Writing an expression for Kp N2 (g) + 3H2 (g) 2 NH3 (g) p 2 NH3 p N2 p 3 H2 Kp= p means the partial pressure of that gas Kp = equilibrium constant Only include gases in the Kp expression. Ignore solids, liquids, and aqueous substances. p1 = x1 P If a reaction contains gases an alternative equilibrium expression can be set up using the partial pressures of the gases instead of concentrations Partial Pressures and Kp mole fraction Total moles of gas = 0.5 + 1.2+ 0.2 = 1.9 mole fraction of N2 = 0.2/1.9 =0.105 mole fraction of O2 = 0.5/1.9 =0.263 mole fraction of CO2 = 1.2/1.9 =0.632 Partial pressure of N2 =0.105x 3 =0.315 Partial pressure of O2 = 0.263 x 3 =0.789 Partial pressure of CO2 = 0.632 x 3 =1.896 Working out the unit of Kp Put the unit of pressure(atm) into the Kp equation atm2 atm atm3 Unit = 1 atm2 Cancel out units Unit = Unit = atm-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 atm2 . It is important therefore to write an equation when quoting values of Kp. p 2 NH3 (g) pN2 (g) p 3H2 (g) Kp= N Goalby chemrevise.org 4 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 2atm what is the value of Kp? N2 (g) + 3H2 (g ) 2 NH3 (g) Example 4 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 Kc 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 atm-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 atm. 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 3 : A mixture contains 0.2 moles N2 , 0.5 moles O2 and 1.2 moles of CO2. If the total pressure is 3atm. What are the partial pressures of the 3 gases? 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

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5.1.2 How far?

Equilibrium (a) use of the terms mole fraction and partial pressure See also 3.2.3 Chemical Equilibrium. (b) calculation of quantities present at equilibrium, given appropriate data M0.2 (c) the techniques and procedures used to determine quantities present at equilibrium Not for Kp. HSW4 Opportunities to carry out experimental and investigative work. (d) expressions for Kc and Kp for homogeneous and heterogeneous equilibria (see also 3.2.3 f) M0.2 Note: liquid and solid concentrations are constant and are omitted in heterogeneous Kc and Kp expressions. (e) calculations of Kc and Kp, or related quantities, including determination of units (see also 3.2.3 f) M0.0, M0.1, M0.2, M0.4, M2.2, M2.3, M2.4 Learners will not be required to solve quadratic equations. (h) application of the above principles in 5.1.2 How far? for Kc , Kp to other equilibrium constants, where appropriate (see also 5.1.3 c etc.).