Equilibrium and Stability Constants Le Chatelier’s Principle A reversible reaction can be approached from either direction In a closed system, no reactants or products can enter or escape For equilibrium to be achieved, a reaction must be reversible, in a closed system, and the rate of the forwards reaction must equal the rate of the reverse reaction. At dynamic equilibrium, there is no change in macroscopic properties like colour. When a system in dynamic equilibrium is subjected to a change, equilibrium will shift to minimise the effect of the change. The Equilibrium Law When a system is at equilibrium, the product of the concentrations of the products raised to the power of their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to the power of their stoichiometric coefficients is a constant at constant temperature. The value of Kc indicates the position of the equilibrium. Value of Kc Position of equilibrium >1010 Very far right >102 Just to right 1 Middle <10-2 Just to left <10-10 Very far left 𝑲 𝒄 = [𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔] 𝒑 [𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔] 𝒓 The units of Kc depend on the equilibrium expression.

Kc Calculations 𝐾 𝑐 = [𝐻𝐼] 2 𝐻 2 [ 𝐼 2 ] Calculation 1: Sub in the equilibrium concentration values given In the following equilibrium: H2 + I2 2HI Kc = 54.1 at a particular temperature. The equilibrium mixture was found to contain H2 at a concentration of 0.48x10-3 moldm-3 and HI at a concentration of 3.53x10-3 moldm-3. What is the equilibrium concentration of I2? 𝐾 𝑐 = [𝐻𝐼] 2 𝐻 2 [ 𝐼 2 ] 54.1= [3.53x10−3] 2 0.48x10−3 [ 𝐼 2 ] [I2 ] = 4.8 x10-4 moldm-3 Calculation 2: Sub in the equilibrium moles values given In this equilibrium system, Kc = 4. In a particular experiment, 0.33 moles of CH3COOH, 0.66 moles of CH3COOC2H5 and 0.66 moles of H2O are found to be present. What amount of C2H5OH is present at equilibrium? Only if volume cancels in Kc expression CH3COOH + C2H5OH CH3COOC2H5 + H2O 𝐾 𝑐 = [CH3COOC2H5][H2O] [CH3COOH][C2H5OH] [C2H5OH]= [0.66][0.66] 4 [0.33] [C2H5OH] = 0.33 moldm-3

Kc Calculations Calculation 3: Finding equilibrium concentrations 6.75g of SO2Cl2 was put into a 2.00moldm3 vessel. The vessel was sealed and its temperature raised to 375°. At equilibrium, the vessel contained 0.0345 moles of chlorine. Calculate the equilibrium constant at 375°. The equation for the reaction: SO2Cl2 SO2 + Cl2 1.Determine the number of starting moles of reactant: 𝑛𝑜. 𝑚𝑜𝑙𝑒𝑠 (SO2Cl2)= 6.75 135 =0.05 𝑚𝑜𝑙𝑒𝑠 2. Use a table to calculate the equilibrium moles of the reactants and products. SO2Cl2 SO2 Cl2 Stoichiometry 1 Initial moles 0.05 Change -0.0345 +0.0345 Equilibrium moles 0.0155 0.0345 3. Calculate concentrations at equilibrium (as volumes do not cancel) by dividing the number of moles at equilibrium by the volume of the vessel. [SO2Cl2 ] = 7.75x10-3 moldm-3 [SO2 ] = 0.01725 moldm-3 [Cl2 ]= 0.01725 moldm-3 4. Substitute the values for equilibrium concentration into the Kc equation: 𝑲 𝒄 = 𝑺𝑶𝟐 [𝑪𝒍𝟐] 𝑺𝑶 𝟐 𝑪𝒍𝟐 𝐾 𝑐 = 0.07125 ×[0.01725] 7.75x10−3 𝑲 𝒄 =𝟑.𝟖𝟒 × 𝟏𝟎 −𝟐

Changing temperature and Kc A change in temperature changes the value of the equilibrium constant. Kc is only a constant at constant temperature. A B ΔH = -ve For an exothermic reaction, increasing the temperature decreases the value of the equilibrium constant. Equilibrium moves to decrease the temperature Equilibrium shifts left as reverse reaction is endothermic On top line of Kc, [B] decreases. On bottom line of Kc, [A] increases – more A is produced. 𝑲 𝒄 = [𝑩] 𝑨 For an endothermic reaction, increasing the temperature increases the value of the equilibrium constant. A B ΔH = +ve Equilibrium moves to decrease the temperature Equilibrium shifts right as forwards reaction is endothermic On top line of Kc, [B] increases. On bottom line of Kc, [A] decreases – more B is produced. 𝑲 𝒄 = [𝑩] 𝑨

Concentration, Pressure and catalysts: Kc Although changes in concentration and pressure may affect the composition of the equilibrium mixture, the value of Kc is not affected by pressure, concentration or use of a catalyst. 2SO2 (g) + O2 (g) 2SO3 (g) For this equilibrium, explain, in terms of Le Chatelier’s Principle, the effect on the equilibrium yield of sulphur trioxide when the pressure is increased. If pressure is increased, equilibrium will shift right to reduce the pressure, as the right has fewer gaseous moles More SO3 will be produced. Explain this in terms of Kc Kc does not change System is no longer at equilibrium Top line of Kc will increase and bottom line of Kc will decrease until Kc is restored. Catalysts increase the rate of the forward and backwards reactions by the same amount. The position of equilibrium does not change, and the value of the equilibrium constant does not change.

Stability Constants of Complex Ions: Kstab The stability constant is the equilibrium constant for an equilibrium existing between a transition metal ion surrounded by water ligands and the complex ion formed during a ligand substitution reaction. [Cu(H20)6]2+ + 4NH3  [Cu(NH3)4 (H20)2]2+ + 4H2O This is the Kc expression for the ligand substitution reaction above. 𝑲 𝒄 = [Cu(NH3)4 (H20)2]2+ 𝐇𝟐𝐎 𝟒 [[Cu(H20)6]2+][ NH3] 𝟒 𝑲 𝒔𝒕𝒂𝒃 = [Cu(NH3)4 (H20)2]2+ [[Cu(H20)6]2+][ NH3] 𝟒 Because the water is the solvent and its concentration can be assumed to be constant, it can be left out of the expression for the stability constant, Kstab. The value of a stability complex depends upon the ligands. For example, a complex of copper is most stable with a multidentate ligand such as edta4- and least stable when water is its ligand. The greater tha value of Kstab, the more stable the complex ion.