## Equilibrium constant Kc

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.).

## Partial pressure and Kp

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.).

## Changing conditions on Kc and Kp

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. Catalysts have no effect on the value of Kc or Kp or the position of equilibrium as they speed up both forward and backward rates by the same amount. Effect of catalysts on position of equilibrium and Kc and Kp 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 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 Increasing pressure does not change Kc. The increased pressure increases concentration terms on bottom of Kc expression more than the top. The system is now no longer in equilibrium so the equilibrium shifts to the right increasing concentrations of products and decreases the concentrations of reactants. The top of Kc expression therefore increases and the bottom decreases until the original value of Kc is restored 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. [NH3 (g)] 2 [N2 (g) ] [H2 (g)]3. 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. Effect of Concentration on position of equilibrium and Kc H2 (g) +Cl2 (g) 2HCl (g) Increasing concentration of H2 would move equilibrium to the right lowering concentration of H2 and Cl2 and increasing concentration of HCl. The new concentrations would restore the equilibrium to the same value of Kc Changing concentration would shift the position of equilibrium but the value of Kc would not change.

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5.1.2 How far?

f) (i) the qualitative effect on equilibrium constants of changing temperature for exothermic and endothermic reactions (ii) the constancy of equilibrium constants with changes in concentration, pressure or in the presence of a catalyst M0.3 (g) explanation of how an equilibrium constant controls the position of equilibrium on changing concentration, pressure and temperature

Credits: Neil Goalby