Equilibrium constant Kc [ C]p [D]q [ A]m [B]n Kc= [ ] means the equilibrium concentration Kc = equilibrium constant Example 1 N2 (g) + 3H2 (g) 2 NH3 (g) [NH3 (g)] 2 [N2 (g)] [H2 (g)] 3 Kc= The unit of Kc changes and depends on the equation. Working out the unit of Kc Put the unit of concentration (mol dm-3) into the Kc equation [NH3 (g)] 2 [N2 (g) ] [H2 (g)]3 Kc= [mol dm-3] 2 [mol dm-3] [mol dm-3] 3 Unit = Cancel out units 1 [mol dm-3] Unit = 2 Unit = [mol dm-3] -2 Unit = mol-2 dm+6 Example 2: writing Kc expression H2 (g) +Cl2 (g) 2HCl (g) [mol dm-3] 2 [mol dm-3] [mol dm-3] Kc= [HCl (g)] 2 [H2 (g) ] [Cl2 (g)] Unit Kc= Working out the unit = no unit Calculating Kc Most questions first involve having to work out the equilibrium moles and then concentrations of the reactants and products. Usually the question will give the initial amounts (moles) of the reactants, and some data that will help you work out the equilibrium amounts. moles of reactant at equilibrium = initial moles – moles reacted moles of product at equilibrium = initial moles + moles formed Calculating the moles at equilibrium In a container of volume 600cm3 there were initially 0.5mol of H2 and 0.6 mol of Cl2 . At equilibrium there were 0.2 moles of HCl. Calculate Kc H2 (g) +Cl2 (g) 2HCl (g) Example 1 For the following equilibrium H2 Cl2 HCl Initial moles 0.5 0.6 0 Equilibrium moles 0.2 It is often useful to put the mole data in a table. Work out the moles at equilibrium for the reactants Using the balanced equation if 0.2 moles of HCl has been formed it must have used up 0.1 of Cl2 and 0.1 moles of H2 (as 1:2 ratio) moles of reactant at equilibrium = initial moles – moles reacted moles of hydrogen at equilibrium = 0.5 – 0.1 = 0.4 moles of chlorine at equilibrium = 0.6 – 0.1 = 0.5 Work out the equilibrium concentrations H2 Cl2 HCl Initial moles 0.5 0.6 0 Equilibrium moles 0.4 0.5 0.2 Equilibrium concentration (M) 0.4/0.6 =0.67 0.5/0.6 =0.83 0.2/0.6 =0.33 conc = moles/ vol (in dm3 ) = 0.332 0.67×0.83 Kc = 0.196 no unit Finally put concentrations into Kc expression If the Kc has no unit then there are equal numbers of reactants and products. In this case you do not have to divide by volume to work out concentration and equilibrium moles could be put straight into the kc expression Kc= [HCl (g)] 2 [H2 (g) ] [Cl2 (g)] N Goalby chemrevise.org 1 2 Initially there were 1.5 moles of N2 and 4 mole of H2, in a 1.5 dm3 container. At equilibrium 30% of the Nitrogen had reacted. Calculate Kc N2 (g) + 3H2 (g ) 2 NH3 (g) Example 2 For the following equilibrium N2 H2 NH3 Initial moles 1.5 4.0 0 Equilibrium moles Work out the moles at equilibrium for the reactants and products 30% of the nitrogen had reacted = 0.3 x1.5 = 0.45 moles reacted. Using the balanced equation 3 x 0.45 moles of H2 must have reacted and 2x 0.45 moles of NH3 must have formed moles of reactant at equilibrium = initial moles – moles reacted moles of nitrogen at equilibrium = 1.5 – 0.45 = 1.05 moles of hydrogen at equilibrium =4.0 – 0.45 x3 = 2.65 Work out the equilibrium concentrations N2 H2 NH3 Initial moles 1.5 4.0 0 Equilibrium moles 1.05 2.65 0.9 Equilibrium concentration (M) 1.05/1.5 =0.7 2.65/1.5 =1.77 0.9/1.5 =0.6 conc = moles/ vol (in dm3 ) = 0.62 0.7×1.773 Kc Finally put concentrations into Kc expression moles of product at equilibrium = initial moles + moles formed moles of ammonia at equilibrium = 0 + (0.45 x 2) = 0.9 [NH3 (g)] 2 [N2 (g) ] [H2 (g)]3 Kc= = 0.0927 mol-2 dm+6. For a generalised reaction mA + nB pC + qD m,n,p,q are the stoichiometric balancing numbers A,B,C,D stand for the chemical formula

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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.).