Revision notes on chemical equilibrium - partial pressure equilibrium expression/constant (Kp) calculations Advanced Level Theoretical-Physical Chemistry

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Doc Brown's Chemistry Advanced A Level Notes - TheoreticalPhysical Advanced Level Chemistry  Equilibria  Chemical Equilibrium Revision Notes PART 2

2c. Kp chemical equilibrium expression, Kp equilibrium constant and Kp equilibrium calculations

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

The equilibrium constant Kp is deduced from the balanced chemical equation for a reversible reaction, NOT experimental data as is the case for rate expressions in kinetics.

• Some 'VERY rough rules of thumb' for an equilibrium Kp value 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 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 Kp is approx. 1 the equilibrium is more evenly distributed between the RHS and LHS.

• If 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. likely to be less thermodynamically feasible.

• BUT remember, Kp 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'.

• Finally a catalyst may speed up getting to the equilibrium but a catalyst cannot affect the position of the equilibrium constant or the value of the equilibrium constant K (Kc or Kp).

2.c Partial pressure equilibrium law expressions

• Kp 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.
• 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 'rule' for the trend in Kp value change is provided by Le Chatelier's Principle.

• (i) If the forward reaction is endothermic, then Kp increases with increase in temperature (so Kp decreases if temperature decreased).

• (ii) If the forward reaction is exothermic, then Kp decreases with increase in temperature. (so Kp increases if temperature decreased).

• 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) 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 sulfur(VI) oxide in the manufacture of sulfuric acid

• The Contact Process

• 2SO2(g) + O2(g) 2SO3(g) (sulfur 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) 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) 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) 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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2.2b 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) 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 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.

• Example Q 2.2b.2 Ammonia synthesis

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

• N2(g) + 3H2(g) 2NH3(g) (ΔH = 92 kJ mol1)

• In an experiment starting with a 1:3 ratio N2:H2 mixture, at 400oC, the final total equilibrium pressure was 200 atm. (See graph of yields in section)

• The final equilibrium mixture was found to contain 36% by volume of ammonia.

• Assume the gases behave ideally.

• (a) Write out the equilibrium expression for Kp and quote some typical units.

•  Kp = pNH32     (units atm2 or Pa2) pN2 pH23
• (b) What was the % by volume of nitrogen in the original mixture and how will it change?

• from the 1:3 ratio, % ammonia = 100 x 1/4 = 25% N2

• Since some nitrogen will combine with the hydrogen, its percentage will decrease.

• (c) Predict and explain any difference between the initial and final pressures of the system.

• Since the equilibrium has moved to the right i.e. ammonia formed, 4 molecules of gaseous reactants gives 2 molecules of product.

• Since the total equilibrium number of molecules is reduced, the initial total pressure must have been greater than 200 atm at the start.

• (d) Calculate the partial pressures of nitrogen, hydrogen and ammonia and the % nitrogen and hydrogen in the final equilibrium mixture.

• 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 N2:H2 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,

• nitrogen is 16% of the equilibrium mixture (1/4 of 64), compared to 25% in the original mixture.

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

• hydrogen is 48% of the equilibrium mixture (3/4 of 64).

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

• (e) 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
• (f) 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.
• Example 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) 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 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

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WHAT NEXT?

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principlerules * Part 2. Kc and Kp equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4 Partition between two phases, solubility product Ksp, common ion effect, ionexchange systems * 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 watch out for sub-indexes to multiple sections or pages

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