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Theoretical-Physical Advanced Level Chemistry - Equilibria - Chemical Equilibrium Revision Notes PART 2

 2. Chemical equilibrium expressions and calculations

How do we write out chemical equilibrium expressions. What is the equilibrium constant? How do you do equilibrium calculations? All is explained with examples including the units for concentrations (mol dm-3) or partial pressures (e.g. atm, Pa or kPa) and how to solve equilibrium problems using concentration or partial pressure units.

(c) doc b KS4 Science GCSE/IGCSE reversible reactions & chemical equilibrium notes

Part 2 sub-index: 2.1 Kc and Kp equilibrium expressions and Le Chatelier's Principle * 2.2 Exemplar calculations using Kc expressions or Kp expressions

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle-rules * Part 3. Equilibria and industrial processes * Part 4. Partition, solubility product and ion-exchange * Part 5. pH, weak-strong acid-base theory and calculations * Part 6. Salt hydrolysis, Acid-base titrations-indicators, pH curves and buffers * Part 7. Redox equilibria, half-cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria-vapour pressure, boiling point and intermolecular forces


2.1 Equilibrium expressions and applying Le Chatelier's Principle

IT IS IMPORTANT and ESSENTIAL that Equilibria Part 1 is studied before working through this page

2.1a Molar concentration expressions

  • Kc concentration equilibrium expression INTRODUCTION

    • It is found experimentally that the concentrations at the equilibrium point are related by a simple mathematical equation known as an equilibrium expression which is governed by an equilibrium constant K, at constant temperature (K only varies with temperature, and nothing else).

      • These equilibrium expressions (many on this page) were originally derived from experimental analysis of mixtures at equilibrium.

      • Patterns in these equilibrium concentrations were 'spotted' and the resulting mathematical expression of these concentrations were considered to conform to what was called the 'law of mass action' ( a term not really used these days!).

      • A classic study of the ester formation equilibrium is often described in textbooks.

        • ALCOHOL + ACID (c) doc b ESTER +WATER

        • Knowing the initial amounts of alcohol and acid, it was easy to titrate the remaining free carboxylic acid and then logically work out the amounts of the water, alcohol and ester left in the equilibrium mixture.

        • They would also have found out that it took some time to reach equilibrium, but eventually all the concentrations remained constant i.e. in the mixture, the point of equilibrium was reached.

          • Later studies were producing graphs of concentration versus time, and these showed that eventually all the component concentrations levelled out i.e. became constant at the first point in time that a true equilibrium was established.

        • See example 2.2a.1 ester equilibrium calculation and 2.a.5 too.

      • The theoretical justification for K expressions came later in chemical history and this need not concern us at this level because it involves some pretty advanced thermodynamics theory!

      • From a student's point of view, here at this level, you are using K in terms of concentrations or partial pressures only and therefore all other equilibrium terms, apart from K itself, should be quoted in either ...

        • ... for liquid mixtures or solutions

          • [x] = concentration of x in mol dm-3 in a Kc equilibrium expression

        • ... or for gaseous mixtures

          • px = partial pressure of x in eg. Pa or atm. in a Kp equilibrium expression

          • Note

          • (i) Partial pressure is effectively a concentration term e.g. double a partial pressure and you effectively double the concentration of the gaseous component.

          • (ii) The subscript c in Kc means a concentration equilibrium expression using mol dm-3 and the subscript p in Kp means a partial pressure equilibrium using partial pressures.

          • (iii) The units of K depend on which units for concentration or partial pressure you have used for the quantities inserted in the equilibrium expression.

            • This means that sometimes the concentration or partial pressure units cancel each other out, hence K can be dimensionless!

    • For any reaction in solution or a gaseous mixture:

      • aA + bB + cC etc. (c) doc b tT + uU + wW etc.

      • in terms of concentrations ...

      • Kc =

        [T]t [U]u [V]v etc.
        --------------------------
        [A]a [B]b [C]c etc.
      • [x(?)] square brackets indicates concentration of x e.g. in mol dm-3 and the state(?) should be quoted too.
      • Note that all the equilibrium expressions you will deal with at this level only involve concentrations or partial pressures.
    • By convention, the arithmetical product of the product concentrations* of the forward reaction (RHS) are on the top line and the arithmetical product of the reactant concentrations* from the backward reaction (LHS) are on the bottom line.

      • * In all cases the product concentrations are raised to the appropriate power (a, b, c, .. t, u, w, ..) given by the stoichiometric mole ratios of the balanced equation.

      • AND again note that Kc (like Kp) is only constant for a specific constant temperature at which the concentrations of the equilibrium components might vary from one dynamic equilibrium situation to another (e.g. in reacting liquid mixtures or in solutions).

    • For heterogeneous equilibria, K expressions do not normally include values for solid phases, since their chemical potential cannot change since the concentration of a solid cannot change.

    • The equilibrium constant, Kc (or Kp later), is governed by temperature, which is the only factor that can alter the internal potential energy of the reactants or products. The 'rule' for the trend in K value change is provided by Le Chatelier's Principle.

      • If the forward reaction is exothermic, Kc (or Kp later) will decrease in value with increase in temperature.

        • The equilibrium position will shift more to the left in the endothermic direction and mathematically, by convention, the top line function will numerically decrease and the bottom line function must numerically increase.

      • If the forward reaction is endothermic, Kc (or Kp) will increase in value with increase in temperature.

        • The equilibrium position will shift more to the right in the endothermic direction and mathematically, by convention, the top line function must numerically increase and the bottom line function must numerically decrease.

      • Changes in pressure or concentration have no effect on a K value for ideal mixtures of gases/liquids or solutions.

      • Application of a catalyst to a reaction also has no effect on a K value.

      • Why are the values of equilibrium constants affected by temperature?

        • As we have noted, it is very important to quote the specific constant temperature that applies to any Kc or Kp value BECAUSE equilibrium constants vary with temperature AND ONLY temperature.

        • This is because changes in concentration, partial pressure (see section 2.1b) or employing a catalyst do NOT affect the energy content of the molecules themselves (referred to in thermodynamics as the internal energy of the molecule).

          • internal energy (symbol U) = chemical energy + thermal energy

        • However, if you change the temperature, you also change the fundamental energy content of a molecule, and we are now talking about the thermodynamic property values of the equilibrium components.

        • Electronic energy is stored in the chemical bonds of the molecules (chemical energy) plus their thermal energies of translation (kinetic energy of movement), rotation (of the whole or parts of a molecule) and vibration (of bonded atoms).

          • internal energy  = chemical energy + thermal energy

        • If you change the temperature, all the internal energy contents of the molecules are also changed, but they don't change to the same amount for each molecule for the same rise in temperature.

        • If you think of the internal energy as a sort of chemical potential to effect a chemical change, the formation of 'reactants' or 'products' may be favoured one way or the other (l to r or r to l as you write the equilibrium equation).

        • Hence the equilibrium constant not only changes, but may increase or decrease depending on whether the reaction is endothermic or exothermic.

          • The enthalpy content of a molecule (H) is related to its internal energy (U), but that's as far as we need to go here!

        • Therefore the change in the internal energy (caused by change in temperature), a fundamental thermodynamic property of a molecule, explains why equilibrium constants are ONLY affected by temperature and NOT by concentration, pressure or indeed, the presence of a catalyst.

        • It is possible to calculate equilibrium constant from thermodynamic Gibbs free energy change data for the reaction, but this may not be part of your course and certainly not appropriate to study here.

    • See later specific examples for the units of Kc (or Kp later) and if K has no units you should state so.

  • TOP and LINKSEquilibrium example 2.1a.1 The formation of hydrogen iodide

    • H2(g) + I2(g) (c) doc b 2HI(g)

    • Kc =

        [HI(g)]2
      ------------------- (no units)
      [H2(g)] [I2(g)]
      • units = (mol dm-3)2/[(mol dm-3) x (mol dm-3)], all cancel out, no units
    • Kc has no units as all the concentration units cancel out.
    • An example of the quantitative connection between kinetics (rates of reaction) and equilibrium expressions.
      • This, historically, has been one of the most studied reactions in terms of kinetics and equilibrium and is a good example to study for comparing and amalgamating two important conceptual frameworks in chemistry. (If you haven't studied kinetics - rate expressions etc. then just miss out this paragraph.)
      • The concentrations of reactants and products have been followed quantitatively by starting with either hydrogen iodide or a hydrogen iodine gas mixture at temperatures of 250-450oC.
        • The graph below show in principle what happens ...
        • (c) doc b
        • Whatever mixture you start with, eventually all the concentrations level out.
      • Both the forward (f) and backward (b) reactions occur via a simple one step mechanism i.e. via a single bimolecular collision and this simple reaction mechanism leads to simple and verifiable second order kinetics rate expressions.
      • ratef = kf [H2(g)] [I2(g)] and rateb = kb [HI(g)]2
      • Now at the point of dynamic equilibrium, with no net change in concentrations, the rate of the forward reaction = rate of the backward reaction, so
      • ratef = kf [H2(g)] [I2(g)] = rateb = kb [HI(g)]2
      • therefore [H2(g)] [I2(g)] = ratef / kf  and  [HI(g)]2 = rateb / kb
      • and substituting into the equilibrium expression, with the 'rates' cancelling out, gives
      • Kc =

          rateb x kf        kf
        ------------------ = -----
         kb x ratef         kb
      • So the equilibrium constant is equal to the ratio of the two rate constants of the forward and backward reaction.
    • Also see calculation example 2.2a.2
  • Equilibrium example 2.1a.2 The formation of the ester ethyl ethanoate

    • CH3COOH(l) + CH3CH2OH(l) (c) doc b CH3COOCH2CH3(l) + H2O(l)

    • Kc =

      [CH3COOCH2CH3(l)] [H2O(l)]
      ------------------------------------------- (no units)
      [CH3COOH(l)] [CH3CH2OH(l)]
      • units = [(mol dm-3) x (mol dm-3)]/[(mol dm-3) x (mol dm-3)], all cancel out, no units
    • See also calculations 2.2a.1 and 2.2a.5
  • Equilibrium example 2.1a.3 The formation of phosphorus(V) chloride (gaseous phase)

    • PCl3(g) + Cl2(g) (c) doc b PCl5(g)

    • Kc =

         [PCl5(g)]
      ------------------------ (units mol-1 dm3)
      [PCl3(g)] [Cl2(g)]
      • units = (mol dm-3)/[(mol dm-3) x (mol dm-3)] = mol-1 dm3
  • Equilibrium example 2.1a.4 The synthesis of ammonia

    • N2(g) + 3H2(g) (c) doc b 2NH3(g)

    • Kc =

        [NH3(g)]2
      --------------------- (units mol-2 dm6)
      [N2(g)] [H2(g)]3
      • units = [(mol dm-3)2]/[(mol dm-3) x (mol dm-3)3] = mol-2 dm6
  • Equilibrium example 2.1a.5 The synthesis of phosphorus(V) fluoride

    • P4(g) + 10F2(g) (c) doc b 4PF5(g)

    • Kc =

        [PF5(g)]4
      ----------------------- (units mol-7 dm21)
      [P4(g)] [F2(g)]10
      • units = ](mol dm-3)4}/[(mol dm-3) x (mol dm-3)10] = mol-7 dm21
  • Equilibrium example 2.1a.7 A cobalt(II) complex ion ligand exchange reaction

    • [Co(H2O)6]2+(aq) + 4Cl-(aq) (c) doc b [CoCl4]2-(aq) + 6H2O(l)

    • Kc =

             [[CoCl4]2-(aq)]
      ----------------------------------------- (units mol-4 dm12)
      [[Co(H2O)6]2+(aq)] [Cl-(aq)]4
      • units = (mol dm-3)/[(mol dm-3) x (mol dm-3)4] = mol-4 dm12
    • Note that the concentration of water is effectively constant in an aqueous solution and omitted from the equilibrium expression, but is effectively subsumed into the Kc value which in complex ion chemistry is called the stability constant denoted by Kstab.

  • Equilibrium example 2.1a.8 The acid behaviour of high oxidation state hexa-aqua ion

    • The hexaaquiron(III) ion is quite acidic in aqueous solution due to proton transfer giving the oxonium ion.

    • [Fe(H2O)6]3+(aq) + H2O(l) (c) doc b [Fe(H2O)5OH]2+(aq) + H3O+(aq)

      Kc =

      [[Fe(H2O)5OH]2+(aq)] [H3O+(aq)]
      ------------------------------------------------ (units mol dm-3)
            [[Fe(H2O)6]3+(aq)]
      • units = [(mol dm-3) x (mol dm-3)]/(mol dm-3) = mol dm-3
    • Again [H2O(l)] incorporated into Kc.

  • Some 'VERY rough rules of thumb' for an equilibrium K value (Kc or Kp in section 2.1b) and the 'position' of the equilibrium in terms of LHS (e.g. original reactants or products of backward reaction) and the RHS (products of the forward reaction):

    • For: LHS (c) doc b RHS

    • (for A + B (c) doc b C + D the rules below work ok BUT once the ratios of reactants or products are not 1:1, things are not so simple)

    • If Kc (or Kp) is >> 1 the equilibrium is mainly on the RHS, maybe virtually 100% completion of the forward reaction i.e. a very large RHS yield i.e. and likely to be very thermodynamically feasible.

    • If Kc (or Kp) is approx. 1 the equilibrium is about 50% RHS and about 50% LHS.

    • If Kc (or Kp) is << 1 the equilibrium is mainly on the LHS, maybe virtually 0% of products of the forward reaction i.e. a very low RHS yield i.e. NOT likely to be thermodynamically feasible).

    • BUT remember K changes with temperature considerably changing the position of an equilibrium, AND, at constant temperature, and therefore constant K, the position of an equilibrium can change significantly depending on relative concentrations/pressures of 'reactants' and 'products'.

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2.1b Partial pressure equilibrium law expressions

  • Kp partial equilibrium expression INTRODUCTION

  • For any gaseous reaction: aA(g) + bB(g) + cC(g) etc. (c) doc b tT(g) + uU(g) + wW(g) etc.

  • Kp =

    pTt pUu pVv etc.
    --------------------------- (for units see later)
    pAa pBb pCc etc.
  • px indicates the partial pressure of x, usually in atm (atmospheres) or Pa (pascals).

    • Do NOT put square brackets in Kp expressions!

    • AND again note that Kp (like Kc) is only constant for a specific constant temperature at which the partial pressures of the component gases might vary from one equilibrium situation to another at the same temperature.

  • The partial pressure of a gas is defined as the pressure a gaseous component in a mixture would exert, if it alone occupied the space/volume in question.

    • e.g. in air, 21% by volume is oxygen and 79% is nitrogen etc.

    • Therefore the fraction of molecules which are oxygen and nitrogen is 0.21 and 0.79 respectively.

    • Since air has a total pressure of 1 atm. or 101325 Pa, the partial pressures of oxygen and nitrogen in air are:

      • pO2 = 0.21 x 1 = 0.21 atm or pO2 = 0.21 x 101325 = 21278 Pa,

      • pN2 = 0.21 x 1 = 0.79 atm or pN2 = 0.21 x 101325 = 80047 Pa,

      • in other words, the partial pressure of a gas in a mixture pgas = its % x ptot / 100

    • In a gaseous mixture the total pressure equals the sum of all the partial pressures:

      • ptot = p1 + p2 + p3 etc. so for air ...

      • ptot-air = pO2 + pN2 + pAr + pCO2 etc. etc. = 1 atm or 101325 Pa

  • Some other useful mathematical ideas and expressions for gas mixtures:

    • The % by volume ratio is also the mole ratio of the gases in a mixture.

      • This derives from Avogadro's Law that "equal volumes of gases at the same temperature and pressure contain the same number of molecules".

    • If we call x the mole fraction of gas in a mixture then:

      • The mole fraction of a gas A in a mixture = xA = mol of A / total moles of all gases

      • Therefore the partial pressure of gas A, pA = xA x ptot

  • Equilibrium example 2.1b.1 The formation of nitrogen(II) oxide (e.g. in car engines)

    • N2(g) + O2(g) (c) doc b 2NO(g)

    • Kp =

        pNO2
      ----------------- (no units)
      pN2 pO2
      • imagine p = partial pressure units
      • units = p2/(p x p), all partial pressure units cancel out, no units
  • Equilibrium example 2.1b.2 The synthesis of sulphur(VI) oxide in the manufacture of sulphuric acid

    • 2SO2(g) + O2(g) (c) doc b 2SO3(g) (sulphur trioxide)

    • Kp =

        pSO32
      ----------------- (units e.g. atm-1 or Pa-1)
      pSO22 pO2
      • imagine p = partial pressure units
      • units = p/(p2 x p) = p/p3 = p-1
  • Equilibrium example 2.1b.3 The synthesis of ammonia (Haber Process)

    • N2(g) + 3H2(g) (c) doc b 2NH3(g)

    • Kp =

        pNH32
      ----------------- (units atm-2 or Pa-2)
      pN2 pH23
      • imagine p = partial pressure units
      • units = p2/(p x p3) = p2/p4 = p-2
    • See calculation 2.2b.2
  • Equilibrium example 2.1b.4 The manufacture of hydrogen (e.g. for ammonia synthesis)

    • CH4(g) + H2O(g) (c) doc b 3H2(g) + CO(g)

    • Kp =

       pH23 pCO
      ------------------ (units atm2 or Pa2)
      pCH4 pH2O
      • imagine p = partial pressure units
      • units = (p3 x p)/(p x p) = p4/p2 = p2
  • Equilibrium example 2.1b.5 The synthesis-manufacture of methanol (gas phase)
    • CO(g) + 2H2(g) (c) doc b CH3OH(g)
    • Kp =

       pCH3OH
      ------------------  (units atm-2 or Pa-2)
      pCO pH22
      • imagine p = partial pressure units
      • units = p/(p x p2) = p/p3 = p-2

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2.2 Exemplar calculations and concept Questions

These exemplar questions involve both numerical calculations and application of Le Chatelier's Principle.

2.2a Kc and concentration calculations

  • Kc Example Q 2.2a.1 Esterification

    • Given the esterification reaction: ethanoic acid + ethanol  ethyl ethanoate + water

    • CH3COOH(l) + CH3CH2OH(l) (c) doc b CH3COOCH2CH3(l) + H2O(l)

    • A mixture of 1.0 mol of ethanoic acid and 1.0 mol of ethanol was left to reach equilibrium at 25oC.

    • On analysis of the equilibrium mixture it was found that by titration with standard sodium hydroxide solution 0.333 mol of the ethanoic acid was left unreacted.

    • Calculate the value of the equilibrium constant Kc and give its units.

    • The reactant-product mole ratios are 1:1 ==> 1:1
    • If 0.333 mol ethanoic acid was left unreacted, then 0.667 mol of the acid had reacted.
    • Therefore 0.667 mol of ethanol must also have reacted, leaving 0.333 unreacted.
    • If 0.667 of the acid/ethanol reacted, 0.667 mol of  ethyl ethanoate and 0.667 mol of water must be formed.
    • The concentrations = mol / volume, but the volume terms cancel each other out, so substitution in the equilibrium expression can be done in terms of final numbers of moles reactants and products
    • Kc =

      [CH3COOCH2CH3(l)] [H2O(l)]
      --------------------------------------------
      [CH3COOH(l)] [CH3CH2OH(l)]
    • Kc =

      0.667 x 0.667
      ---------------------- = 4.01 (no units)
      0.333 x 0.333
  • Kc Example Q 2.2a.2 Formation of hydrogen iodide or the decomposition of hydrogen iodide

    • In these examples, don't forget that all the V's cancel, so you can work your logic in moles when solving these equilibrium problems. In the first example, concentrations are given, BUT after that its all moles, logic and maybe algebra!

    • Example (a) For the reaction: H2(g) + I2(g) (c) doc b 2HI(g)

    • An equimolar mixture of hydrogen and iodine was heated in a sealed flask at 491oC and the concentration of iodine was found to be 2.5 x 10-2 mol dm-3 and that of hydrogen iodide 1.71 x 10-1 mol dm-3.

    • (a) Calculate the value of the equilibrium constant Kc.

    • The concentration of the remaining hydrogen and iodine must be the same since they started at a 1:1 ratio and react in a 1:1 ratio.

    • Therefore substituting in the equilibrium expression:

    • Kc =

        [HI(g)]2
      -------------------
      [H2(g)] [I2(g)]
    • Kc =

            (1.71 x 10-1)2
      ---------------------------------------- = 46.8 (no units)
      (2.5 x 10-2) x (2.5 x 10-2)
    • Examples (b) More logic-mathematical solutions to the hydrogen, iodine and hydrogen iodide equilibrium

      • HI/H2/I2 moles logic to the fore!

      • (i) For the equilibrium reaction: 2HI(g) (c) doc bH2(g) + I2(g)

        • Suppose we start with 1 mole of hydrogen iodide and fraction x of it decomposes into hydrogen and iodine (x x 100 = % decomposition).

        • For every mole of HI that decomposes, you will form 0.5 moles of hydrogen and 0.5 moles of iodine.

        • TOP and LINKSThis arises from the stoichiometry of the equation i.e. the 2 : 1 : 1 mole ratio, so we can tabulate this logic as follows ...

        •   HI H2 I2
          initial moles 1.0 0.0 0.0
          moles at equilibrium 1-x 0.5x 0.5x
        • Noting that all the V's cancel, so you can write the equilibrium expression in terms of the relative moles of hydrogen iodide, hydrogen and iodine, giving .....
        • Kc =

           [H2(g)] [I2(g)]
          -----------------------
              [HI(g)]2
        • Kc =

           (0.5x)(0.5x)       0.25x2
          ------------------- = -------------- = ..... (no units)
              (1-x)2             (1-x)2
        • So, if for example, the hydrogen iodide was 20% decomposed, then obviously the proportion decomposed x = 0.20, and simple substitution into the equation enables you to calculate the equilibrium constant K without any difficulty.
          • Incidentally, since K is dimensionless, Kc = Kp even if you were working in partial pressures.
        • However, if you were given K, to solve the equation for x, the proportion decomposed, this requires some carefully worked our algebra and the formula for solving quadratic equations (shown on the right).
          • Rearranging the equilibrium expression gives
          • K(1-x)2 = 0.25x2
          • K(x2 - 2x + 1) = 0.25x2
          • Kx2 - 2Kx + K = 0.25x2
          • (K - 0.25)x2 - 2Kx + K = 0
          • So, in the quadratic equation formula
          • a = K - 0.25, b = -2K and c = K, to solve for x.
          • bon voyage and watch the sign!
      • (ii) For the equilibrium reaction: H2(g) + I2(g) (c) doc b 2HI(g)

        • Suppose we start with 1 mole of hydrogen and 1 mole of iodine and no hydrogen iodide.

        • For every x moles of hydrogen or iodine that react, 2x moles of hydrogen iodide will be formed, so we can tabulate this argument as follows ...

            H2 I2 HI
          initial moles 1.0 1.0 0.0
          moles at equilibrium 1-x 1-x 2x

          Kc =

            [HI(g)]2
          -------------------
          [H2(g)] [I2(g)]

          Kc =

           (2x)(2x)             4x2
          ------------------- = ------------- = .... (no units)
             (1-x)2             (1-x)2
        • Again, if you know the amount of HI formed OR the amount of iodine or hydrogen reacted, you can readily calculate K via the molar logic.
        • BUT if you are given K, to solve for x, this requires solving a quadratic equation.
          • Rearranging the equilibrium expression
          • K(1-x)2 = 4x2
          • K(x2 -2x + 1) = 4x2
          • Kx2 - 2Kx + K = 4x2
          • (K - 4)x2 - 2Kx + K = 0
          • So, in the quadratic equation formula
          • a = K-4, b = -2K and c = K, to solve for x.
          • watch the signs and enjoy!
      • (iii) For the equilibrium reaction: H2(g) + I2(g) (c) doc b 2HI(g)

        • A general solution for ANY combination of hydrogen and iodine forming hydrogen iodide.

        • Suppose we start with A moles of hydrogen and B moles of iodine and no hydrogen iodide.

        • For every x moles of hydrogen OR iodine that react, 2x moles of hydrogen iodide will be formed, so we can tabulate this argument as follows ...

            H2 I2 HI
          initial moles A B 0.0
          moles at equilibrium A-x B-x 2x

          Kc =

            [HI(g)]2
          -------------------
          [H2(g)] [I2(g)]

          Kc =

             (2x)2
          ------------------------ = .... (no units)
          (A-x)(B-x)
        • Again, if you know the amount of HI formed (2x) OR the amount of iodine/hydrogen reacted (x), you can readily calculate K.
          • Note that x is NOT a fraction here!
          • Its the actual moles of hydrogen or iodine that have reacted to form hydrogen iodide.
        • BUT if you are given K, to solve for x, this requires solving a quadratic equation.
          • Rearranging the equilibrium expression
          • K(A-x)(B-x) = 4x2
          • K(x2 -Ax - Bx + AB) = 4x2
          • Kx2 -AKx -BKx + ABK = 4x2
          • (K-4)x2 -(A + B)Kx + ABK = 0
          • So, in the quadratic equation formula
          • a = K-4, b = -(A+B)K and c = ABK, to solve for x.
          • This methodology is not designed for late night working! and watch the signs!
      • (iv) For the equilibrium reaction: 2HI(g) (c) doc bH2(g) + I2(g)

        • A general solution for the decomposition of hydrogen iodide.

        • Suppose we start with A moles of hydrogen iodide and x moles of it decomposes into 0.5x moles of hydrogen and 0.5x moles of iodine.

          • Remember, for every mole of HI that decomposes, you will form 0.5 moles of hydrogen and 0.5 moles of iodine.

        • TOP and LINKSThis arises from the stoichiometry of the equation i.e. the 2 : 1 : 1 mole ratio, so we can tabulate this logic as follows ...

        •   HI H2 I2
          initial moles A 0.0 0.0
          moles at equilibrium A-x 0.5x 0.5x
        • Noting that all the V's cancel, so you can write the equilibrium expression in terms of the relative moles of hydrogen iodide, hydrogen and iodine, giving .....
        • Kc =

           [H2(g)] [I2(g)]
          -----------------------
              [HI(g)]2
        • Kc =

          (0.5x)(0.5x)            0.25x2
          ---------------------- = ------------- = ..... (no units)
              (A-x)2                 (A-x)2
        • So, if you know how much HI reacted-decomposed, OR, the amount of iodine formed, then simple substitution into the equilibrium equation enables you to calculate the equilibrium constant K without any difficulty.
          • Note that x is NOT a fraction here!
          • Its the actual moles of hydrogen iodide that have decomposed to form hydrogen and iodine.
        • However, if you were given K, to solve the equation for x, the amount decomposed, this requires some carefully worked our algebra and the formula for solving quadratic equations (shown on the right).
          • Rearranging the equilibrium expression
          • K(A-x)2 = 0.25x2
          • K(x2 -2Ax + A2) = 0.25x2
          • Kx2 -2KAx + KA2 = 0.25x2
          • (K - 0.25)x2 -2KAx + KA2 = 0
          • So, in the quadratic equation formula
          • a = K-0.25, b = -2KA and c = KA2, to solve for x.
          • keep a clear head and watch the signs!
  • Kc Example Q 2.2a.3 Formation of complex ion

    • Aqueous iron(II) ions can complex with chloride ions to form the tetrachloroferrate(III) ion.

    • For the equilibrium: Fe3+(aq) + 4Cl-(aq) (c) doc b FeCl4-(aq)

    • Kc = 8.0 x 10-2 mol-1 dm3 at 298K.

    • (a) If the concentration of the free chloride ion is 0.80 mol dm-3 and that of the free iron(III) ion 0.20 mol dm-3, calculate the concentration of the tetrachloroferrate(III) complex ion.

    • Kc =

        [FeCl4-(aq)]
      -------------------------------
      [Fe3+(aq)] [Cl-(aq)]4
      • rearranging the equilibrium expression gives ...
      • [FeCl4-(aq)] = Kc x [Fe3+(aq)] x [Cl-(aq)]4
      • [FeCl4-(aq)]  = 8.0 x 10-2 x 0.20 x (0.80)4 = 6.55 x 10-3 mol dm-3
    • (b) Equal volumes of 5 molar sodium chloride solution and 0.02 molar iron(III) nitrate Fe(NO3)3 solution were mixed together.
      • (i) Assuming the chloride ion concentration changes very little and c represents the concentration of the tetrachlorateferrate(III) ion, show how the concentration of the complex ion can be approximately calculated.
        • The nitrate ion is a spectator ion and can be ignored.
        • Let [FeCl4-(aq)]equilib = c, and
        • since 1 mole of Fe3+ forms 1 mole of FeCl4- complex
        • then [Fe3+(aq)]equilib = [Fe3+(aq)]init - c, and assuming
        • [Cl-(aq)]equilib ~ [Cl-(aq)]init since [Cl-(aq)]  >> [Fe3+(aq)] + [FeCl4-(aq)]
      • (ii) Calculate the concentration of the [FeCl4-(aq)] ion.
        • The dilution factor on mixing equal volumes is 2, therefore
        • [Fe3+(aq)]init = 0.02/2 = 0.01 mol dm-3 and
        • Cl-(aq)]equilib ~ [Cl-(aq)]init = 5.0/2 = 2.5 mol dm-3
        • from (a) [FeCl4-(aq)] = Kc x [Fe3+(aq)] x [Cl-(aq)]4
        • c = 8.0 x 10-2 x (0.01 - c) x (2.5)4
        • c = 3.125 x (0.01 - c) = 0.03125 - 3.125c
        • 4.125c = 0.03125
        • [FeCl4-(aq)] = c =  0.03125/4.125 = 0.7575 = 7.58 x 10-3 mol dm-3
      • (iii) What percentage of the original Fe3+ ion is converted into the complex?
        • If all of the Fe3+ ion had been converted to the chloro complex, the concentration of the complex would be 0.01 mol dm-3
        • Therefore the % conversion = 7.58 x 10-3 x 100/0.01 = 75.8%
     
  • Kc Example Q 2.2a.4 Iodine-iodide equilibrium

    • Iodine is much more soluble in potassium iodide solution than pure water because of the equilibrium:

    • I-(aq) + I2(aq) (c) doc b I3-(aq) for which Kc = 7.10 x 102 mol-1 dm3 at 298K.

    • If the concentration of the I- ion is 0.122 mol dm-3, and that of the I3- ion is 0.153 mol dm-3, calculate the concentration of free iodine.

    • Kc =

        [I3-(aq)]
      ------------------------
      [I-(aq)] [I2(aq)]
    • Rearranging gives ...

    • [I2(aq)] =

        [I3-(aq)]
      --------------------
      Kc x [I-(aq)]
    • [I2(aq)] = 0.153 / (7.10 x 102 x 0.122) = 1.77 x 10-3 mol dm-3

  • TOP and LINKSKc Example Q 2.2a.5 Ester equilibrium - titrimetric analysis

    • A titration method for determining the equilibrium constant for an esterification reaction.

    • 12.0g of pure ethanoic acid was mixed with 11.5g of pure ethanol and left to stand for a week at room temperature (298K/25oC). The mixture was then mixed with deionised water and made up to 250 cm3 in a calibrated volumetric flask. When a 25.00 cm3 aliquot of the mixture was titrated with 0.50 mol dm-3 sodium hydroxide solution using phenolphthalein indicator (colourless ==> 1st permanent pink end-point), 10.60 cm3 of the alkali was needed for complete neutralisation.

      • (i) CH3COOH(l) + CH3CH2OH(l) (c) doc b CH3COOCH2CH3(l) + H2O(l) (esterification reaction)

      • (ii) CH3COOH(l) + NaOH(aq) ==> CH3COO-Na+(aq) + H2O(l) (neutralisation titration reaction)

    • (a) Calculate the moles of ethanoic acid unreacted in the original mixture.

      • 1 mole CH3COOH : 1 mole NaOH from equation (i) above.

      • moles = molarity x vol(dm3), so moles acid = 0.50 x 10.6/1000 = 0.0053 mol

      • since the 25.00 cm3 aliquot titrated is equal to 1/10th of the total mixture,

      • the total moles of unreacted acid = 10 x 0.0053 = 0.053 mol CH3COOH left

    • (b) Calculate the moles of ethanoic acid and ethanol in the starting mixture.

      • Mr(CH3COOH) = 60, mol ethanoic acid = 12/60 = 0.20

      • Mr(CH3CH2OH) = 46, mol ethanol = 11.5/46 = 0.25

    • (c) Calculate the moles of ethanol left unreacted and the moles of ethyl ethanoate ester and water formed.

      • This requires a little bit of logic, best appreciated with a little table showing the 'logical thinking' and the final mol numbers.

    • equation CH3COOH(l) + CH3CH2OH(l) (c) doc b CH3COOCH2CH3(l) + H2O(l)
      moles at start 0.20                0.25                    0                             0
      mol at equilibrium 0.20 - x           0.25 - x               x                             x
      mol at equilibrium 0.053         0.103            0.147              0.147
      • x = moles of ester or water formed,

      • so x mol of acid or alcohol must be used to reach equilibrium,

      • since mol ratios are 1:1 ==> 1:1 in the reaction equation.

      • Therefore 0.20 - x = 0.053 from calculation (a), therefore x = 0.20 - 0.053 = 0.147

      • so all the molar quantities can be logically deduced and are shown in the final line of the table.

      • TIP In an exam, doing the working of this part of the Q under the equation is not a bad idea.

    • (d) Calculate the equilibrium constant Kc for this esterification.

      • Kc =

        [CH3COOCH2CH3(l)] [H2O(l)]
        ---------------------------------------------
        [CH3COOH(l)] [CH3CH2OH(l)]
      • Kc =

        (0.147/V) x (0.147/V)
        ----------------------------------- = 3.96 (no units)
        (0.053/V) x (0.103/V)
      • Note that all the volume terms cancel out, so you can work purely in moles to calculate Kc.
    • (e) Suggest several reasons why it may take a week for the equilibrium point to be reached and suggest ways of speeding up the reaction.

      • (i) The reaction will be slow at room temperature, refluxing the mixture will speed things up!

      • (ii) The reaction is catalysed by hydrogen ions (H+) and since ethanoic acid is a weak acid, the hydrogen ion concentration will be very low. Adding a strong mineral acid e.g. conc. sulphuric acid.

      • (iii) Not surprisingly, esters are often prepared by refluxing the acid and alcohol with a few drops of conc. sulphuric acid added to the mixture.

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2.2b Kp, partial pressure calculations and applying Le Chatelier's Principle

  • Example Q 2.2b.1 Nitrogen(IV) oxide-dinitrogen tetroxide equilibrium (nitrogen dioxide <=> dimer)

    • For the equilibrium between dinitrogen tetroxide and nitrogen dioxide:

      • N2O4(g) (c) doc b 2NO2(g) (ΔH = +58 kJ mol-1)

    • the value of the equilibrium constant Kp = 0.664 atm at 45oC.

    • (a) If the partial pressure of dinitrogen tetroxide was 0.449 atm, what would be the equilibrium partial pressure of nitrogen dioxide and the total pressure of the gases? and what % of the dinitrogen tetroxide is dissociated?

      • Kp =

        pNO22
        -------------
        pN204
      • Rearranging the

      • pNO22 = Kp x pN204, so, pNO2 = √(Kp x pN204 ) = √(0.664 X 0.449) = 0.546 atm

      • ptot = pNO2 + pN2O4 = 0.546 + 0.449 = 0.995 atm

      • The pressure of NO2 is 0.546, which is equivalent to 0.273 atm of N2O4 before dissociation.

      • Therefore the % N2O4 dissociated = 0.273 x 100 / (0.449 + 0.273) = 37.8%

    • (b) In another experiment at 77oC and a total pressure of 1.000 atm, the partial pressure of dinitrogen dioxide was found to be 0.175 atm. Calculate the value of the equilibrium constant at 77oC and the % dissociation of the dinitrogen tetroxide.

      • ptot = pNO2 + pN2O4, so pNO2 = ptot - pN2O4 = 1.000 - 0.175 = 0.825 atm

      • From the equilibrium expression above:

      • Kp = 0.8252 / 0.175 = 3.89 atm

      • The pressure of NO2 is 0.825, which is equivalent to 0.4125 atm of N2O4 before dissociation.

      • Therefore the % N2O4 dissociated = 0.4125 x 100 / (0.175 + 0.4125) = 70.2%

    • (c) At a total pressure of 102000 Pa, dinitrogen tetroxide is 40% dissociated at 50oC.

      • (i) Calculate the partial pressures of the gases present.

        • Each mole of N2O4 dissociated gives 2 moles of NO2.

        • Lets assume we start with 100 mol N2O4.

        • An initial 100 mol of N2O4 gives, at equilibrium,

        • 60 mol of undissociated N2O4, and 40 x 2 = 80 mol of NO2.

        • mole fraction of NO2 = 80 / (60 + 80) = 0.571,

        • and pNO2 =  0.571 x 102000 = 58242 Pa

        • mole fraction of N2O4 = 60 / (60 + 80) = 0.429,

        • and pN2O4 =  0.429 x 102000 = 43758 Pa

        • (see also example Q 2.2b.3 for another way of setting out a method to calculate the partial pressures)

      • (ii) Calculate the value of the equilibrium constant.

        • Kp =

          pNO22    582422
          ------- = ------------- = 7.75 x 104 Pa
          pN204     43758
    • (d) Explain the effect, if any, of increasing the total pressure of the system on the position of the N2O4/NO2 equilibrium.

      • Increasing pressure will favour the LHS, i.e. more N2O4 and less NO2, because the system will move to the side with the least number of gaseous molecules to try to minimise the increase in pressure (1 mol gas <== 2 mol gas).

    • (e) From information from parts (a)-(c) is the dissociation exothermic or endothermic? and give your reasoning.

      • The dissociation is endothermic because on raising the temperature from 45oC to 77oC the value of Kp has increased.

      • i.e. pNO2 has increased in value and pN2O4 decreased in value from more dissociation.

      • An equilibrium system moves in the endothermic direction on raising the temperature to absorb the 'added' heat to try to minimise the temperature rise.

  • TOP and LINKSExample Q 2.2b.2 Ammonia synthesis

    • Ammonia is synthesised by combing nitrogen and hydrogen, but the reaction is reversible.

    • N2(g) + 3H2(g) (c) doc b 2NH3(g) (ΔH = -92 kJ mol-1)

    • In an experiment starting with a 1:3 ratio H2:N2 mixture, at 400oC and total pressure of 200 atm, the equilibrium mixture was found to contain 36.3% by volume of ammonia.

    • (a) Write out the equilibrium expression for Kp.

      • Kp =

          pNH32
        ---------------- (units atm-2 or Pa-2)
        pN2 pH23
    • (b) Calculate the partial pressures of nitrogen, hydrogen and ammonia.

      • 36% of ammonia means its mole fraction is 0.36, so pNH3 = 0.36 x 200 = 72 atm

      • The other 64% of gases must be split on a 1:3 ratio between nitrogen and hydrogen.

        • If you think of the 1:3 ratio as 1 and 3 parts out of 4 the logic becomes easy.

      • So mole fraction of nitrogen x ptot = pN2 = 64/100 x 1/4 x 200 = 32 atm,

      • and mole fraction of hydrogen x ptot = pH2 = 64/100 x 3/4 x 200 = 96 atm

      • check: ptot = pN2 + pH2 + pNH3 = 32 + 96 + 72 = 200 atm!

    • (c) Calculate the value of the equilibrium constant at 400oC and give its units.

      • Kp =

          pNH32
        ------------------
        pN2 pH23
      • Kp =

          722
        ---------------- = 1.83 x 10-4 atm-2
        32 x 963
    • (d) What will be the effect on ammonia yields and the value of Kp by (i) raising the pressure and (ii) raising the temperature. Give reasons for your answers.
      • (i) The reaction to form ammonia is exothermic, so higher temperature will favour its endothermic decomposition back to nitrogen and hydrogen. So the yield of ammonia and the value of Kp will be reduced as the system will absorb heat energy to attempt to minimise temperature rise.
      • (ii) Higher pressure will favour a higher yield of ammonia because 4 mol gaseous reactants ==> 2 mol gaseous products, so if the system is subjected to higher pressure it reduces the number of gas molecules to attempt to minimise the enforced increase in pressure. Assuming the gases behave ideally, the value of the equilibrium constant Kp is unaffected by pressure changes, only temperature affects the value of Kp in an ideal gas mixture.
  • TOP and LINKSExample Q 2.2b.3 Phosphorus(V) chloride (phosphorus pentachloride) dissociation

    • At high temperatures vapourised phosphorus(V) chloride dissociates into gaseous phosphorus(III) chloride (phosphorus trichloride) and chlorine.

    • PCl5(g) (c) doc b PCl3(g) + Cl2(g)

      (a) Write the equilibrium expression for this reaction in terms of Kp and partial pressures.

      Kp =

        PPCl3 x PCl2
      --------------------- (units atm, Pa, kPa etc.)
          PPCl5
    • (b) At a particular temperature 40% of the PCl5 dissociates into PCl3 and Cl2. If the total equilibrium pressure is 210 kPa calculate the value of Kp at this temperature quoting appropriate units. The solution is set out below in a series of logical steps, either thinking from the point of view of starting with n/1 moles of PCl5 or starting with 100/100% of PCl5 molecules.

    • PCl5(g (c) doc b PCl3(g) Cl2(g) comments, z refers to any component in the mixture
      n(1-x)

      (1-x)

      can think of as (100 - x%)

        nx

      x

      can think of as  x% dissociated

      nx

      x

      can think of as  x% dissociated

      n = initial moles of undissociated PCl5 for a general solution, but consider 1 mole of PCl5 (n = 1) of which fraction x dissociates giving a total of (1 + x) moles of gas from 1 mol of PCl5

      This is the logic thinking in terms of 100/100% undissociated PPCl5 molecules and x% is the percentage of PCl5 molecules dissociated.

      (1-x)/(1 + x)   x/(1+x) x/(1+x) calculation of mole fractions

      mole fraction z = moles z/total moles

      0.6/1.4 = 0.4286

      can think of as 60/140 = 0.4286

        0.4/1.4 = 0.2857

      can think of as 40/140 = 0.2857

      0.4/1.4 = 0.2857

      can think of as 40/140 = 0.2857

      since x = 0.4 (= 40% dissociated)

      It works out the same in terms of the original 100/100% undissociated PCl5

      PPCl5 = 0.4286 x 210 = 90.0   PPCl3 = 0.2857 x 210 = 60.0 PCl2 = 0.2857 x 210 = 60 partial pressure of component z

      Pz = mole fraction z x Ptot

      note 90 + 60 + 60 = 210 check!

    • Kp =

        60.0 x 60.0
      --------------------- = 40.0 kPa
          90.0
    • -

  • Example Q

    • -

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Part 2 sub-index: 2.1 Kc and Kp equilibrium expressions and Le Chatelier's Principle * 2.2 Exemplar calculations using Kc expressions or Kp expressions

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle-rules * Part 3. Equilibria and industrial processes * Part 4. Partition, solubility product and ion-exchange * Part 5. pH, weak-strong acid-base theory and calculations * Part 6. Salt hydrolysis, Acid-base titrations-indicators, pH curves and buffers * Part 7. Redox equilibria, half-cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria-vapour pressure, boiling point and intermolecular forces

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