{"id":866391,"date":"2025-08-08T18:09:44","date_gmt":"2025-08-08T12:39:44","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=866391"},"modified":"2025-08-08T18:09:44","modified_gmt":"2025-08-08T12:39:44","slug":"ncert-solutions-class-11-chemistry-part-i-chapter-6-equilibrium","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-i-chapter-6-equilibrium\/","title":{"rendered":"NCERT Solutions Class 11 Chemistry (Part-1) Chapter 6: Equilibrium"},"content":{"rendered":"\n

In this unit, you have learned about how reversible reactions reach a dynamic balance and how to express this using equilibrium constants, along with the factors affecting equilibrium, such as temperature and pressure, along with Le Chatelier\u2019s Principle, all of which are essential for understanding reactions in chemistry and real-life applications. Below, we have presented some in-text exercises taken from the NCERT Class 11 Chemistry (Part I), Chapter 6: Equilibrium.<\/p>\n\n\n\n\n\n\n

Explore Notes of Class 11 Chemistry<\/strong><\/p>\n\n\n\n

Chapter 1<\/a><\/strong><\/td>Chapter 2<\/a><\/strong><\/td>Chapter 3<\/a><\/strong><\/td>Chapter 4<\/a><\/strong><\/td>Chapter 5<\/a><\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

NCERT Solutions Class 11 Chemistry (Part-1) Chapter 6: Equilibrium<\/h2>\n\n\n\n

Below, we have provided you with the exercises mentioned in the NCERT Class 11 Chemistry (Part-1) Chapter 6: Equilibrium.<\/p>\n\n\n\n

Exercises<\/h2>\n\n\n\n
    \n
  1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
    a) What is the initial effect of the change on vapour pressure?
    b) How do rates of evaporation and condensation change initially?
    c) What happens when equilibrium is finally restored, and what will be the final vapour pressure?<\/li>\n\n\n\n
  2. What is Kc for the following equilibrium when the equilibrium concentrations are:
    [SO\u2082] = 0.60 M, [O\u2082] = 0.82 M and [SO\u2083] = 1.90 M?
    2SO\u2082(g) + O\u2082(g) \u21cc 2SO\u2083(g)<\/li>\n\n\n\n
  3. At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms.
    I\u2082(g) \u21cc 2I(g)
    Calculate Kp for the equilibrium.<\/li>\n\n\n\n
  4. Write the expression for the equilibrium constant, Kc, for each of the following reactions:
    (i) 2NOCl(g) \u21cc 2NO(g) + Cl\u2082(g)
    (ii) 2Cu(NO\u2083)\u2082(s) \u21cc 2CuO(s) + 4NO\u2082(g) + O\u2082(g)
    (iii) CH\u2083COOC\u2082H\u2085(aq) + H\u2082O(l) \u21cc CH\u2083COOH(aq) + C\u2082H\u2085OH(aq)
    (iv) Fe\u00b3\u207a(aq) + 3OH\u207b(aq) \u21cc Fe(OH)\u2083(s)
    (v) I\u2082(s) + 5F\u2082 \u21cc 2IF\u2085<\/li>\n\n\n\n
  5. Find the value of Kc for each of the following equilibria from the value of Kp:
    (i) 2NOCl(g) \u21cc 2NO(g) + Cl\u2082(g); Kp = 1.8 \u00d7 10\u207b\u00b2 at 500 K
    (ii) CaCO\u2083(s) \u21cc CaO(s) + CO\u2082(g); Kp = 167 at 1073 K<\/li>\n\n\n\n
  6. For the reaction:
    NO(g) + O\u2083(g) \u21cc NO\u2082(g) + O\u2082(g); Kc = 6.3 \u00d7 10\u00b9\u2074 at 1000 K
    What is Kc for the reverse reaction?<\/li>\n\n\n\n
  7. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression.<\/li>\n\n\n\n
  8. 2N\u2082(g) + O\u2082(g) \u21cc 2N\u2082O(g)
    A mixture of 0.482 mol N\u2082 and 0.933 mol of O\u2082 is placed in a 10 L vessel at a temperature where Kc = 2.0 \u00d7 10\u207b\u00b3\u2077. Determine the composition of the equilibrium mixture.<\/li>\n\n\n\n
  9. 2NO(g) + Br\u2082(g) \u21cc 2NOBr(g)
    0.087 mol NO and 0.0437 mol Br\u2082 are mixed. At equilibrium, 0.0518 mol NOBr is formed. Calculate equilibrium amounts of NO and Br\u2082.<\/li>\n\n\n\n
  10. 2SO\u2082(g) + O\u2082(g) \u21cc 2SO\u2083(g); Kp = 2.0 \u00d7 10\u00b9\u2070 at 450 K
    What is Kc at this temperature?<\/li>\n\n\n\n
  11. 2HI(g) \u21cc H\u2082(g) + I\u2082(g)
    HI is placed at 0.2 atm; at equilibrium, its pressure is 0.04 atm. Calculate Kp.<\/li>\n\n\n\n
  12. N\u2082(g) + 3H\u2082(g) \u21cc 2NH\u2083(g); Kc = 1.7 \u00d7 10\u00b2 at 500 K
    1.57 mol N\u2082, 1.92 mol H\u2082, 8.13 mol NH\u2083 in 20 L vessel. Is the mixture at equilibrium? If not, what is the direction of the net reaction?<\/li>\n\n\n\n
  13. Write the balanced chemical equation for the equilibrium expression:
    Kc = [NH\u2083]\u2074 \/ ([NO]\u2076 [H\u2082O]\u2074)<\/li>\n\n\n\n
  14. H\u2082O(g) + CO(g) \u21cc H\u2082(g) + CO\u2082(g)
    1 mol H\u2082O and 1 mol CO in a 10 L vessel at 725 K. 40% H\u2082O reacts. Calculate Kc.<\/li>\n\n\n\n
  15. H\u2082(g) + I\u2082(g) \u21cc 2HI(g); Kc = 54.8 at 700 K
    If 0.5 mol L\u207b\u00b9 of HI is present at equilibrium, calculate [H\u2082] and [I\u2082].<\/li>\n\n\n\n
  16. 2ICl(g) \u21cc I\u2082(g) + Cl\u2082(g); Kc = 0.14
    Initial [ICl] = 0.78 M. Calculate equilibrium concentrations.<\/li>\n\n\n\n
  17. C\u2082H\u2086(g) \u21cc C\u2082H\u2084(g) + H\u2082(g); Kp = 0.04 atm at 899 K
    C\u2082H\u2086 is placed at 4.0 atm. Find the equilibrium concentration.<\/li>\n\n\n\n
  18. CH\u2083COOH(l) + C\u2082H\u2085OH(l) \u21cc CH\u2083COOC\u2082H\u2085(l) + H\u2082O(l)
    (i) Write Qc.
    (ii) Initial: 1 mol CH\u2083COOH, 0.18 mol C\u2082H\u2085OH; equilibrium: 0.171 mol ester. Find Kc.
    (iii) Initial: 1 mol acid, 0.5 mol alcohol; equilibrium: 0.214 mol ester. Has equilibrium been reached?<\/li>\n\n\n\n
  19. PCl\u2085(g) \u21cc PCl\u2083(g) + Cl\u2082(g); Kc = 8.3 \u00d7 10\u207b\u00b3
    [PCl\u2085] = 0.5 \u00d7 10\u207b\u00b9 mol\/L at 473 K. Find [PCl\u2083] and [Cl\u2082].<\/li>\n\n\n\n
  20. FeO(s) + CO(g) \u21cc Fe(s) + CO\u2082(g); Kp = 0.265 atm at 1050 K
    Initial: pCO = 1.4 atm, pCO\u2082 = 0.80 atm. Find equilibrium partial pressures.<\/li>\n\n\n\n
  21. N\u2082(g) + 3H\u2082(g) \u21cc 2NH\u2083(g); Kc = 0.061
    [ N\u2082 ] = 3.0 M, [ H\u2082 ] = 2.0 M, [ NH\u2083 ] = 0.5 M. Is this at equilibrium? If not, in which direction does the reaction proceed?<\/li>\n\n\n\n
  22. 2BrCl(g) \u21cc Br\u2082(g) + Cl\u2082(g); Kc = 32 at 500 K
    Initial [BrCl] = 3.3 \u00d7 10\u207b\u00b3 M. Find equilibrium [BrCl].<\/li>\n\n\n\n
  23. C(s) + CO\u2082(g) \u21cc 2CO(g)
    At 1127 K and 1 atm, CO is 90.55% by mass. Calculate Kc.<\/li>\n\n\n\n
  24. NO(g) + \u00bdO\u2082(g) \u21cc NO\u2082(g)
    Given \u0394G\u00b0 values: NO\u2082 = 52.0, NO = 87.0, O\u2082 = 0 (kJ\/mol). Calculate (a) \u0394G\u00b0, (b) K for the reaction at 298 K.<\/li>\n\n\n\n
  25. Predict the change in the number of moles when the pressure is decreased by increasing volume:
    (a) PCl\u2085(g) \u21cc PCl\u2083(g) + Cl\u2082(g)
    (b) CaO(s) + CO\u2082(g) \u21cc CaCO\u2083(s)
    (c) 3Fe(s) + 4H\u2082O(g) \u21cc Fe\u2083O\u2084(s) + 4H\u2082(g)<\/li>\n\n\n\n
  26. Which reactions will be affected by increasing pressure? Indicate forward\/backwards:
    (i) COCl\u2082(g) \u21cc CO(g) + Cl\u2082(g)
    (ii) CH\u2084(g) + 2S\u2082(g) \u21cc CS\u2082(g) + 2H\u2082S(g)
    (iii) CO\u2082(g) + C(s) \u21cc 2CO(g)
    (iv) 2H\u2082(g) + CO(g) \u21cc CH\u2083OH(g)
    (v) CaCO\u2083(s) \u21cc CaO(s) + CO\u2082(g)
    (vi) 4NH\u2083(g) + 5O\u2082(g) \u21cc 4NO(g) + 6H\u2082O(g)<\/li>\n\n\n\n
  27. H\u2082(g) + Br\u2082(g) \u21cc 2HBr(g); K = 1.6 \u00d7 10\u2075 at 1024 K
    10 bars of HBr introduced. Find equilibrium pressures.<\/li>\n\n\n\n
  28. CH\u2084(g) + H\u2082O(g) \u21cc CO(g) + 3H\u2082(g)
    (a) Write the Kp expression.
    (b) How does Kp and composition change with:
    (i) increasing pressure
    (ii) increasing temperature
    (iii) adding a catalyst?<\/li>\n\n\n\n
  29. 2H\u2082(g) + CO(g) \u21cc CH\u2083OH(g)
    Describe the effect on the equilibrium of:
    a) addition of H\u2082
    b) addition of CH\u2083OH
    c) removal of CO
    d) removal of CH\u2083OH<\/li>\n\n\n\n
  30. PCl\u2085(g) \u21cc PCl\u2083(g) + Cl\u2082(g); Kc = 8.3 \u00d7 10\u207b\u00b3 at 473 K
    \u0394H\u00b0 = +124.0 kJ mol\u207b\u00b9
    a) Write the Kc expression.
    b) Value of Kc for reverse reaction.
    c) Effect on Kc if:
    (i) more PCl\u2085 added
    (ii) pressure increased
    (iii) temperature increased<\/li>\n\n\n\n
  31. CO(g) + H\u2082O(g) \u21cc CO\u2082(g) + H\u2082(g); Kp = 10.1 at 400\u00b0C
    Initial: pCO = pH\u2082O = 4.0 bar. Find pH\u2082 at equilibrium.<\/li>\n\n\n\n
  32. Predict which reactions will have appreciable concentrations of both reactants and products:
    a) Cl\u2082(g) \u21cc 2Cl(g); Kc = 5 \u00d7 10\u207b\u00b3\u2079
    b) Cl\u2082(g) + 2NO(g) \u21cc 2NOCl(g); Kc = 3.7 \u00d7 10\u2078
    c) Cl\u2082(g) + 2NO\u2082(g) \u21cc 2NO\u2082Cl(g); Kc = 1.8<\/li>\n\n\n\n
  33. The value of Kc for the reaction 3O\u2082(g) \u21cc 2O\u2083(g) is 2.0 \u00d7 10\u207b\u2075\u2070 at 25\u00b0C.
    If the equilibrium concentration of O\u2082 in air at 25\u00b0C is 1.6 \u00d7 10\u207b\u00b2 M, what is the concentration of O\u2083?<\/li>\n\n\n\n
  34. The reaction, CO(g) + 3H\u2082(g) \u21cc CH\u2084(g) + H\u2082O(g) is at equilibrium at 1300 K in a 1 L flask.
    It contains 0.30 mol CO, 0.10 mol H\u2082, 0.02 mol H\u2082O and an unknown amount of CH\u2084.
    Determine the concentration of CH\u2084 in the mixture. Kc = 3.90 at 1300 K.<\/li>\n\n\n\n
  35. What is meant by the conjugate acid-base pair?
    Find the conjugate acid\/base for the following species:
    HNO\u2082, CN\u207b, HClO\u2084, F\u207b, OH\u207b, CO\u2083\u00b2\u207b and S\u00b2\u207b<\/li>\n\n\n\n
  36. Which of the following are Lewis acids? H\u2082O, BF\u2083, H\u207a, and NH\u2084\u207a<\/li>\n\n\n\n
  37. What will be the conjugate bases for the Br\u00f6nsted acids: HF, H\u2082SO\u2084, and HCO\u2083\u207b?<\/li>\n\n\n\n
  38. Write the conjugate acids for the following Br\u00f6nsted bases: NH\u2082\u207b, NH\u2083, and HCOO\u207b.<\/li>\n\n\n\n
  39. The species: H\u2082O, HCO\u2083\u207b, HSO\u2084\u207b, and NH\u2083 can act both as Br\u00f6nsted acids and bases.
    For each, give the corresponding conjugate acid and conjugate base.<\/li>\n\n\n\n
  40. Classify the following species into Lewis acids and Lewis bases and show how they act:
    (a) OH\u207b (b) F\u207b (c) H\u207a (d) BCl\u2083<\/li>\n\n\n\n
  41. The concentration of hydrogen ions in a sample of soft drink is 3.8 \u00d7 10\u207b\u00b3 M.
    What is its pH?<\/li>\n\n\n\n
  42. The pH of a sample of vinegar is 3.76.
    Calculate the concentration of hydrogen ions in it.<\/li>\n\n\n\n
  43. Theionisation constants at 298 K are:
    HF = 6.8 \u00d7 10\u207b\u2074, HCOOH = 1.8 \u00d7 10\u207b\u2074, HCN = 4.8 \u00d7 10\u207b\u2079
    Calculate the ionisation constants of their conjugate bases.<\/li>\n\n\n\n
  44. The ionisation constant of phenol is 1.0 \u00d7 10\u207b\u00b9\u2070.
    What is the concentration of phenolate ion in a 0.05 M phenol solution?
    What will be its degree of ionisation if the solution also contains 0.01 M sodium phenolate?<\/li>\n\n\n\n
  45. The first ionisation constant of H\u2082S is 9.1 \u00d7 10\u207b\u2078.
    Calculate the concentration of HS\u207b in a 0.1 M H\u2082S solution.
    How is it affected if the solution also contains 0.1 M HCl?
    If the second ionisation constant of H\u2082S is 1.2 \u00d7 10\u207b\u00b9\u00b3, calculate the concentration of S\u00b2\u207b in both cases.<\/li>\n\n\n\n
  46. The ionisation constant of acetic acid is 1.74 \u00d7 10\u207b\u2075.
    Calculate the degree of dissociation of acetic acid in a 0.05 M solution.
    Also, calculate the concentration of the acetate ion and the pH of the solution.<\/li>\n\n\n\n
  47. A 0.01 M solution of an organic acid has a pH of 4.15.
    Calculate the concentration of anion, the ionisation constant of the acid, and its pKa.<\/li>\n\n\n\n
  48. Assuming complete dissociation, calculate the pH of:
    (a) 0.003 M HCl
    (b) 0.005 M NaOH
    (c) 0.002 M HBr
    (d) 0.002 M KOH<\/li>\n\n\n\n
  49. Calculate the pH of the following solutions:
    (a) 2 g of TlOH in 2 L of solution
    (b) 0.3 g of Ca(OH)\u2082 in 500 mL solution
    (c) 0.3 g of NaOH in 200 mL of solution
    (d) 1 mL of 13.6 M HCl diluted to 1 L<\/li>\n\n\n\n
  50. The degree of ionisation of 0.1 M bromoacetic acid is 0.132.
    Calculate the pH and pKa of the acid.<\/li>\n\n\n\n
  51. The pH of 0.005 M codeine (C\u2081\u2088H\u2082\u2081NO\u2083) is 9.95.
    Calculate its ionisation constant and pKb.<\/li>\n\n\n\n
  52. What is the pH of a 0.001 M aniline solution?
    Using the ionisation constant of aniline, calculate:
    Degree of ionisation
    Ionisation constant of its conjugate acid<\/li>\n\n\n\n
  53. Calculate the degree of ionisation of 0.05 M acetic acid, if pKa = 4.74.
    How is it affected when the solution contains:
    (a) 0.01 M HCl
    (b) 0.1 M HCl<\/li>\n\n\n\n
  54. The ionisation constant of dimethylamine is 5.4 \u00d7 10\u207b\u2074.
    Calculate its degree of ionisation in a 0.02 M solution.
    What percentage is ionised if the solution also contains 0.1 M NaOH?<\/li>\n\n\n\n
  55. Calculate the hydrogen ion concentration in the following fluids:
    (a) Human muscle fluid, pH = 6.83
    (b) Human stomach fluid, pH = 1.2
    (c) Human blood, pH = 7.38
    (d) Human saliva, pH = 6.4<\/li>\n\n\n\n
  56. The pH values are:
    Milk = 6.8, Black coffee = 5.0, Tomato juice = 4.2, Lemon juice = 2.2, Egg white = 7.8
    Calculate the hydrogen ion concentration in each.<\/li>\n\n\n\n
  57. 0.561 g of KOH is dissolved in water to make 200 mL of solution at 298 K.
    Calculate the concentrations of potassium, hydrogen, and hydroxyl ions.
    What is the pH?<\/li>\n\n\n\n
  58. The solubility of Sr(OH)\u2082 at 298 K is 19.23 g\/L.
    Calculate the concentrations of Sr\u00b2\u207a and OH\u207b ions and the pH.<\/li>\n\n\n\n
  59. The ionisation constant of propanoic acid is 1.32 \u00d7 10\u207b\u2075.
    Calculate the degree of ionisation and pH of its 0.05 M solution.
    What will be the degree of ionisation if the solution also contains 0.01 M HCl?<\/li>\n\n\n\n
  60. The pH of 0.1 M cyanic acid (HCNO) is 2.34.
    Calculate the ionisation constant and the degree of ionisation.<\/li>\n\n\n\n
  61. The ionisation constant of nitrous acid is 4.5 \u00d7 10\u207b\u2074.
    Calculate the pH of a 0.04 M sodium nitrite solution and its degree of hydrolysis.<\/li>\n\n\n\n
  62. A 0.02 M pyridinium hydrochloride solution has a pH = 3.44.
    Calculate the ionisation constant of pyridine.<\/li>\n\n\n\n
  63. Predict whether the solutions are neutral, acidic, or basic:
    NaCl, KBr, NaCN, NH\u2084NO\u2083, NaNO\u2082, KF<\/li>\n\n\n\n
  64. The ionisation constant of chloroacetic acid is 1.35 \u00d7 10\u207b\u00b3.
    Calculate the pH of its 0.1 M acid solution and its 0.1 M sodium salt solution.<\/li>\n\n\n\n
  65. Ionic product of water at 310 K is 2.7 \u00d7 10\u207b\u00b9\u2074.
    What is the pH of neutral water at this temperature?<\/li>\n\n\n\n
  66. Calculate the pH of the resulting mixtures:
    (a) 10 mL of 0.2 M Ca(OH)\u2082 + 25 mL of 0.1 M HCl
    (b) 10 mL of 0.01 M H\u2082SO\u2084 + 10 mL of 0.01 M Ca(OH)\u2082
    (c) 10 mL of 0.1 M H\u2082SO\u2084 + 10 mL of 0.1 M KOH<\/li>\n\n\n\n
  67. Determine the solubilities of:
    Ag\u2082CrO\u2084, BaCrO\u2084, Fe(OH)\u2083, PbCl\u2082, Hg\u2082I\u2082 at 298 K from their Ksp values.
    Find the molarities of individual ions.<\/li>\n\n\n\n
  68. Ksp of Ag\u2082CrO\u2084 = 1.1 \u00d7 10\u207b\u00b9\u00b2; Ksp of AgBr = 5.0 \u00d7 10\u207b\u00b9\u00b3
    Calculate the ratio of molarities of their saturated solutions.<\/li>\n\n\n\n
  69. Equal volumes of 0.002 M sodium iodate and cupric chlorate are mixed.
    Will Cu(IO\u2083)\u2082 precipitate? (Ksp = 7.4 \u00d7 10\u207b\u2078)<\/li>\n\n\n\n
  70. Ka of benzoic acid = 6.46 \u00d7 10\u207b\u2075; Ksp of silver benzoate = 2.5 \u00d7 10\u207b\u00b9\u00b3
    How much more soluble is silver benzoate in a buffer of pH 3.19 compared to pure water?<\/li>\n\n\n\n
  71. What is the maximum concentration of equimolar FeSO\u2084 and Na\u2082S so that no FeS precipitates? (Ksp = 6.3 \u00d7 10\u207b\u00b9\u2078)<\/li>\n\n\n\n
  72. What is the minimum volume of water required to dissolve 1 g of CaSO\u2084 at 298 K? (Ksp = 9.1 \u00d7 10\u207b\u2076)<\/li>\n\n\n\n
  73. The concentration of S\u00b2\u207b in 0.1 M HCl saturated with H\u2082S is 1.0 \u00d7 10\u207b\u00b9\u2079 M.
    If 10 mL of this is added to 5 mL of 0.04 M FeSO\u2084, MnCl\u2082, ZnCl\u2082, and CdCl\u2082, in which solution will precipitation occur?<\/li>\n<\/ol>\n\n\n\n

    Also Read: NCERT Notes Class 11 Geography Fundamentals of Physical Geography Chapter 12: Water (Oceans) (Free PDF)<\/strong><\/a><\/strong><\/strong><\/p>\n\n\n\n

    Solutions<\/h2>\n\n\n\n
      \n
    1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
      a) The initial effect of an increase in volume at constant temperature is a decrease in pressure. Since the vapour pressure depends only on temperature and not on volume, the vapour pressure remains initially unchanged.
      b) The rate of condensation decreases immediately because fewer vapour molecules now strike the liquid surface per unit time. The rate of evaporation remains unchanged at first.
      c) As the rate of evaporation is now greater than the rate of condensation, more molecules pass into the vapour phase until a new equilibrium is established. The final vapour pressure will be equal to the original vapour pressure at the same temperature.<\/li>\n\n\n\n
    2. What is Kc for the following equilibrium when the equilibrium concentrations are:
      [SO\u2082] = 0.60 M, [O\u2082] = 0.82 M and [SO\u2083] = 1.90 M
      2SO\u2082(g) + O\u2082(g) \u21cc 2SO\u2083(g)
      Kc = [SO\u2083]\u00b2 \/ ([SO\u2082]\u00b2 \u00d7 [O\u2082])
      = (1.90)\u00b2 \/ [(0.60)\u00b2 \u00d7 (0.82)] = 3.61 \/ (0.36 \u00d7 0.82) = 3.61 \/ 0.2952 \u2248 12.23<\/strong><\/li>\n\n\n\n
    3. At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms.
      I\u2082(g) \u21cc 2I(g)
      Let the initial pressure of I\u2082 = 100 Pa.
      At equilibrium, 40% dissociates to atoms \u2192 40 Pa of I atoms and 60 Pa of I\u2082 remain.
      I\u2082 \u21cc 2I
      Initial: 100 Pa \u2192 0
      Change: \u2013x \u2192 +2x
      Equilibrium: 100 \u2013 x \u2192 2x
      Given: 2x = 40 \u2192 x = 20 \u2192 [I\u2082] = 80 Pa
      Kp = (PI)\u00b2 \/ PI\u2082 = (40)\u00b2 \/ 65 = 24.6 Pa<\/strong><\/li>\n\n\n\n
    4. Write the expression for the equilibrium constant, K,c, for the following reactions:
      (i) Kc = [NO]\u00b2[Cl\u2082] \/ [NOCl]\u00b2
      (ii) Kc = [NO\u2082]^4[O\u2082] \/ 1 (pure solids are omitted)
      (iii) Kc = [CH\u2083COOH][C\u2082H\u2085OH] \/ [CH\u2083COOC\u2082H\u2085]
      (iv) Kc = 1 \/ ([Fe\u00b3\u207a][OH\u207b]\u00b3)
      (v) Kc = [IF\u2085]\u00b2 \/ [F\u2082]\u2075 (I\u2082 is a solid and not included)<\/li>\n\n\n\n
    5. Find the value of Kc for each of the following equilibria from the value of Kp:
      (i) 2NOCl(g) \u21cc 2NO(g) + Cl\u2082(g); Kp = 1.8 \u00d7 10\u207b\u00b2 at 500 K
      \u0394n = 3 \u2013 2 = 1
      Kc = Kp \u00d7 (RT)^\u2013\u0394n = 1.8 \u00d7 10\u207b\u00b2 \/ (0.0821 \u00d7 500) = 1.8 \u00d7 10\u207b\u00b2 \/ 41.05 \u2248 4.38 \u00d7 10\u207b\u2074<\/strong><\/li>\n\n\n\n
    6. (ii) CaCO\u2083(s) \u21cc CaO(s) + CO\u2082(g); Kp = 167 at 1073 K
      \u0394n = 1 \u2013 0 = 1
      Kc = Kp \/ (RT) = 167 \/ (0.0821 \u00d7 1073) = 167 \/ 88.09 \u2248 1.896 mol L\u207b\u00b9<\/strong><\/li>\n\n\n\n
    7. For the reaction: NO(g) + O\u2083(g) \u21cc NO\u2082(g) + O\u2082(g); Kc = 6.3 \u00d7 10\u00b9\u2074
      Kc (reverse) = 1 \/ Kc = 1 \/ (6.3 \u00d7 10\u00b9\u2074) \u2248 1.59 \u00d7 10\u207b\u00b9\u2075<\/strong><\/li>\n\n\n\n
    8. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression.
      In the expression for the equilibrium constant, the concentrations (or activities) of pure solids and pure liquids are taken as constant because their molar concentration remains unchanged. Therefore, they are not included in the equilibrium constant expression.<\/li>\n\n\n\n
    9. 2N\u2082(g) + O\u2082(g) \u21cc 2N\u2082O(g)
      Initial moles: [N\u2082] = 0.482 \/ 10 = 0.0482 M, [O\u2082] = 0.933 \/ 10 = 0.0933 M
      Let x be the amount of N\u2082O formed at equilibrium:
      Change: \u2013x for N\u2082 and \u2013\u00bdx for O\u2082
      Kc = [N\u2082O]\u00b2 \/ ([N\u2082]\u00b2[O\u2082]) = (x\/10)\u00b2 \/ ((0.0482 \u2013 x\/10)\u00b2 \u00d7 (0.0933 \u2013 x\/20)) = 2.0 \u00d7 10\u207b\u00b3\u2077
      Due to extremely small Kc, x \u2248 0, i.e., negligible formation of N\u2082O \u2192 composition remains essentially unchanged.<\/li>\n\n\n\n
    10. 2NO(g) + Br\u2082(g) \u21cc 2NOBr(g)
      Initial: [NO] = 0.087 mol, [Br\u2082] = 0.0437 mol
      At equilibrium: [NOBr] = 0.0518 mol
      From the balanced equation:
      2NO + Br\u2082 \u21cc 2NOBr
      0.087 \u2013 x + 0.0437 \u2013 x + x = 0.0518 \u00d7 2 \u2192 x = 0.0518
      [NO] = 0.087 \u2013 0.0518 = 0.0352 mol<\/strong>, [Br\u2082] = 0.0437 \u2013 0.0259 = 0.0178 mol<\/strong><\/li>\n\n\n\n
    11. 2SO\u2082(g) + O\u2082(g) \u21cc 2SO\u2083(g); Kp = 2.0 \u00d7 10\u00b9\u2070 at 450 K
      \u0394n = \u20131 \u2192 Kc = Kp \/ (RT)^\u0394n = Kp \/ (RT)\u207b\u00b9 = Kp \u00d7 RT
      = 2.0 \u00d7 10\u00b9\u2070 \u00d7 0.0821 \u00d7 450 = 7.389 \u00d7 10\u00b9\u00b9 mol L\u207b\u00b9<\/strong><\/li>\n\n\n\n
    12. 2HI(g) \u21cc H\u2082(g) + I\u2082(g)
      Initial pressure of HI = 0.2 atm; Equilibrium pressure = 0.04 atm
      Let dissociation = x
      2HI \u21cc H\u2082 + I\u2082
      Initial: 0.2, 0, 0
      Change: \u20132x, +x, +x
      At eq: 0.2 \u2013 2x = 0.04 \u2192 x = 0.08
      Kp = (x)\u00b2 \/ (0.2 \u2013 2x)\u00b2 = (0.08)\u00b2 \/ (0.04)\u00b2 = 0.0064 \/ 0.0016 = 4.0<\/strong><\/li>\n\n\n\n
    13. N\u2082(g) + 3H\u2082(g) \u21cc 2NH\u2083(g); Kc = 1.7 \u00d7 10\u00b2
      [ N\u2082 ] = 1.57 \/ 20 = 0.0785 M
      [ H\u2082 ] = 1.92 \/ 20 = 0.096 M
      [ NH\u2083 ] = 8.13 \/ 20 = 0.4065 M
      Qc = [NH\u2083]\u00b2 \/ ([N\u2082][H\u2082]\u00b3) = (0.4065)\u00b2 \/ (0.0785 \u00d7 (0.096)\u00b3)
      Qc = 0.1652 \/ (0.0785 \u00d7 0.0008847) = 0.1652 \/ 6.94 \u00d7 10\u207b\u2075 \u2248 2380
      Since Qc > Kc, the reaction will proceed in the reverse direction to attain equilibrium.<\/li>\n\n\n\n
    14. Kc = [NH\u2083]\u2074 \/ ([NO]\u2076[H\u2082O]\u2074)
      Balanced equation: 4NH\u2083 + 6NO \u21cc 5N\u2082 + 6H\u2082O<\/li>\n\n\n\n
    15. 1 mol H\u2082O and 1 mol CO in 10 L \u2192 0.1 M each
      40% of H\u2082O reacts \u2192 0.04 mol \u2192 [H\u2082] = [CO\u2082] = 0.04 mol \/ 10 = 0.004 M
      [H\u2082O] = [CO] = 0.06 M
      Kc = [H\u2082][CO\u2082] \/ [H\u2082O][CO] = (0.004 \u00d7 0.004) \/ (0.06 \u00d7 0.06) = 0.000016 \/ 0.0036 = 4.44 \u00d7 10\u207b\u00b3<\/strong><\/li>\n\n\n\n
    16. H\u2082(g) + I\u2082(g) \u21cc 2HI(g); Kc = 54.8
      [HI] = 0.5 M
      Let x be [H\u2082] = [I\u2082] at equilibrium
      Then Kc = [HI]\u00b2 \/ ([H\u2082][I\u2082]) = 0.25 \/ x\u00b2 \u2192 x\u00b2 = 0.25 \/ 54.8 = 0.00456
      x = \u221a0.00456 \u2248 0.068 M<\/strong>
      <\/strong> [H\u2082] = [I\u2082] \u2248 0.068 M<\/li>\n\n\n\n
    17. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
      a) Vapour pressure initially decreases due to an increase in volume.
      b) Rate of condensation decreases; rate of evaporation exceeds rate of condensation.
      c) Eventually, a new equilibrium is established at the same temperature, but the final vapour pressure remains the same as before, since vapour pressure is temperature dependent.<\/li>\n\n\n\n
    18. What is Kc for the following equilibrium when the equilibrium concentrations are: [SO\u2082] = 0.60 M, [O\u2082] = 0.82 M and [SO\u2083] = 1.90 M?
      2SO\u2082(g) + O\u2082(g) \u21cc 2SO\u2083(g)
      Kc = [SO\u2083]\u00b2 \/ ([SO\u2082]\u00b2 \u00d7 [O\u2082])
      = (1.90)\u00b2 \/ ((0.60)\u00b2 \u00d7 0.82)
      = 3.61 \/ (0.36 \u00d7 0.82)
      = 3.61 \/ 0.2952
      \u2248 12.23<\/li>\n\n\n\n
    19. At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms.
      I\u2082(g) \u21cc 2I(g)
      Let the initial amount of I\u2082 = 1 mol (by volume)
      At equilibrium: I\u2082 = 0.60 mol, I = 0.40 mol
      Total moles = 0.60 + 0.40 = 1.00 mol
      Mole fraction of I = 0.40
      Partial pressure of I = 0.40 \u00d7 105 Pa = 42 Pa
      Partial pressure of I\u2082 = 0.60 \u00d7 105 Pa = 63 Pa
      Kp = (PI)\u00b2 \/ PI\u2082 = (42)\u00b2 \/ 63 = 1764 \/ 63 \u2248 28 Pa<\/li>\n\n\n\n
    20. Write the expression for the equilibrium constant, Kc, for each of the following reactions:
      (i) Kc = [NO]\u00b2[Cl\u2082] \/ [NOCl]\u00b2
      (ii) Kc = [NO\u2082]\u2074[O\u2082] (since solids are not included)
      (iii) Kc = [CH\u2083COOH][C\u2082H\u2085OH] \/ [CH\u2083COOC\u2082H\u2085]
      (iv) Kc = 1 \/ [Fe\u00b3\u207a][OH\u207b]\u00b3 (or omit solids)
      (v) Kc = [IF\u2085]\u00b2 \/ [F\u2082]\u2075 (omit I\u2082 solid)<\/li>\n\n\n\n
    21. Find the value of Kc for each of the following equilibria from the value of Kp:
      (i) 2NOCl(g) \u21cc 2NO(g) + Cl\u2082(g); Kp = 1.8 \u00d7 10\u207b\u00b2 at 500 K
      \u0394ng = (2 + 1) – 2 = 1
      Kc = Kp \u00d7 (RT)^(-\u0394ng)
      Kc = 1.8 \u00d7 10\u207b\u00b2 \/ (0.0821 \u00d7 500)
      = 1.8 \u00d7 10\u207b\u00b2 \/ 41.05 \u2248 4.4 \u00d7 10\u207b\u2074<\/li>\n\n\n\n
    22. (ii) CaCO\u2083(s) \u21cc CaO(s) + CO\u2082(g); Kp = 167 at 1073 K
      \u0394ng = 1 – 0 = 1
      Kc = 167 \/ (0.0821 \u00d7 1073)
      = 167 \/ 88.1 \u2248 1.90 mol L\u207b\u00b9<\/li>\n\n\n\n
    23. For the reaction: NO(g) + O\u2083(g) \u21cc NO\u2082(g) + O\u2082(g); Kc = 6.3 \u00d7 10\u00b9\u2074 at 1000 K
      Kc for reverse = 1 \/ Kc = 1 \/ 6.3 \u00d7 10\u00b9\u2074 \u2248 1.59 \u00d7 10\u207b\u00b9\u2075<\/li>\n\n\n\n
    24. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression.
      The concentration of pure solids and pure liquids remains constant and does not change with the extent of reaction. Therefore, they are incorporated in the equilibrium constant and are not explicitly written in the expression.<\/li>\n\n\n\n
    25. 2HI(g) \u21cc H\u2082(g) + I\u2082(g)
      Initial pressure of HI = 0.2 atm; equilibrium pressure = 0.04 atm
      Decrease in HI = 0.2 \u2013 0.04 = 0.16 atm
      Since 2 mol HI \u21cc 1 mol H\u2082 + 1 mol I\u2082,
      H\u2082 = I\u2082 = 0.16 \/ 2 = 0.08 atm
      Kp = (PH\u2082 \u00d7 PI\u2082) \/ (PHI)\u00b2 = (0.08 \u00d7 0.08) \/ (0.04)\u00b2 = 0.0064 \/ 0.0016 = 4.0<\/li>\n\n\n\n
    26. N\u2082(g) + 3H\u2082(g) \u21cc 2NH\u2083(g); Kc = 1.7 \u00d7 10\u00b2 at 500 K
      [N\u2082] = 1.57 mol \/ 20 L = 0.0785 M
      [H\u2082] = 1.92 mol \/ 20 L = 0.096 M
      [NH\u2083] = 8.13 mol \/ 20 L = 0.4065 M
      Q = [NH\u2083]\u00b2 \/ ([N\u2082][H\u2082]\u00b3) = (0.4065)\u00b2 \/ (0.0785 \u00d7 (0.096)\u00b3)
      \u2248 0.165 \/ (0.0785 \u00d7 8.84\u00d710\u207b\u2074) \u2248 0.165 \/ 6.94\u00d710\u207b\u2075 \u2248 2378.38
      Since Q > Kc, the reaction will proceed in the reverse direction (toward reactants).<\/li>\n\n\n\n
    27. Write the balanced chemical equation for the equilibrium expression:
      Kc = [NH\u2083]\u2074 \/ ([NO]\u2076 [H\u2082O]\u2074)
      Balanced equation: 6NO(g) + 4H\u2082O(g) \u21cc 4NH\u2083(g) + 3O\u2082(g)<\/li>\n\n\n\n
    28. H\u2082O(g) + CO(g) \u21cc H\u2082(g) + CO\u2082(g)
      1 mol of H\u2082O and CO in a 10 L vessel \u2192 0.1 M each
      40% of H\u2082O reacts \u2192 0.04 mol reacts
      H\u2082O = 0.06 mol = 0.006 M
      CO = 0.06 mol = 0.006 M
      H\u2082 = CO\u2082 = 0.04 mol = 0.004 M
      Kc = [H\u2082][CO\u2082] \/ [H\u2082O][CO] = (0.004 \u00d7 0.004) \/ (0.006 \u00d7 0.006) = 1.78 \u00d7 10\u207b\u00b9<\/li>\n\n\n\n
    29. H\u2082(g) + I\u2082(g) \u21cc 2HI(g); Kc = 54.8 at 700 K
      [HI] = 0.5 M
      Let [H\u2082] = [I\u2082] = x
      Kc = [HI]\u00b2 \/ ([H\u2082][I\u2082]) = (0.5)\u00b2 \/ x\u00b2 = 0.25 \/ x\u00b2
      54.8 = 0.25 \/ x\u00b2 \u2192 x\u00b2 = 0.25 \/ 54.8 = 0.00456 \u2192 x = \u221a0.00456 \u2248 0.068 M
      So, [H\u2082] = [I\u2082] = 0.068 M<\/li>\n\n\n\n
    30. 2ICl(g) \u21cc I\u2082(g) + Cl\u2082(g); Kc = 0.14
      Initial [ICl] = 0.78 M, let x be the amount dissociated
      At equilibrium: [ICl] = 0.78 \u2013 2x, [I\u2082] = x, [Cl\u2082] = x
      Kc = (x \u00d7 x) \/ (0.78 \u2013 2x)\u00b2
      Let\u2019s solve using approximation: assume x is small.
      Kc \u2248 x\u00b2 \/ (0.78)\u00b2 = 0.14
      x\u00b2 \u2248 0.14 \u00d7 0.6084 = 0.08518
      x \u2248 \u221a0.08518 \u2248 0.292 M
      [ICl] \u2248 0.78 \u2013 2(0.292) \u2248 0.196 M
      [I\u2082] = [Cl\u2082] \u2248 0.292 M<\/li>\n\n\n\n
    31. C\u2082H\u2086(g) \u21cc C\u2082H\u2084(g) + H\u2082(g); Kp = 0.04 atm at 899 K
      Initial pressure of C\u2082H\u2086 = 4.0 atm
      Let x atm decompose
      At equilibrium:
      PC\u2082H\u2086 = 4.0 \u2013 x
      PC\u2082H\u2084 = PH\u2082 = x
      Kp = x\u00b2 \/ (4.0 \u2013 x) = 0.04
      Solve: x\u00b2 = 0.04(4.0 \u2013 x) = 0.16 \u2013 0.04x
      x\u00b2 + 0.04x \u2013 0.16 = 0
      Using the quadratic formula:
      x = [-0.04 \u00b1 \u221a(0.0016 + 0.64)] \/ 2 \u2248 0.38 atm
      PC\u2082H\u2086 = 4.0 \u2013 0.38 = 3.62 atm<\/li>\n\n\n\n
    32. CH\u2083COOH(l) + C\u2082H\u2085OH(l) \u21cc CH\u2083COOC\u2082H\u2085(l) + H\u2082O(l)
      (i) Qc = [CH\u2083COOC\u2082H\u2085][H\u2082O] \/ [CH\u2083COOH][C\u2082H\u2085OH]
      (ii) Initial: CH\u2083COOH = 1 mol, C\u2082H\u2085OH = 0.18 mol
      At equilibrium, ester = 0.171 mol \u2192 [ester] = 0.171
      So, [CH\u2083COOH] = 1 \u2013 0.171 = 0.829
      [C\u2082H\u2085OH] = 0.18 \u2013 0.171 = 0.009
      Kc = (0.171 \u00d7 0.171) \/ (0.829 \u00d7 0.009) \u2248 0.0292 \/ 0.007461 \u2248 3.91
      (iii) Initial acid = 1 mol, alcohol = 0.5 mol; ester = 0.214 mol
      [acid] = 1 \u2013 0.214 = 0.786
      [alcohol] = 0.5 \u2013 0.214 = 0.286
      Qc = (0.214 \u00d7 0.214) \/ (0.786 \u00d7 0.286) = 0.0458 \/ 0.2248 \u2248 0.204
      Since Qc < Kc (3.91), the reaction is not at equilibrium and will proceed forward<\/li>\n\n\n\n
    33. PCl\u2085(g) \u21cc PCl\u2083(g) + Cl\u2082(g); Kc = 8.3 \u00d7 10\u207b\u00b3
      [PCl\u2085] = 0.05 mol\/L
      Let x mol\/L dissociate.
      At equilibrium: [PCl\u2085] = 0.05 \u2013 x, [PCl\u2083] = x, [Cl\u2082] = x
      Kc = x\u00b2 \/ (0.05 \u2013 x)
      Assume x is small:
      Kc \u2248 x\u00b2 \/ 0.05 = 8.3 \u00d7 10\u207b\u00b3
      x\u00b2 = 8.3 \u00d7 10\u207b\u00b3 \u00d7 0.05 = 4.15 \u00d7 10\u207b\u2074
      x \u2248 \u221a4.15 \u00d7 10\u207b\u2074 \u2248 0.0204 mol\/L
      [PCl\u2083] = [Cl\u2082] \u2248 0.0204 mol\/L
      [PCl\u2085] \u2248 0.05 \u2013 0.0204 = 0.0296 mol\/L<\/li>\n\n\n\n
    34. FeO(s) + CO(g) \u21cc Fe(s) + CO\u2082(g); Kp = 0.265 atm
      Initial: pCO = 1.4 atm, pCO\u2082 = 0.80 atm
      Qp = pCO\u2082 \/ pCO = 0.80 \/ 1.4 \u2248 0.571
      Since Qp > Kp, the reaction will shift in the reverse direction.
      Let x be the change
      At equilibrium: pCO = 1.4 + x, pCO\u2082 = 0.80 \u2013 x
      Kp = (0.80 \u2013 x) \/ (1.4 + x) = 0.265
      0.80 \u2013 x = 0.265 \u00d7 (1.4 + x)
      0.80 \u2013 x = 0.371 + 0.265x
      0.429 = 1.265x \u2192 x \u2248 0.339 atm
      pCO = 1.4 + 0.339 = 1.739 atm
      pCO\u2082 = 0.80 \u2013 0.339 = 0.461 atm<\/li>\n\n\n\n
    35. N\u2082(g) + 3H\u2082(g) \u21cc 2NH\u2083(g); Kc = 0.061
      [ N\u2082 ] = 3.0 M, [ H\u2082 ] = 2.0 M, [ NH\u2083 ] = 0.5 M
      Qc = [NH\u2083]\u00b2 \/ ([N\u2082][H\u2082]\u00b3) = (0.5)\u00b2 \/ (3.0 \u00d7 8) = 0.25 \/ 24 = 0.0104
      Since Qc < Kc, the reaction proceeds in the forward direction<\/li>\n\n\n\n
    36. 2BrCl(g) \u21cc Br\u2082(g) + Cl\u2082(g); Kc = 32 at 500 K
      Initial [BrCl] = 3.3 \u00d7 10\u207b\u00b3 M
      Let x be the amount dissociated:
      [Br\u2082] = x, [Cl\u2082] = x, [BrCl] = 3.3 \u00d7 10\u207b\u00b3 \u2013 2x
      Kc = x\u00b2 \/ (3.3 \u00d7 10\u207b\u00b3 \u2013 2x)\u00b2 = 32
      Use approximation: if x is small, (3.3 \u00d7 10\u207b\u00b3)\u00b2 \u2248 1.089 \u00d7 10\u207b\u2075
      So x\u00b2 \u2248 32 \u00d7 1.089 \u00d7 10\u207b\u2075 \u2248 3.485 \u00d7 10\u207b\u2074 \u2192 x \u2248 \u221a(3.485 \u00d7 10\u207b\u2074) \u2248 0.0187
      But this value is greater than the initial concentration, so no valid solution using approximation. Solve using full quadratic or assume 100% dissociation. In this case, [BrCl] at equilibrium is negligible, and [Br\u2082] \u2248 [Cl\u2082] \u2248 0.00165 M each (approx).<\/li>\n\n\n\n
    37. C(s) + CO\u2082(g) \u21cc 2CO(g)
      Given 90.55% CO by mass at 1127 K and 1 atm
      Let total gas = 100 g
      CO = 90.55 g, CO\u2082 = 9.45 g
      Moles CO = 90.55 \/ 28 = 3.23 mol
      Moles CO\u2082 = 9.45 \/ 44 = 0.215 mol
      Total moles = 3.445 mol
      Mole fraction CO = 3.23 \/ 3.445 = 0.9375 \u2192 pCO = 0.9375 atm
      pCO\u2082 = 0.0625 atm
      Kp = (pCO)\u00b2 \/ pCO\u2082 = (0.9375)\u00b2 \/ 0.0625 = 0.879 \/ 0.0625 = 14.06<\/li>\n\n\n\n
    38. NO(g) + \u00bdO\u2082(g) \u21cc NO\u2082(g)
      \u0394G\u00b0 = \u03a3\u0394G\u00b0(products) \u2013 \u03a3\u0394G\u00b0(reactants)
      = [52.0] \u2013 [87.0 + 0] = \u201335.0 kJ\/mol
      K = e^(\u2013\u0394G\u00b0\/RT) = e^[35,000 \/ (8.314 \u00d7 298)] \u2248 e^14.1 \u2248 1.34 \u00d7 10\u2076<\/li>\n\n\n\n
    39. Change in number of moles with pressure decrease (volume increase):
      (a) PCl\u2085(g) \u21cc PCl\u2083(g) + Cl\u2082(g): forward reaction increases moles (1 \u2192 2), so forward reaction favoured
      (b) CaO(s) + CO\u2082(g) \u21cc CaCO\u2083(s): gas \u2192 solid, decrease in moles, backward favoured
      (c) 3Fe(s) + 4H\u2082O(g) \u21cc Fe\u2083O\u2084(s) + 4H\u2082(g): 4 mol gas \u2192 4 mol gas \u2192 no change<\/li>\n\n\n\n
    40. (i) COCl\u2082(g) \u21cc CO(g) + Cl\u2082(g) \u2014 1 mole \u21cc 2 moles \u2192 increasing pressure favours backward
      (ii) CH\u2084(g) + 2S\u2082(g) \u21cc CS\u2082(g) + 2H\u2082S(g) \u2014 3 \u21cc 3 \u2192 no effect
      (iii) CO\u2082(g) + C(s) \u21cc 2CO(g) \u2014 1 \u21cc 2 \u2192 increasing pressure favours backward
      (iv) 2H\u2082(g) + CO(g) \u21cc CH\u2083OH(g) \u2014 3 \u21cc 1 \u2192 increasing pressure favours forward
      (v) CaCO\u2083(s) \u21cc CaO(s) + CO\u2082(g) \u2014 solids ignored, gas on RHS \u2192 increasing pressure favours backward
      (vi) 4NH\u2083(g) + 5O\u2082(g) \u21cc 4NO(g) + 6H\u2082O(g) \u2014 9 \u21cc 10 \u2192 increasing pressure favours backward<\/li>\n\n\n\n
    41. H\u2082(g) + Br\u2082(g) \u21cc 2HBr(g); K = 1.6 \u00d7 10\u2075 at 1024 K
      Total pressure of HBr introduced = 10 bar.
      Let dissociation be x \u2192 [HBr] = 10 \u2013 2x, [H\u2082] = x, [Br\u2082] = x
      K = (10 \u2013 2x)\u00b2 \/ (x\u00b2) = 1.6 \u00d7 10\u2075
      Solving: (10 \u2013 2x)\u00b2 = 1.6 \u00d7 10\u2075 x\u00b2
      (100 \u2013 40x + 4x\u00b2) = 1.6 \u00d7 10\u2075 x\u00b2
      1.6 \u00d7 10\u2075 x\u00b2 \u2013 4x\u00b2 + 40x \u2013 100 = 0
      Use quadratic formula or approximate: x \u2248 0.0226 ba.r
      Then, [H\u2082] = [Br\u2082] = 0.0226 bar, [HBr] = 10 \u2013 2(0.0226) \u2248 9.9548 bar<\/li>\n\n\n\n
    42. CH\u2084(g) + H\u2082O(g) \u21cc CO(g) + 3H\u2082(g)
      (a) Kp = (PCO)(PH\u2082)\u00b3 \/ (PCH\u2084)(PH\u2082O)
      (b) (i) Increasing pressure \u2192 decrease volume \u2192 reaction shifts toward fewer gas molecules \u2192 backward
      (ii) Increasing temperature \u2192 endothermic reaction \u2192 forward reaction favoured \u2192 Kp increases
      (iii) Adding catalyst \u2192 no effect on Kp, only speeds up equilibrium attainment<\/li>\n\n\n\n
    43. 2H\u2082(g) + CO(g) \u21cc CH\u2083OH(g)
      a) Addition of H\u2082 \u2192 reaction shifts forward
      b) Addition of CH\u2083OH \u2192 reaction shifts backward
      c) Removal of CO \u2192 reaction shifts backwards
      d) Removal of CH\u2083OH \u2192 reaction shifts forward<\/li>\n\n\n\n
    44. PCl\u2085(g) \u21cc PCl\u2083(g) + Cl\u2082(g); Kc = 8.3 \u00d7 10\u207b\u00b3 at 473 K; \u0394H\u00b0 = +124.0 kJ mol\u207b\u00b9
      a) Kc = [PCl\u2083][Cl\u2082] \/ [PCl\u2085]
      b) Kc for reverse = 1 \/ 8.3 \u00d7 10\u207b\u00b3 \u2248 120.5
      c) (i) Adding PCl\u2085 \u2192 no change in Kc
      (ii) Increasing pressure \u2192 reaction shifts to fewer moles \u2192 backward \u2192 no change in Kc
      (iii) Increasing temperature \u2192 endothermic \u2192 favours forward \u2192 Kc increases<\/li>\n\n\n\n
    45. CO(g) + H\u2082O(g) \u21cc CO\u2082(g) + H\u2082(g); Kp = 10.1 at 400\u00b0C
      Initial pCO = pH\u2082O = 4.0 bar
      Let x be the change in pressure:
      pCO = 4 \u2013 x, pH\u2082O = 4 \u2013 x, pCO\u2082 = x, pH\u2082 = x
      Kp = (x \u00d7 x) \/ ((4 \u2013 x)\u00b2) = 10.1
      x\u00b2 \/ (4 \u2013 x)\u00b2 = 10.1 \u2192 solving gives x \u2248 3.2
      Thus, pH\u2082 = 3.2 bar<\/li>\n\n\n\n
    46. Predict which reactions will have appreciable concentrations of both reactants and products:
      a) Cl\u2082(g) \u21cc 2Cl(g); Kc = 5 \u00d7 10\u207b\u00b3\u2079 \u2192 lies far to the left (reactants dominate)
      b) Cl\u2082(g) + 2NO(g) \u21cc 2NOCl(g); Kc = 3.7 \u00d7 10\u2078 \u2192 lies far to the right (products dominate)
      c) Cl\u2082(g) + 2NO\u2082(g) \u21cc 2NO\u2082Cl(g); Kc = 1.8 \u2192 value near 1, appreciable concentrations of both<\/li>\n\n\n\n
    47. 3O\u2082(g) \u21cc 2O\u2083(g); Kc = 2.0 \u00d7 10\u207b\u2075\u2070 at 25\u00b0C
      [O\u2082] = 1.6 \u00d7 10\u207b\u00b2 M
      Kc = [O\u2083]\u00b2 \/ [O\u2082]\u00b3 \u2192 2.0 \u00d7 10\u207b\u2075\u2070 = x\u00b2 \/ (1.6 \u00d7 10\u207b\u00b2)\u00b3
      x\u00b2 = 2.0 \u00d7 10\u207b\u2075\u2070 \u00d7 4.096 \u00d7 10\u207b\u2076 = 8.192 \u00d7 10\u207b\u2075\u2076
      x = \u221a(8.192 \u00d7 10\u207b\u2075\u2076) = 9.05 \u00d7 10\u207b\u00b2\u2078 M
      [O\u2083] = 9.05 \u00d7 10\u207b\u00b2\u2078 M<\/li>\n\n\n\n
    48. CO(g) + 3H\u2082(g) \u21cc CH\u2084(g) + H\u2082O(g); Kc = 3.90 at 1300 K
      Initial: CO = 0.30 mol, H\u2082 = 0.10 mol, H\u2082O = 0.02 mol, CH\u2084 = x mol
      At equilibrium: CO = 0.30 \u2013 x, H\u2082 = 0.10 \u2013 3x, H\u2082O = 0.02 + x
      Kc = x(0.02 + x) \/ [(0.30 \u2013 x)(0.10 \u2013 3x)\u00b3] = 3.90
      Solving this equation gives x \u2248 0.016 M (approximate from the textbook)<\/li>\n\n\n\n
    49. Conjugate acid-base pairs:
      A conjugate acid-base pair consists of two species that differ by a proton (H\u207a)
      Examples:
      HNO\u2082 \u2192 conjugate base: NO\u2082\u207b
      CN\u207b \u2192 conjugate acid: HCN
      HClO\u2084 \u2192 conjugate base: ClO\u2084\u207b
      F\u207b \u2192 conjugate acid: HF
      OH\u207b \u2192 conjugate acid: H\u2082O
      CO\u2083\u00b2\u207b \u2192 conjugate acid: HCO\u2083\u207b
      S\u00b2\u207b \u2192 conjugate acid: HS\u207b<\/li>\n\n\n\n
    50. Lewis Acids: BF\u2083 and H\u207a are Lewis acids because they can accept a pair of electrons. H\u2082O and NH\u2084\u207a are not Lewis acids.<\/li>\n\n\n\n
    51. Conjugate Bases for Br\u00f8nsted Acids:
      HF \u2192 F\u207b
      H\u2082SO\u2084 \u2192 HSO\u2084\u207b
      HCO\u2083\u207b \u2192 CO\u2083\u00b2\u207b<\/li>\n\n\n\n
    52. Conjugate Acids for Br\u00f8nsted Bases:
      NH\u2082\u207b \u2192 NH\u2083
      NH\u2083 \u2192 NH\u2084\u207a
      HCOO\u207b \u2192 HCOOH<\/li>\n\n\n\n
    53. Amphoteric Species and their Conjugates:
      H\u2082O \u2192 conjugate acid: H\u2083O\u207a; conjugate base: OH\u207b
      HCO\u2083\u207b \u2192 conjugate acid: H\u2082CO\u2083; conjugate base: CO\u2083\u00b2\u207b
      HSO\u2084\u207b \u2192 conjugate acid: H\u2082SO\u2084; conjugate base: SO\u2084\u00b2\u207b
      NH\u2083 \u2192 conjugate acid: NH\u2084\u207a; conjugate base: NH\u2082\u207b<\/li>\n\n\n\n
    54. Classification of Species:
      OH\u207b: Lewis base (donates an electron\u207b pair)
      F\u207b: Lewis base (donates an electron\u207b pair)
      H\u207a: Lewis acid (accepts an electron\u207b pair)
      BCl\u2083: Lewis acid (electron-deficient)<\/li>\n\n\n\n
    55. pH from [H\u207a] = 3.8 \u00d7 10\u207b\u00b3 M (soft drink):
      pH = \u2013log(3.8 \u00d7 10\u207b\u00b3) \u2248 2.42<\/li>\n\n\n\n
    56. [H\u207a] from pH = 3.76 (vinegar):
      [H\u207a] = 10\u207b\u00b3.\u2077\u2076 \u2248 1.74 \u00d7 10\u207b\u2074 M<\/li>\n\n\n\n
    57. Ionisation Constants of Conjugate Bases:
      K_b = K_w \/ K_a
      HF: 1.0 \u00d7 10\u207b\u00b9\u2074 \/ 6.8 \u00d7 10\u207b\u2074 = 1.47 \u00d7 10\u207b\u00b9\u00b9
      HCOOH: 1.0 \u00d7 10\u207b\u00b9\u2074 \/ 1.8 \u00d7 10\u207b\u2074 = 5.56 \u00d7 10\u207b\u00b9\u00b9
      HCN: 1.0 \u00d7 10\u207b\u00b9\u2074 \/ 4.8 \u00d7 10\u207b\u2079 = 2.08 \u00d7 10\u207b\u2076<\/li>\n\n\n\n
    58. Phenol Solution:
      Ka = 1 \u00d7 10\u207b\u00b9\u2070, [Phenol] = 0.05 M
      x\u00b2 \/ (0.05 \u2013 x) = 1 \u00d7 10\u207b\u00b9\u2070 \u2192 x \u2248 7.07 \u00d7 10\u207b\u2076 M
      With 0.01 M sodium phenolate: [C\u2086H\u2085O\u207b] = 0.01 M
      \u03b1 = x \/ (0.05) = 1.41 \u00d7 10\u207b\u2074 without salt; suppressed with salt<\/li>\n\n\n\n
    59. Ionisation of H\u2082S:
      Ka\u2081 = 9.1 \u00d7 10\u207b\u2078
      x\u00b2 \/ (0.1 \u2013 x) = 9.1 \u00d7 10\u207b\u2078 \u2192 x \u2248 9.5 \u00d7 10\u207b\u2075 M (HS\u207b)
      With HCl, the common ion effect suppresses ionisation.
      Ka\u2082 = 1.2 \u00d7 10\u207b\u00b9\u00b3 \u2192 [S\u00b2\u207b] = Ka\u2082 \u00d7 [HS\u207b] \/ [H\u207a]
      Without HCl: [S\u00b2\u207b] \u2248 1.14 \u00d7 10\u207b\u00b9\u00b2 M
      With HCl: [S\u00b2\u207b] is even lower<\/li>\n\n\n\n
    60. 0.05 M Acetic Acid (Ka = 1.74 \u00d7 10\u207b\u2075):
      x\u00b2 \/ (0.05 \u2013 x) = 1.74 \u00d7 10\u207b\u2075 \u2192 x \u2248 9.3 \u00d7 10\u207b\u00b3
      \u03b1 = 0.186; [CH\u2083COO\u207b] = 9.3 \u00d7 10\u207b\u00b3 M
      pH = \u2013log(9.3 \u00d7 10\u207b\u00b3) \u2248 2.03<\/li>\n\n\n\n
    61. 0.01 M Organic Acid, pH = 4.15:
      [H\u207a] = 7.1 \u00d7 10\u207b\u2075; Ka = (7.1 \u00d7 10\u207b\u2075)\u00b2 \/ 0.01 \u2248 5.0 \u00d7 10\u207b\u2077
      pKa = \u2013log(5.0 \u00d7 10\u207b\u2077) \u2248 6.30<\/li>\n\n\n\n
    62. Strong Acids\/Bases (Complete dissociation):
      HCl (0.003 M): pH = 2.52
      NaOH (0.005 M): pOH = 2.3 \u2192 pH = 11.7
      HBr (0.002 M): pH = 2.70
      KOH (0.002 M): pOH = 2.7 \u2192 pH = 11.3<\/li>\n\n\n\n
    63. pH from Mass + Volume:
      (a) TlOH: 2 g \/ 204.38 g\/mol = 0.00979 mol in 2 L \u2192 0.0049 M \u2192 pOH \u2248 2.31 \u2192 pH = 11.69
      (b) Ca(OH)\u2082: 0.3 g \/ 74.09 g\/mol = 0.00405 mol = 0.0081 mol OH\u207b in 0.5 L \u2192 0.0162 M \u2192 pOH \u2248 1.79 \u2192 pH = 12.21
      (c) NaOH: 0.3 g \/ 40 = 0.0075 mol in 200 mL \u2192 0.0375 M \u2192 pOH \u2248 1.42 \u2192 pH = 12.58
      (d) 1 mL of 13.6 M HCl in 1 L = 0.0136 M \u2192 pH = 1.87<\/li>\n\n\n\n
    64. 0.1 M Bromoacetic Acid, \u03b1 = 0.132:
      [H\u207a] = 0.0132 M \u2192 pH = 1.88
      Ka = \u03b1\u00b2 \u00d7 C = (0.132)\u00b2 \u00d7 0.1 = 1.74 \u00d7 10\u207b\u00b3
      pKa = 2.76<\/li>\n\n\n\n
    65. Codeine, 0.005 M, pH = 9.95:
      [OH\u207b] = 1.12 \u00d7 10\u207b\u2075 M
      Kb = x\u00b2 \/ (0.005 \u2013 x) = 2.5 \u00d7 10\u207b\u2078
      pKb = 7.6<\/li>\n\n\n\n
    66. Aniline, 0.001 M:
      Kb = 4.27 \u00d7 10\u207b\u00b9\u2070
      x\u00b2 \/ 0.001 = Kb \u2192 x \u2248 2.07 \u00d7 10\u207b\u2076 \u2192 pOH = 5.68 \u2192 pH = 8.32
      \u03b1 = x \/ C = 0.00207
      Ka = Kw \/ Kb = 2.34 \u00d7 10\u207b\u2075<\/li>\n\n\n\n
    67. 0.05 M Acetic Acid, pKa = 4.74:
      Ka = 1.82 \u00d7 10\u207b\u2075 \u2192 x = 9.5 \u00d7 10\u207b\u00b3
      With 0.01 M HCl, [H\u207a] from HCl suppresses dissociation
      With 0.1 M HCl, even more suppression; \u03b1 decreases further<\/li>\n\n\n\n
    68. Dimethylamine, 0.02 M, Kb = 5.4 \u00d7 10\u207b\u2074:
      x\u00b2 \/ 0.02 = 5.4 \u00d7 10\u207b\u2074 \u2192 x \u2248 3.3 \u00d7 10\u207b\u00b3
      \u03b1 = x \/ 0.02 = 0.165
      With 0.1 M NaOH: OH\u207b common ion \u2192 \u03b1 suppressed<\/li>\n\n\n\n
    69. [H\u207a] from pH of fluids:
      pH = \u2013log[H\u207a] \u2192 [H\u207a] = 10^\u2013pH
      Muscle: 1.48 \u00d7 10\u207b\u2077
      Stomach: 6.3 \u00d7 10\u207b\u00b2
      Blood: 4.17 \u00d7 10\u207b\u2078
      Saliva: 3.98 \u00d7 10\u207b\u2077<\/li>\n\n\n\n
    70. pH from Food pH Values:
      Milk: 1.58 \u00d7 10\u207b\u2077
      Coffee: 1.0 \u00d7 10\u207b\u2075
      Tomato: 6.3 \u00d7 10\u207b\u2075
      Lemon: 6.3 \u00d7 10\u207b\u00b3
      Egg: 1.58 \u00d7 10\u207b\u2078<\/li>\n\n\n\n
    71. KOH: 0.561 g in 200 mL:
      Moles = 0.561 \/ 56.1 = 0.01 mol \u2192 0.05 M
      [OH\u207b] = 0.05 \u2192 pOH = 1.3 \u2192 pH = 12.7
      [K\u207a] = 0.05 M, [H\u207a] = 2 \u00d7 10\u207b\u00b9\u00b3<\/li>\n\n\n\n
    72. Sr(OH)\u2082 Solubility = 19.23 g\/L:
      Moles = 0.227 mol
      [Sr\u00b2\u207a] = 0.227 M, [OH\u207b] = 2 \u00d7 0.227 = 0.454 M
      pOH \u2248 0.34 \u2192 pH \u2248 13.66<\/li>\n\n\n\n
    73. 0.05 M Propanoic Acid, Ka = 1.32 \u00d7 10\u207b\u2075:
      x \u2248 8.1 \u00d7 10\u207b\u00b3 \u2192 pH = 2.09
      With 0.01 M HCl, \u03b1 decreases<\/li>\n<\/ol>\n\n\n\n

      Also Read: NCERT Solutions Class 11 Geography Fundamentals of Physical Geography Chapter 11: World Climate and Climate Change (Free PDF)<\/strong><\/a><\/strong><\/strong><\/strong><\/p>\n\n\n\n

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