CHEMICAL EQUILIBRIUM condensed version

At equilibrium, the rate at which NO 2 forms in the forward reaction equals the rate at which N 2 O 4 forms in the reverse reaction: N 2 O 4 2 NO 2 colorless brown Forward Reaction: N 2 O 4 2 NO 2 rate f = k f [N 2 O 4 ] Reverse Reaction: 2NO 2 N 2 O 4 rate r = k r [NO 2 ] 2

Products raised to stoichiometric powers… …divided by reactants raised to their stoichiometric powers The EQUILIBRIUM CONSTANT EXPRESSION For a general equilibrium equation: a A + b B d D + e E

For a general equilibrium equation: When the reactants and products in a chemical reaction are all gases, we can formulate the equilibrium-constant expression in terms of partial pressures. Where p is pressure Relating K p and K c : a A + b B d D + e E

For a heterogeneous equilibria: (substances are in different phases) Whenever a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium-constant expression. PbCl 2 (s) Pb 2+ (aq) + Cl - (aq) CaCO 3 (s) CaO(s) + CO 2 (g) H 2 O(l) + CO 3 2- (aq ) OH - (aq) + HCO 3 - (aq)

Magnitude of the equilibrium constant: K c >> 1 (large K c )  Reaction is strongly product favored.  very little reactant remains  often written as a forward reaction only.  assume reaction goes to completion. K c << 1 (small K c )  Reaction is strongly reactant favored.  very little product forms  usually written as “no reaction” or NR (large K c ) (small K c )

8 Write expressions for K c, and K P if applicable, for the following reversible reactions at equilibrium: (a) HF(aq) + H 2 O(l) H 3 O + (aq) + F − (aq) (b) 2NO(g) + O 2 (g) 2NO 2 (g) (c) CH 3 COOH(aq) + C 2 H 5 OH(aq) CH 3 COOC 2 H 5 (aq) + H 2 O(l)

Magnitude of the Equilibrium Constant The larger the value of the equilibrium constant, K c, the farther the reaction proceeds to the right before reaching the equilibrium state Very large Very small Reaction proceeds nearly to completion Reaction proceeds hardly at all 1 Appreciable concentrations of both reactants and products present at equilibrium

Methanol (CH 3 OH) is manufactured industrially by the reaction CO(g) + 2H 2 (g) CH 3 OH(g) The equilibrium constant (K c ) for the reaction is 10.5 at 220°C. What is the value of K P at this temperature?

Strategy The relationship between K c and K P is given by Equation (15.5). What is the change in the number of moles of gases from reactants to product? Recall that Δn = moles of gaseous products - moles of gaseous reactants What unit of temperature should we use?

Solution The relationship between K c and K P is K P = K c (0.0821T ) Δn Because T = = 493 K and Δn = = -2, we have K P = (10.5) ( x 493) -2 = 6.41 x 10 -3

If Q = K c, the system is at equilibrium. increase forward If Q < K c, Q must increase to reach equilibrium, the system shifts forward. decrease backwards If Q > K c, Q must decrease to reach equilibrium, the system shifts backwards. Reaction quotient a A + b B c C + d D

1.Fuel engineers use the extent of the change from CO and H 2 O to CO 2 and H 2 to regulate the proportions of synthetic fuel mixtures. If mol of CO and mol of H 2 O are placed in a 125-mL flask at 900 K. What is the composition of the equililbrium mixture? At this temperature, K c is 1.56 for the equation: CO(g) + H 2 O(g)  CO 2 (g) + H 2 (g) 2. Phosgene is a potent chemical warfare agent that is now outlawed by internation agreement. It decomposes by the reaction: COCl 2 (g)  CO(g) + Cl 2 (g) K c = 8.3 x (at 360 o C) Calculate [CO], [Cl 2 ] and [COCl 2 ] when each of the following amounts of phosgene decomposes and reaches equilibrium in a 10.0 –L flask: a) 5.00 mol of CoCl 2 b) mol of COCl 2

Consider the reaction: N 2 O 4 2 NO 2 ΔH o = 58 kJ ChangeSHIFT Addition of N 2 O 4 (g)right Addition of NO 2 (g)left Removal of N 2 O 4 (g)left Removal of NO 2 (g)right Addition of He(g)none Decrease in container volumeleft Increase in container volumeright Increase in Tempright Decrease in Templeft