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 Equilibria Part 2

 "Chemical equilibrium expressions and calculations"

GCSE Notes on reversible reactions-equilibrium * Advanced Part 1. Equilibrium, Le Chatelier's Principle-rules * Part 2 sub-index: 2.1 K_c and K_p equilibrium expressions and Le Chatelier's Principle * 2.2 Exemplar calculations using K_c expressions or K_p expressions * 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 * The K and ΔS-ΔG connection with E^Ø_cell will be dealt with via new thermodynamics pages later, but an example of a ΔS-ΔG calculation is given at the end of the advanced kinetics pages and ΔG for cells is mentioned in Equilibria Part 7. M = old fashioned shorthand for mol dm^-3 * EMAIL query?comment

2.1a Molar concentration expressions

• K_c 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. This used to be called the 'law of mass action'.

    • The theoretical justification for K expressions came later in chemical history and need not concern us at this level.

  • For any reaction in solution or a gaseous mixture:

    • aA + bB + cC etc. tT + uU + wW etc.

  • 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.

  • 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.

  • 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, K_c (or K_p 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, K_c (or K_p 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, K_c (or K_p) 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.

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

• Equilibrium example 2.1a.1 The formation of hydrogen iodide

  • H_2(g) + I_2(g) 2HI_(g)

  • Kc =	[HI(g)]2
    	----------- (no units)
    	[H2(g)] [I2(g)]

  • K_c 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 250-450^oC. The graphs below show in principle what happens.
    • 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.
    • rate_f = k_f [H_2(g)] [I_2(g)] and rate_b = k_b [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
    • rate_f = k_f [H_2(g)] [I_2(g)] = rate_b = k_b [HI_(g)]²
    • therefore [H_2(g)] [I_2(g)] = rate_f / k_f  and  [HI_(g)]² = rate_b / k_b
    • 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) CH₃COOCH₂CH_3(l) + H₂O_(l)

  • Kc =	[CH3COOCH2CH3(l)] [H2O(l)]
    	----------------------- (no units)
    	[CH3COOH(l)] [CH3CH2OH(l)]

  • See also calculations 2.2a.1 and 2.2a.5

• Equilibrium example 2.1a.3 The formation of phosphorus(V) chloride (gaseous phase)

  • PCl_3(g) + Cl_2(g) PCl_5(g)

  • Kc =	[PCl5(g)]
    	------------ (units mol-1 dm3)
    	[PCl3(g)] [Cl2(g)]

• Equilibrium example 2.1a.4 The synthesis of ammonia

  • N_2(g) + 3H_2(g) 2NH_3(g)

  • Kc =	[NH3(g)]2
    	----------- (units mol-2 dm6)
    	[N2(g)] [H2(g)]3

• Equilibrium example 2.1a.5 The synthesis of phosphorus(V) fluoride

  • P_4(g) + 10F_2(g) 4PF_5(g)

  • Kc =	[PF5(g)]4
    	----------- (units mol-7 dm21)
    	[P4(g)] [F2(g)]10

• Equilibrium example 2.1a.7 A cobalt(II) complex ion ligand exchange reaction

  • [Co(H₂O)₆]²⁺_(aq) + 4Cl^-_(aq) [CoCl₄]^2-_(aq) + 6H₂O_(l)

  • Kc =	[[CoCl4]2-(aq)]
    	---------------------- (units mol-4 dm12)
    	[[Co(H2O)6]2+(aq)] [Cl-(aq)]4

  • Note that the concentration of water is effectively constant in an aqueous solution and ommited from the equilibrium expression, but is effectively subsumed into the K_c value which in complex ion chemistry is called the stability constant denoted by K_stab.

• 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) [Fe(H₂O)₅OH]²⁺_(aq) + H₃O⁺_(aq)

    Kc =	[[Fe(H2O)5OH]2+(aq)] [H3O+(aq)]
    	------------------------- (units mol dm-3)
    	[[Fe(H2O)6]3+(aq)]

  • Again [H₂O_(l)] incorporated into K_c.

• Some 'VERY rough rules of thumb' for an equilibrium K value (K_c or K_p 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 RHS

  • (for A + 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 K_c (or K_p) 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 K_c (or K_p) is approx. 1 the equilibrium is about 50% RHS and about 50% LHS.

  • If K_c (or K_p) 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'.

2.1b Partial pressure expressions

• K_p partial equilibrium expression INTRODUCTION

• For any gaseous reaction: aA_(g) + bB_(g) + cC_(g) etc. tT_(g) + uU_(g) + wW_(g) etc.

• Kp =	pTt pUu pVv etc.
  	-------------- (for units see later)
  	pAa pBb pCc etc.

• p_x indicates the partial pressure of x, usually in atm (atmospheres) or Pa (pascals).

  • Do NOT put square brackets in K_p expressions!

• 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:

    • p_O2 = 0.21 x 1 = 0.21 atm or p_O2 = 0.21 x 101325 = 21278 Pa,

    • p_N2 = 0.21 x 1 = 0.79 atm or p_N2 = 0.21 x 101325 = 80047 Pa,

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

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

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

    • p_tot-air = p_O2 + p_N2 + p_Ar + p_CO2 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 = x_A = mol of A / total moles of all gases

    • Therefore the partial pressure of gas A, p_A = x_A x p_tot

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

  • N_2(g) + O_2(g) 2NO_(g)

  • Kp =	pNO2
    	------- (no units)
    	pN2 pO23

• Equilibrium example 2.1b.2 The synthesis of sulphur(VI) oxide in the manufacture of sulphuric acid

  • 2SO_2(g) + O_2(g) 2SO_3(g) (sulphur trioxide)

  • Kp =	pSO32
    	-------- (units atm-1 or Pa-1)
    	pSO22 pO2

• Equilibrium example 2.1b.3 The synthesis of ammonia (Haber Process)

  • N_2(g) + 3H_2(g) 2NH_3(g)

  • Kp =	pNH32
    	------- (units atm-2 or Pa-2)
    	pN2 pH23

  • See calculation 2.2b.2

• Equilibrium example 2.1b.4 The manufacture of hydrogen (e.g. for ammonia synthesis)

  • CH_4(g) + H₂O_(g) 3H_2(g) + CO_(g)

  • Kp =	pH23 pCO
    	-------- (units atm2 or Pa2)
    	pCH4 pH2O

• Equilibrium example 2.1b.5 The synthesis-manufacture of methanol (gas phase)
  • CO_(g) + 2H_2(g) CH₃OH_(g)

  • Kp =	pCH3OH
    	-------  (units atm-2 or Pa-2)
    	pCO pH22

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

• K_c Example Q 2.2a.1 Esterification

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

  • CH₃COOH_(l) + CH₃CH₂OH_(l) CH₃COOCH₂CH_3(l) + H₂O_(l)

  • A mixture of 0.5 mol of ethanoic acid and 0.5 mol of ethanol was left to reach equilibrium at 25^oC.

  • 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 K_c 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

• K_c Example Q 2.2a.2 Formation of hydrogen iodide

  • For the reaction: H_2(g) + I_2(g) 2HI_(g)

  • An equimolar mixture of hydrogen and iodine was heated in a sealed flask at 491^oC 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 K_c.

  • 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)

  • ?

• K_c 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) FeCl₄^-_(aq)

  • K_c = 8.0 x 10^-2 mol^-1 dm³ at 298.

  • If the concentration of the 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)] = K_c x [Fe³⁺_(aq)] x [Cl^-_(aq)]⁴  = 8.0 x 10^-2 x 0.80 x (0.20)⁴ = 1.02 x 10^-4 mol dm^-3

• K_c 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) + I_2(aq) I₃^-_(aq) for which K_c = 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)]

  • [I_2(aq)] = 0.153 / (7.10 x 10² x 0.122) = 1.77 x 10^-3 mol dm^-3

• K_c 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/25^oC). 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) CH₃COOCH₂CH_3(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 unreated acid = 10 x 0.0053 = 0.053 mol CH₃COOH left

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

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

    • M_r(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) 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 K_c 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 K_c.

  • (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.

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

  • For the equilibrium between dinitrogen tetroxide and nitrogen dioxide:

    • N₂O_4(g) 2NO_2(g) (ΔH = +58 kJ mol^-1)

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

  • (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

    • p_NO2² = K_p x p_N204, so, p_NO2 = √(K_p x p_N204 ) = √(0.664 X 0.449) = 0.546 atm

    • p_tot = p_NO2 + p_N2O4 = 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 77^oC 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 77^oC and the % dissociation of the dinitrogen tetroxide.

    • p_tot = p_NO2 + p_N2O4, so p_NO2 = p_tot - p_N2O4 = 1.000 - 0.175 = 0.825 atm

    • From the equilibrium expression above:

    • K_p = 0.825² / 0.175 = 3.89 atm

    • The pressure of NO₂ is 0.802, which is equivalent to 0.401 atm of N₂O₄ before dissociation.

    • Therefore the % N₂O₄ dissociated = 0.401 x 100 / (0.175 + 0.401) = 69.6%

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

    • (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 p_NO2 =  0.571 x 102000 = 58242 Pa

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

      • and p_N2O4 =  0.429 x 102000 = 43758 Pa

    • (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 45^oC to 77^oC the value of K_p has increased.

    • i.e. p_NO2 has increased in value and p_N2O4 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.

• Example Q 2.2b.2 Ammonia synthesis

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

  • N_2(g) + 3H_2(g) 2NH_3(g) (ΔH = -92 kJ mol^-1)

  • In an experiment starting with a 1:3 ratio H₂:N₂ mixture, at 400^oC 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 K_p.

    • 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 p_NH3 = 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 p_tot = p_N2 = ⁶⁴/₁₀₀ x ¹/₄ x 200 = 32 atm,

    • and mole fraction of hydrogen x p_tot = p_H2 = ⁶⁴/₁₀₀ x ³/₄ x 200 = 96 atm

    • check: p_tot = p_N2 + p_H2 + p_NH3 = 32 + 96 + 72 = 200 atm!

  • (c) Calculate the value of the equilibrium constant at 400^oC 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 K_p 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 K_p 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 K_p is unaffected by pressure changes, only temperature affects the value of K_p in an ideal gas mixture.

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