3 Equilibrium equations

Equilibrium Equations 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 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 3 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 3 4 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 4 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 The larger the Kc the greater the amount of products. If Kc is small we say the equilibrium favours the reactants Kc only changes with temperature. It does not change if pressure or concentration is altered. A catalyst also has no effect on Kc Effect of changing conditions on value of Kc Effect of Temperature on position of equilibrium and Kc Both the position of equilibrium and the value of Kc 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. Effect of Pressure on position of equilibrium and Kc The position of equilibrium will change it pressure is altered but the value of Kc stays constant as Kc 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 Kc stays the same though as only temperature changes the value of Kc. Catalysts have no effect on the value of Kc or the position of equilibrium as they speed up both forward and backward rates by the same amount. N Goalby chemrevise.org N Goalby chemrevise.org 5 Calculating the amounts of the equilibrium mixture from Kc Using algebra it is possible to work out the amounts of each component in an equilibrium mixture using the value of Kc H2 (g) + Br2 (l) 2HBr (g) Example 5 For the following equilibrium If 0.200 mol of H2 and 0.200 mol of Br2 are mixed and allowed to reach equilibrium. If Kc = 0.210 what are the equilibrium amounts of each substance. Kc= [HBr (g)]2 [H2 (g) ] [Br2 (g)] Make x = moles of H2 that have reacted at equilibrium V = volume of container This reaction is equimolar (same number of moles of reactant as products) so it is possible to cancel out the volume V (2x/V)2 (0.2-x)/V . (0.2-x)/V (2x)2 (0.2-x) . (0.2-x) 0.21 = 0.21 = (2x)2 (0.2-x)2 0.21 = Square root both sides 2x 0.2-x √0.21 = Rearrange to give x 0.458(0.2-x) = 2x 0.0917 = 2x + 0.458x x = 0.0917/2.458 x = 0.0373 So at equilibrium Moles of H2 = 0.2000- 0.0373 = 0.163 mol Moles of Br2 = 0.2000- 0.0373 = 0.163mol Moles of HBr = 2x 0.0373 = 0.0746mol