5.1.2 How Far

• How is the mole fraction calculated?

  Mole fraction = number of moles of gas / total number of moles of gas

  • The sum of mole fractions is 1
• What is partial pressure and how is it calculated?

  • Contribution made to the total pressure

  Partial pressure p(A) = mole fraction of A x total pressure

  • Units can be any from: Pa, kPa, atm
• What does the value of Kp suggest about equilibrium?

  • A large Kp value indicates equilibrium lies to the right
  • A small Kp value indicates equilibrium lies to the left
• What is each part of the Kp constant expression?

  .

  A) pressure

  B) partial pressure

  C) moles

  D) top

  E) bottom

  
• How are the units for Kp determined?

  .

  A) -2

  
• Explain why changing the temperature changes the equilibrium position in an endothermic reaction

  • Kc / Kp is temperature dependent
  • Increasing the temperature increases the value of Kc / Kp in an endothermic reaction
• Explain why changing the temperature changes the equilibrium position in an exothermic reaction

  • Kc / Kp is temperature dependent
  • Increasing the temperature decreases the value of Kc / Kp for an exothermic reaction
• Explain how changes in concentration affect how equilibrium is controlled

  • Changes in concentration do not change the value of Kc / Kp
  • If the concentration is changed the system is no longer in dynamic equilibrium and the calculated expression is no longer obeyed
  • The equilibrium adjusts itself so the concentration matches the value of Kc / Kp
• Explain how changes in pressure affect how equilibrium is controlled

  • Changes in pressure does not change the value of Kc / Kp
  • If the pressure is changed the system is no longer in dynamic equilibrium and the calculated expression is no longer obeyed
  • The equilibrium adjusts itself so the pressure matches the value of Kc / Kp
• Does a catalyst affect the value of Kc / Kp?

  • No
• Does the system remain in equilibrium when there are the same number of moles on each side?

  • Yes, therefore Kc remains constant