Kp chemical equilibrium expression partial pressures Kp equilibrium constant how to do Kp equilibrium partial pressure calculations problem solving Advanced A level chemistry GCE AS A2 revision notes KS5

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

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

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

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.

Chemical Equilibrium Notes Index

Part 2a. How to write equilibrium expressions and the effect of temperature on the equilibrium constant K

Part 2b. Writing K_c equilibrium expressions, units and exemplar K_c concentration calculations including esterification

Part 2c. Writing K_p expressions, units, and exemplar K_p partial pressure equilibrium calculations

Part 3. Applying Le Chatelier's Principle and equilibrium expressions to Industrial Processes

The equilibrium constant K_p 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 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_p is approx. 1 the equilibrium is more evenly distributed between the RHS and LHS.
- If 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. likely to be less thermodynamically feasible.
- BUT remember, K_p 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 (K_c or K_p).

2.c Partial pressure equilibrium law 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.

K_p =	p_T^t p_U^u p_V^v etc.
	––––––––––––––––– (for units see later)
	p_A^a p_B^b p_C^c 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!
- AND again note that K_p (like K_c) 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 K_p value change is provided by Le Chatelier's Principle.
- (i) If the forward reaction is endothermic, then K_p increases with increase in temperature (so K_p decreases if temperature decreased).
- (ii) If the forwards reaction is exothermic, then K_p decreases with increase in temperature. (so K_p 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:
  - 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)

K_p =	p_NO²
	–––––––––– (no units)
	p_N2 p_O2

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
2SO_2(g) + O_2(g) 2SO_3(g) (sulfur trioxide)

K_p =	p_SO3²
	–––––––––– (units e.g. atm^–1 or Pa^–1)
	p_SO2² p_O2

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)

N_2(g) + 3H_2(g) 2NH_3(g)

K_p =	p_NH3²
	–––––––––––– (units atm^–2 or Pa^–2)
	p_N2 p_H2³

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)

CH_4(g) + H₂O_(g) 3H_2(g) + CO_(g)

K_p =	p_H2³ p_CO
	––––––––––– (units atm² or Pa²)
	p_CH4 p_H2O

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) + 2H_2(g) CH₃OH_(g)

K_p =	p_CH3OH
	––––––––––––– (units atm^–2 or Pa^–2)
	p_CO p_H2²

imagine p = partial pressure units
units = p/(p x p²) = p/p³ = p^–2

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2.2b K_p, 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:
- 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?

K_p =	p_NO2²
	––––––
	p_N204

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 0.825, 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 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
- 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.

K_p =	p_NO2² 58242²
	––––––– = ––––––– = 7.75 x 10⁴ Pa
	p_N204 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 N₂:H₂ mixture, at 400^oC, the final total equilibrium pressure was 200 atm. (See graph of yields in Haber synthesis 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 K_p and quote some typical units.

K_p =	p_NH3²
	––––––––– (units atm^–2 or Pa^–2)
	p_N2 p_H2³

(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 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 N₂:H₂ 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,
  - nitrogen is 16% of the equilibrium mixture (1/4 of 64), compared to 25% in the original mixture.
- and mole fraction of hydrogen x p_tot = p_H2 = ⁶⁴/₁₀₀ x ³/₄ x 200 = 96 atm
  - hydrogen is 48% of the equilibrium mixture (3/4 of 64).
- check: p_tot = p_N2 + p_H2 + p_NH3 = 32 + 96 + 72 = 200 atm !

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

K_p =	p_NH3²
	––––––––
	p_N2 p_H2³

K_p =	72²
	–––––––– = 1.83 x 10^–4 atm^–2
	32 x 96³

(f) 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.

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.
PCl_5(g) PCl_3(g) + Cl_2(g)

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

K_p =
P_PCl3 x P_Cl2

–––––––––––– (units atm, Pa, kPa etc.)

P_PCl5
(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 K_p 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.

PCl_5(g	PCl_3(g)	Cl_2(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 P_PCl5 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
P_PCl5 = 0.4286 x 210 = 90.0	P_PCl3 = 0.2857 x 210 = 60.0	P_Cl2 = 0.2857 x 210 = 60	partial pressure of component z P_z = mole fraction z x P_tot note 90 + 60 + 60 = 210 check!

K_p =
60.0 x 60.0

–––––––––––– = 40.0 kPa

90.0
–

Example Q
- –

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Part 2a. How to write equilibrium expressions and the effect of temperature on the equilibrium constant K

Part 2b. Writing K_c equilibrium expressions, units and exemplar K_c concentration calculations including esterification

Part 2c. Writing K_p expressions, units, and exemplar K_p partial pressure equilibrium calculations

Part 3. Applying Le Chatelier's Principle and equilibrium expressions to Industrial Processes

Chemical Equilibrium Notes Index

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Chemical Equilibrium Notes Index