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Temperature, pressure and Kp - Equilibria and Kp, Lecture notes of Chemistry

Blackburn CollegeChemistry

Characteristics of the dynamic equilibrium and Le Chatelier’s Principle are explained

Typology: Lecture notes

2020/2021

Uploaded on 05/24/2021

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1.10 Equilibria and Kp
AS review
• Consider the reaction:
2SO2(g)
+
O2(g)
2SO3(g)
Characteristics of the dynamic equilibrium
1) The rate of the forward reaction is equal to the rate of the reverse reaction
2) The concentration of the reactants and products are unchanged under stable conditions
Le Chatelier’s Principle
When a reaction at equilibrium is subject to a change in concentration, pressure
or temperature, the position of the equilibrium will move to counteract the change.
1) Changing concentration – Position of equilibria shifts, Kc unchanged
H2(g)
+
I2(g)
2HI(g)
Purple Colourless
Increasing concentration:
• Adding H2 – increases the concentration of H2
• Equilibrium shifts to the products
• To reduce the concentration of the H2 – counteracting the change
• The mixture will become less purple
Decreasing concentration:
• Removing I2 – decreases the concentration of I2
• Equilibrium shifts to the reactants
• To increase the concentration of the I2 – counteracting the change
• The mixture will become more purple
The equilibrium moves to oppose the change in concentration
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff

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Download Temperature, pressure and Kp - Equilibria and Kp and more Lecture notes Chemistry in PDF only on Docsity!

1.10 Equilibria and Kp

AS review

  • Consider the reaction: 2SO2(g) + O2(g) 2SO3(g) Characteristics of the dynamic equilibrium 1 ) The rate of the forward reaction is equal to the rate of the reverse reaction
  1. The concentration of the reactants and products are unchanged under stable conditions **Le Chatelier’s Principle When a reaction at equilibrium is subject to a change in concentration, pressure or temperature, the position of the equilibrium will move to counteract the change.
  2. Changing concentration** – Position of equilibria shifts, Kc unchanged H2(g) + I2(g) 2 HI(g) Purple Colourless Increasing concentration:
  • Adding H 2 – increases the concentration of H 2
  • Equilibrium shifts to the products
  • To reduce the concentration of the H 2 – counteracting the change
  • The mixture will become less purple Decreasing concentration:
  • Removing I 2 – decreases the concentration of I 2
  • Equilibrium shifts to the reactants
  • To increase the concentration of the I 2 – counteracting the change
  • The mixture will become more purple The equilibrium moves to oppose the change in concentration

2) Changing pressure – gases only - Position of equilibria shifts, Kc unchanged N 2 O 4 2NO 2 Colourless Brown 1 Mole 2 Moles Increasing pressure:

  • Equilibrium shifts to the reactants
  • This is the side with fewer moles of gas
  • This will reduce the pressure – counteracting the change
  • The mixture will become less brown Decreasing pressure:
  • Equilibrium shifts to the products
  • This is the side with more moles of gas
  • This will increase the pressure – counteracting the change
  • The mixture will become more brown **The equilibrium moves to oppose the change in pressure
  1. Changing temperature** - Position of equilibria shifts, Kc changes N 2 O 4 2NO 2 D H = +58 kj mol-^1 Colourless Brown Exothermic Endothermic Increasing temperature:
  • Equilibrium shifts to the products
  • As this is the endothermic direction
  • This will decrease temperature – counteracting the change
  • The mixture will become more brown Decreasing temperature:
  • Equilibrium shifts to the reactants
  • As this is the exothermic direction
  • This will increase temperature – counteracting the change
  • The mixture will become less brown The equilibrium moves to oppose the change in temperature

Kp and equilibria:

  • Kc and Kp are both expressions to show the position of an equilibria
  • c stands for an equilibrium where the individual species are measured using concentrations
  • p stands for an equilibrium where the individual species are measured using partial pressures:

Imagine a mixture of gases at equilibrium with a total pressure, P t = 100 KPa

  • A total of 20 moles are responsible for the 100 KPa of pressure.
  • If we separate the mixture into individual species: 10 / 20 th’s moles of the filled red particles are responsible for a partial amount of the pressure. Partial pressure ‘r’, ppr 6 / 20 th’s moles of the horizontal blue striped particles are responsible for a partial amount of the pressure. Partial pressure ‘b’, ppb 4 / 20 th’s moles of the vertical green striped particles are responsible for a partial amount of the pressure. Partial pressure ‘g’, ppg pp r = 50 KPa pp b = 30 KPa pp g = 20 KPa The partial pressures, pp , are essentially proportional to the concentrations The total pressure, Pt is equal to the sum of all of the partial pressures: P t = P r + P b + P g 100 = 50 + 30 + 20

Writing Kp

  • The units of Kp will depend upon the units that pressure is measured in: Pa KPa mmHg Atm MPa
  • It will also depend on the molar ratio in the equilibria equation. Example: 2SO3(g) 2SO2(g) + O2(g)

Kp = pp products

pp reactants

Kp = (pp so 2 )

2 x pp O 2

(pp so 3 )

2 Units: Kp = Pa 2 x Pa Pa^2 Kp = Pa^2 x Pa Pa 2 Kp = Pa

**The equilibrium position and Kp

  1. Changing pressure – gases only** - Position of equilibria shifts, Kc unchanged N 2 O 4 2NO 2 Colourless Brown 1 Mole 2 Moles Increasing pressure: a) Equilibrium position moves to left:
  • Equilibrium shifts to the reactants
  • This is the side with fewer moles of gas
  • This will reduce the pressure – counteracting the change
  • The mixture will become less brown b) Kp is unchanged:
  • With an increase in pressure both partial pressures increase (smaller volume)
  • Equilibrium shifts to the side with fewer moles of gas (LHS) to relieve pressure.
  • This will increase the partial pressure further of the N 2 O 4
  • The partial pressure of NO 2 decreases slightly due to the shift in equilibrium. The initial increase in the partial pressures at the start outweighs this slight decrease resulting in an overall increase partial pressure of NO 2

Decreasing pressure: a) Equilibrium position moves to right:

  • Equilibrium shifts to the products
  • This is the side with more moles of gas
  • This will increase the pressure – counteracting the change
  • The mixture will become more brown b) Kp is unchanged:
  • With a decrease in pressure both partial pressures decrease (larger volume)
  • Equilibrium shifts to the side with more moles of gas (RHS) to increase pressure.
  • This will decrease the partial pressure further of the N 2 O 4
  • The partial pressure of NO 2 increases slightly due to the shift in equilibrium. The initial decrease in the partial pressures at the start outweighs this slight increase resulting in an overall decrease partial pressure of NO 2

Decreasing temperature: a) Equilibrium shifts to the reactants

  • As this is the exothermic direction
  • This will increase temperature – counteracting the change
  • The mixture will become less brown b) Kp is decreases:
  • As equilibria shifts to the products, pp NO 2 decreases
  • As equilibria shifts to the products, pp N 2 O 4 increases
  • Products decrease / reactants increase, therefore Kp decreases 3 ) The effect of a catalyst on an equilibrium Forward reaction Reverse reaction
  • A catalyst has no effect on the position of the equilibrium.
  • A catalyst speeds up the forward and reverse reaction equally so it will only increase the rate at which equilibrium is achieved.

How are partial pressures calculated:

  • Remember, each individual species exerts their own partial pressure, which, when all added together gives the total pressure.
  • The partial pressure is due to the mole fraction of the total number of moles of that species present in the mixture: Partial pressure = mole fraction x total pressure Mole fraction = number of moles of gas total number of moles of gas Mole fraction: mf = n / nt = 10 / 20 = 0. Partial pressure: ppr = mf x Pt = 0.5 x 100 = 50 Mole fraction: mf = n / nt = 6 / 20 = 0. Partial pressure: ppb = mf x Pt = 0.3 x 100 = 30 Mole fraction: mf = n / nt = 4 / 20 = 0. Partial pressure: ppg = mf x Pt = 0.2 x 100 = 20 ppr = 50 KPa ppb = 30 KPa ppg = 20 KPa Kp = 30 x 20 50 Kp = 1.2 KPa

Questions: Give all answers to 3SF

  1. The following equilibria was found to contain 5 moles of HCl, 10 moles of H 2 and 8 moles of Cl 2. The total pressure of the sealed vessel was 50KPa 2 HCl H 2 + Cl 2 a) Write an expression for Kp b) Calculate the mole fractions for each gas c) Calculate the partial pressures for each gas d) Calculate Kp for the equilibria. Include units in your answer. Ans = 3.
  2. The following equilibria was found to contain 5 moles of NO 2 and 8 moles of N 2 O 4. The total pressure of the sealed vessel was 760 mmHg N 2 O 4 2NO 2 Calculate Kp. Include units and show all working out clearly: Ans = 183
  1. The following equilibria was found to contain 5 moles of SO 2 , 5 moles of O 2 and 8 moles of SO 3. The total pressure of the sealed vessel was 2 atm 2SO 2 + O 2 2SO 3 Calculate Kp. Include units and show all working out clearly: Ans = 4.
  2. An 80 dm 3 reaction vessel contains a mixture of hydrogen and oxygen with a total pressure of 3 00 kPa. The partial pressure of oxygen 20 kPa. a) Calculate the partial pressure of hydrogen. b) Calculate the mole fraction of hydrogen and oxygen.

Challenging:

  1. 3.00 moles of SO3(g) are placed into an 8.00 dm^3 container and heated to 1105 K. 2SO 3 (g) 2SO 2 (g) + O 2 (g)
  • At equilibrium the mixture contains 0.58 mol of O 2 (g).
  • The total pressure of the equilibrium mixture is 3.45 × 10^6 Pa.
  • Use this information to calculate Kp a) Determine equilibrium moles of all gases. b) Determine the mole fractions of each of the gases. c) Calculate the partial pressures. d) Write the expression for Kp. e) Calculate Kp and state its units. Ans = 222000
  1. 9.20 g of N 2 O 4 (g) was heated to a temperature of 340 K at a pressure of 13.3 kPa. N 2 O 4 (g) 2NO 2 (g) Once equilibrium had been reached, 70% of the N 2 O 4 (g) had dissociated. Use this information to calculate Kp Ans = 51.