Partial Pressure The partial pressure of a gas in a mixture is the pressure that the gas would have if it alone occupied the volume occupied by the whole mixture. If a mixture of gases contains 3 different gases then the total pressure will equal the 3 partial pressure added together P =p1 + p2 + p3 partial pressure = mole fraction x total pressure of gas 1 of gas 1 mole fraction = number of moles of a gas total number of moles of all gases For a 3 part mixture x1 = y1 y1+y2 +y3 Example : A mixture contains 0.2 moles N2 , 0.5 moles O2 and 1.2 moles of CO2. If the total pressure is 3kPa. What are the partial pressures of the 3 gases?If a reaction contains gases an alternative equilibrium expression can be set up using the partial pressures of the gases instead of concentrations

3.1.10 Equilibrium constant K p for homogeneous systems (A-level only)

Students should be able to:

• derive partial pressure from mole fraction and total pressure

p means the partial pressure of that gas Kp = equilibrium constant Only include gases in the Kp expression. Ignore solids, liquids, and aqueous substances. Working out the unit of Kp Put the unit of pressure(kPa) into the Kp equation kPa2 kPa kPa3 Unit = 1 kPa2 Cancel out units Unit = Unit = kPa-2 However, if the equation is written the other way round, the value of Kp will be the inverse of above and the units will be kPa2 . It is important therefore to write an equation when quoting values of Kp.1 mole of N2 and 3 moles of H2 are added together and the mixture is allowed to reach equilibrium. At equilibrium 20% of the N2 has reacted. If the total pressure is 2kPa what is the value of Kp? N2 (g) + 3H2 (g ) 2 NH3 (g) Example For the following equilibrium N2 H2 NH3 Initial moles 1.0 3.0 0 Equilibrium moles Work out the moles at equilibrium for the reactants and products 20% of the nitrogen had reacted = 0.2 x1.0 = 0.2 moles reacted. Using the balanced equation 3 x 0.2 moles of H2 must have reacted and 2x 0.2 moles of NH3 must have formed moles of reactant at equilibrium = initial moles – moles reacted moles of nitrogen at equilibrium = 1.0 – 0.2 = 0.8 moles of hydrogen at equilibrium =3.0 – 0.20 x3 = 2.40 N2 H2 NH3 Initial moles 1.0 3.0 0 Equilibrium moles 0.80 2.40 0.40 Mole fractions 0.8/3.6 =0.222 2.40/3.6 =0.667 0.40/3.6 =0.111 Partial pressure 0.222 x2 = 0.444 0.667 x2 =1.33 0.111 x2 = 0.222 = 0.2222 0.444×1.333 Kp Finally put concentrations into Kp expression moles of product at equilibrium = initial moles + moles formed moles of ammonia at equilibrium = 0 + (0.2 x 2) = 0.4 p 2 NH3 (g) pN2 (g) p 3H2 (g) Kp= = 0.0469 kPa-2 CaCO3 (s) CaO (s) + CO2 (g) Kp expressions only contain gaseous substances. Any substance with another state is left out Heterogeneous equilibria for Kp Kp =p CO2 Unit

3.1.10 Equilibrium constant K p for homogeneous systems (A-level only)

The equilibrium constant Kp is deduced from the equation for a reversible reaction occurring in the gas phase.

Kp is the equilibrium constant calculated from partial pressures for a system at constant temperature.

Students should be able to:

• construct an expression for K p for a homogeneous system in equilibrium

• perform calculations involving K p

Effect of changing conditions on value of Kc or Kp Effect of Temperature on position of equilibrium and Kc Both the position of equilibrium and the value of Kc or Kp will change it temperature is altered N2 (g) + 3H2 (g ) 2 NH3 (g) In this equilibrium which is exothermic in the forward direction If temperature is increased the reaction will shift to oppose the change and move in the backwards endothermic direction. The position of equilibrium shifts left. The value of Kc gets smaller as there are fewer products. N Goalby chemrevise.org Increasing pressure does not change Kp. The increased pressure increases the pressure terms on bottom of Kp expression more than the top. The system is now no longer in equilibrium so the equilibrium shifts to the right increasing mole fractions of products and decreases the mole fractions of reactants. The top of Kp expression therefore increases and the bottom decreases until the original value of Kp is restored Effect of Pressure on position of equilibrium and Kp The position of equilibrium will change it pressure is altered but the value of Kp stays constant as Kp only varies with temperature N2 (g) + 3H2 (g ) 2 NH3 (g) In this equilibrium which has fewer moles of gas on the product side If pressure is increased the reaction will shift to oppose the change and move in the forward direction to the side with fewer moles of gas. The position of equilibrium shifts right. The value of Kp stays the same though as only temperature changes the value of Kp. p 2 NH3 p N2 p 3 H2 Kp= x 2 NH3 . P2 x N2 .P x 3 H2 .P3 Kp= x 2 NH3 . P2 x N2 x 3 H2 . P4 Kp= Where P is total pressure and x mole fraction. The larger the Kc the greater the amount of products. If Kc is small we say the equilibrium favours the reactants Kc and Kp only change with temperature. It does not change if pressure or concentration is altered. A catalyst also has no effect on Kc or Kp

3.1.10 Equilibrium constant K p for homogeneous systems (A-level only)

Students should be able to:

• predict the qualitative effects of changes in temperature and pressure on the position of equilibrium

• predict the qualitative effects of changes in temperature on the value of K p

• understand that, whilst a catalyst can affect the rate of attainment of an equilibrium, it does not affect the value of the equilibrium constant.