CHEMICAL EQUILIBRIUM Dynamic Equilibrium Equilibrium constant expression – K c – K p – Q c Le Chatelier’s principle

DYNAMIC EQUILIBRIUM rate (forward reaction) = rate (reverse reaction) overall composition of reaction mixture does not change forward and reverse reactions continues after equilibrium is reached As long as the temperature and pressure remain constant and nothing is added to or taken from the mixture, the equilibrium state remains unchanged. aA + bB ⇌ cC + dD

EQUILIBRIUM CONSTANT K C aA + bB ⇌ cC + dD Substitute EQUILIBRIUM concentrations Exclude solids and liquids which are in heterogenous reactions No unit for K c

K > 1 products favoured, forward reaction favoured K < 1 reactants favoured, reverse reaction favoured K = 1 the equilibrium mixture contains appreciable amounts of both reactants and products.

PRACTICE EXAMPLES a) 2 H 2 S (g) ⇌ 2 H 2(g) + S 2(g) b) PCl 3(g) + 3 NH 3(g) ⇌ P(NH 2 ) 3(g) + 3 HCl (g) c) Na 2 CO 3(s) + SO 2(g) + ½ O 2(g) ⇌ Na 2 SO 4(s) + CO 2(g)

Relating K c to balanced chemical equations  Reverse equation  invert K c  Multiply equation  take corresponding power of K c  Divide equation  take corresponding root of K c  Combining equations (mechanisms)  multiply K c values

PRACTICE EXAMPLE N 2 O (g) + ½ O 2(g) ⇌ 2 NO (g) K c = ? N 2(g) + ½ O 2(g) ⇌ N 2 O (g) K c = 2.7 x N 2(g) + O 2(g) ⇌ 2NO (g) K c = 4.7 x

ICE TABLES FOR EQUILIBRIUM CALCULATIONS ABCD INITIAL[A] 0 [B] 0 00 CHANGES-ax-bx+cx+dx EQUILIBRIUM[A] 0 - ax[B] 0 - bxcxdx aA + bB ⇌ cC + dD ⇌

PRACTICE EXAMPLE A 5.00 L flask is filled with 1.86 mol NOBr. At equilibrium there is mol Br 2 present. Determine K c for the reaction at 25  C. 2 NOBr (g) ⇌ 2 NO (g) + Br 2(g)

PRACTICE EXAMPLE A mol sample of N 2 O 4(g) is allowed to come to equilibrium with NO 2(g) in a L flask at 25  C. Calculate the amount of N 2 O 4(g) at equilibrium. N 2 O 4(g) ⇌ 2 NO 2(g) K c = 4.61 x 10 3 at 25  C

EQUILIBRIUM CONSTANT K P Equilibrium constant expressed in terms of partial pressures wrt gases aA + bB ⇌ cC + dD

Kp  KcKp  Kc PV = nRT P = n/V x RT = [ ] x RT K p = K c (RT)  n  n =(c + d) – (a +b) R = T: Kelvin temp aA + bB ⇌ cC + dD Leave out units

PRACTICE EXAMPLE 2 NH 3(g) ⇌ N 2(g) + 3 H 2(g) K c = 2.8 x at 298K K p = ?

REACTION QUOTIENT Q C aA + bB ⇌ cC + dD To describe changes that occur to establish equilibrium To check if the reaction really is at equilibrium Use initial concentrations

Predicting the direction of reaction Q > K to establish equilibrium reverse reaction is favoured (amount of products at equilibrium is less) Q < K to establish equilibrium forward reaction is favoured (amount of reactants at equilibrium is less) Q = K at equilibrium (initial concentrations same as equilibrium conditions)

PRACTICE EXAMPLE Is a mixture of mol NO 2(g) and mol N 2 O 4(g) in a 5.25 L flask at 25  C at equilibrium? If not, in which direction will the reaction proceed? N 2 O 4(g) ⇌ 2 NO 2(g) K c = 4.61 x at 25  C

LE CHATELIER’S PRINCIPLE If a system at equilibrium is subjected to a stress, the system will react in a way that tends to relieve the stress.

Adding or removing a reactant or product The reaction shifts in a direction that will partially remove a substance that has been added or partially replace a substance that has been removed.

Changing the volume Reducing the volume of a gaseous reaction mixture shifts the equilibrium in direction that results in decrease in the number of gas molecules *Opposite is true

Increase pressure: ↳ pressure increase ∝ volume decrease Adding inert gas (at constant-pressure): ↳ pressure decrease ∝ volume increase

Changing the temperature Increasing the temperature shifts an equilibrium in a direction that produces an endothermic change (which absorbs heat). *Opposite is true

K c changes with temperature If the forward reaction in an equilibrium is exothermic, raising the temperature causes the equilibrium constant to become smaller. *Opposite is true

Agents that do not affect the position of equilibrium catalyst – Speeds up forward and reverse reactions equally and does not affect the equilibrium amounts – causes no change in the value of K adding an inert gas at constant volume

PRACTICE EXAMPLE 4 HCl (g) + O 2(g) ⇌ 2 H 2 O (g) + 2 Cl 2(g) ΔH = 28 kJ Effect on equilibrium and K c ? (a)Addition of oxygen gas (b)An increase in temperature (c)Reduction of the volume of the reaction container (d)Addition of a catalyst (e)Removal of HCl(g) from the reaction vessel.