CHEMICAL EQUILIBRIUM

Chemical Equilibrium Reversible Reactions: A chemical reaction in which the products can react to re-form the reactants Chemical Equilibrium: When the rate of the forward reaction equals the rate of the reverse reaction and the concentration of products and reactants remains unchanged 2HgO(s)  2Hg(l) + O2(g) Arrows going both directions (  ) indicates equilibrium in a chemical equation

2NO2(g)  2NO(g) + O2(g) Remember this from Chapter 12? Why was it so important to measure reaction rate at the start of the reaction (method of initial rates?)

2NO2(g)  2NO(g) + O2(g)

jA + kB  lC + mD Law of Mass Action For the reaction: Where K is the equilibrium constant, and is unitless

Product Favored Equilibrium Large values for K signify the reaction is “product favored” When equilibrium is achieved, most reactant has been converted to product

Reactant Favored Equilibrium Small values for K signify the reaction is “reactant favored” When equilibrium is achieved, very little reactant has been converted to product

Writing an Equilibrium Expression Write the equilibrium expression for the reaction: 2NO2(g)  2NO(g) + O2(g) K = ???

N2(g) + 3H2  2NH3 The following equilibrium concentrations were observed for the above process at 127oC: [NH3] = 3.1 x 10-2 mol/L [N2] = 8.5 x 10-1 mol/L [H2] = 3.1 x 10-3 mol/L Calculate the value of K at 127oC for this reaction. B. Calculate the value of the equilibrium constant at 127oC for the reaction: 2NH3(g)  N2(g) + 3H2(g) Calculate the value of the equilibrium constant at 127oC for the reaction given by the equation: 1/2N2(g) + 3/2H2(g)  NH3(g)

[NH3]2 A. (3.1 x 10-2)2 K = = [N2] [H2]3 (8.5 x 10-1)(3.1 x 10-3)3 Solution [NH3]2 A. (3.1 x 10-2)2 K = = [N2] [H2]3 (8.5 x 10-1)(3.1 x 10-3)3 = 3.8 x 104 What are the units for K?

[N2][H2]3 K’ = [NH3]2 [N2][H2]3 K’ = 1 1 = = [NH3]2 K 3.8 x 104 Solution B. This reaction is written in the reverse order from the equation given in part A. [N2][H2]3 K’ = [NH3]2 Which is the reciprocal of the expression used in part A. [N2][H2]3 K’ = 1 1 = = [NH3]2 K 3.8 x 104 = 2.6 x 10-5

( ) [NH3] K’’ = [N2]1/2[H2]3/2 1/2 [NH3] [NH3]2 = [N2]1/2[H2]3/2 Solution C. We use the law of mass action: [NH3] K’’ = [N2]1/2[H2]3/2 If we compare this expression to that obtained in part A. we see that since: 1/2 ( ) [NH3] [NH3]2 = [N2]1/2[H2]3/2 [N2][H2]3 Thus: K’’ = K1/2 = (3.8 x 104)1/2 = 1.9 x 102

Conclusions about Equilibrium Expressions The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse 2NO2(g)  2NO(g) + O2(g) 2NO(g) + O2(g)  2NO2(g)

Conclusions about Equilibrium Expressions When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power. 2NO2(g)  2NO(g) + O2(g) NO2(g)  NO(g) + ½O2(g)

Equilibrium Expressions Involving Pressure For the gas phase reaction: 3H2(g) + N2(g)  2NH3(g) Δn = the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants.

Heterogeneous Equilibria The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present Write the equilibrium expression for the reaction: PCl5(s)  PCl3(l) + Cl2(g) Pure solid Pure liquid

jA + kB  lC + mD The Reaction Quotient For some time, t, when the system is not at equilibrium, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action. jA + kB  lC + mD

Significance of the Reaction Quotient If Q = K, the system is at equilibrium If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

Try me! For the synthesis of ammonia at 500oC , the equilibrium constant is 6.0 x 10-2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases. [NH3]0 = 1.0 x 10-3M; [N2]0 = 1.0 x 10-5M; [H2]0 = 2.0 x 10-3M [NH3]0 = 2.0 x 10-4M; [N2]0 = 1.5 x 10-5M; [H2]0 = 3.54 x 10-1M [NH3]0 = 1.0 x 10-4M; [N2]0 = 5.0M; [H2]0 = 1.0 x 10-2M

First we calculate the value of Q Q = 1.3 x 107 Since K = 6.0 x 10-2, Q is MUCH greater than K. To attain equilibrium, the concentrations of the products must be decreased and the concentrations of the reactants increased. So…. The system will shift to the LEFT: N2 + 3H2 2NH3

N2 + 3H2 2NH3 B. First we calculate the value of Q Q = 6.01 x 10-2 In this case Q = K, so the system is at equilibrium. NO shift will occur. C. Calculate Q Q = 2.0 x 10-3 Here Q is less than K, so the system will shift to the right to attain equilibrium by increasing the concentration of the product and decreasing the reactant concentrations N2 + 3H2 2NH3

Solving for Equilibrium Concentration Consider this reaction at some temperature: H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0 Assume you start with 8 molecules of H2O and 6 molecules of CO. How many molecules of H2O, CO, H2, and CO2 are present at equilibrium? Here, we learn about “ICE” – the most important problem solving technique in chemistry! Lol (ok…maybe sometimes).

Solving for Equilibrium Concentration H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0 Step #1: We write the law of mass action for the reaction:

Solving for Equilibrium Concentration Step #2: We “ICE” the problem, beginning with the Initial concentrations H2O(g) + CO(g)  H2(g) + CO2(g) Initial: Change: Equilibrium: 8 6 -x -x +x +x 8-x 6-x x x

Solving for Equilibrium Concentration Step #3: We plug equilibrium concentrations into our equilibrium expression, and solve for x H2O(g) + CO(g)  H2(g) + CO2(g) Equilibrium: 8-x 6-x x

Solving for Equilibrium Concentration Step #4: Substitute x into our equilibrium concentrations to find the actual concentrations H2O(g) + CO(g)  H2(g) + CO2(g) Equilibrium: 8-x 6-x x Equilibrium: 8-4=4 6-4=2 4

A Weak Base Equilibrium Problem What are the equilibrium concentrations of NH3 and NH4+ in a 0.5M solution of ammonia? (Kb + 1.8 x 10-5) Balanced Chemical Rxn and…ICE it! NH3 + H2O  NH4+ + OH- I C E 0.50 - x +x +x 0.50 - x x x

A Weak Base Equilibrium Problem Set up the law of mass action NH3 + H2O  NH4+ + OH- E 0.50 - x x x

A Weak Base Equilibrium Problem Solve for x, which is also [OH-] NH3 + H2O  NH4+ + OH- E 0.50 - x x x [OH-] = 3.0 x 10-3 M

Gaseous NOCl decomposes to form the gases NO and Cl2. At 35oC the equilibrium constant is 1.6 x 10-5. In an experiment in which 1.0 mol NOCl is placed in a 2.0 L flask, what are the equilibrium concentrations? 2NOCl(g) 2NO(g) + Cl2(g)

Start the following problems: Work cooperatively! Practice time! Start the following problems: Work cooperatively! Pages 616-18: #42,44,46,52,68

A Weak Base Equilibrium Problem What are the equilibrium concentrations of NH3 and NH4+ in a 0.5M solution of ammonia? (Kb + 1.8 x 10-5) Balanced Chemical Rxn and…ICE it! NH3 + H2O  NH4+ + OH- I C E 0.50 - x +x +x 0.50 - x x x

A Weak Base Equilibrium Problem Set up the law of mass action NH3 + H2O  NH4+ + OH- E 0.50 - x x x

A Weak Base Equilibrium Problem Solve for x, which is also [OH-] NH3 + H2O  NH4+ + OH- E 0.50 - x x x [OH-] = 3.0 x 10-3 M

Start the following problems: Practice time! Start the following problems: Pages 614-15: #42,44,46,52,68

Assume that the reaction foe the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15 x 102 at a certain temperature. In a particular experiment, 3.000 mol of each component was added to 1.500 L flask. Calculate the equilibrium concentrations of all species.

THE HABER PROCESS

LeChatelier’s Principle WHO AM I? Fritz Haber LeChatelier’s Principle When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress. Translated: The system undergoes a temporary shift in order to restore equilibrium.

LeChatelier Example #1 A closed container of ice and water is at equilibrium. Then, the temperature is raised. Ice + Energy  Water The system temporarily shifts to the _______ to restore equilibrium. right

LeChatelier Example #2 A closed container of N2O4 and NO2 is at equilibrium. NO2 is added to the container. N2O4 (g) + Energy  2 NO2 (g) The system temporarily shifts to the _______ to restore equilibrium. left

LeChatelier Example #3 A closed container of water and its vapor is at equilibrium. Vapor is removed from the system. water + Energy  vapor The system temporarily shifts to the _______ to restore equilibrium. right

LeChatelier Example #4 A closed container of N2O4 and NO2 is at equilibrium. The pressure is increased. N2O4 (g) + Energy  2 NO2 (g) The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation. left

LeChatelier Example #5 The preparation of liquid phosphorus trichloride by the below reaction. The volume is decreased. P4 (s) + 6 Cl2 (g)  4 PCl3 (l) The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation. right

LeChatelier Example #6 The preparation of gaseous phosphorus pentachloride by the below reaction. The volume is decreased. PCl3 (g) + Cl2 (g)  PCl5 (g) The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation. right

LeChatelier Example #7 The preparation of phosphorus trichloride with ammonia by the below reaction. The volume is decreased. PCl3 (g) + 3 NH3 (g)  P(NH2)3 (g) + 3 HCl (g) Both sides have four gaseous molecules. A change in volume will have no effect on the equilibrium position. There is NO SHIFT in this case.

LeChatelier Example #8 N2O4 (g)  2 NO2 (g) ΔHo = 58 kJ Change Shift Right Left None Addition of N2O4 (g) Addition of NO2 (g) Removal of N2O4 (g) Removal of NO2 (g) Addition of He (g) Decrease container volume Increase container volume Increase temperature Decrease temperature