.5407 (iii) 84.38 9. 307.46 mm 10. (i) 1.178 × 108 atm (ii) – 35.44 kJ SECTION - II SINGLE CHOICE QUESTIONS
• 39. Chemistry : Chemical Equilibrium 1. (d) 2. (c) 3. (a) 4. (d) 5. (b) 6. (a) 7. (c) 8. (c) 9. (d) 10. (d) 11. (c) 12. (a) 13. (d) 14. (d) 15. (b) 16. (b) 17. (a) 18. (d) 19. (c) 20. (a) SECTION - III MULTIPLE CHOICE QUESTIONS 1. (b, c, d) 2. (a, b, d) 3. (a, b, c, d) 4. (a, b, c) 5. (a, b, c, d) 6. (c) 7. (b, c) 8. (b, c) 9. (a, c, d) 10. (a, b, c, d) SECTION - IV COMPREHENSION TYPE QUESTIONS 1. (a) 2. (c) 3. (b) 4. (b) 5. (a) 6. (b) MATCHING TYPE QUESTIONS 7. (a) 8. (b) ASSERTION – REASONING TYPE QUESTIONS 9. (a) 10. (a) 11. (a) 12. (d) 13. (a) 14. (a) 15. (b)
• 40. Chemistry : Chemical Equilibrium SECTION - V QUESTIONS ASKED IN IIT–JEE OBJECTIVE A. Fill in the Blanks 1. g n p c K K (RT)   2. no change 3. less B. True False 1. T 2. F 3. F C. Objective Questions with one option correct 1. (a) 2. (c) 3. (a) 4. (d) 5. (d) 6. (d) 7. (c) 8. (c) D. Objective Questions with one or more than one option correct 1. (a, b, c, d) 2. (c, d) 3. (d) 4. (c, d, e) SUBJECTIVE 1. p 2K P 2. 1.86 mole 3. H2 = 0.111 atm, I2 = 0.111 atm, HI = 0.778 atm 4. [H2] = 0.316 mole litre–1 , [I2] = 1.316 mol litre–1 , [HI] = 4.368 mol litre–1 5. i) 0.681; ii) CO2 = 0.34 atm, CO = 1.16 atm; iii) H2O = 0.16 atm 6. N2 = 79%, O2 = 21 7. a) CO2 = H2 = 0.338 mole, CO = H2O = 0.112 mole b) CO2 = H2 = 0.593 mole, CO = H2O = 0.197 mole 8. 480 mm 9. N2O4 = 0.095 atm, NO2 = 0.64 atm 10. 0.969 atm 11. 4.53 g/litre
• 41. Chemistry : Chemical Equilibrium 12. 5.845 13. 2% 14. i) KC = 8.1 × 10–5 mol2 litre2 , Kp = 4.90 × 10–2 atm2 15. 1.8 atm 16. a) CO = 0.041 mole litre–1 , CO2 = 0.015 mol litre–1 , (b) 0.145 atm 17. Kc = 279.64 litre2 mol–2 , Kp = 0.115 atm–2 18. G1 0 = 16.178kJ, G2 0 = 12.282 kJ, B > A, B > C, B > C > A 19. (i) 56.276 L-atm. (ii) the reaction will proceed in backward direction
Idioma actualEnglish
Español
Portugues
Français
Deutsche