15.3 LeChatelier Predicting Direction of a Chemical Reaction Dr. Fred Omega Garces Chemistry 201 Miramar College Two equal-weight boys at equilibrium on a teeter totter. The boy at the right is given a five-pound weight thereby disturbing the equilibrium. The boy on the left scoots farther back on the teeter totter to restore the equilibrium.

Quotient: Review For reactions not yet at equilibrium, the Law of Mass action yield information in terms of the Reaction Quotient. Consider the following chemical process not at equilibrium. aA + bB  rR + pP A reaction quotient expression can be written: Where the numerical value of Q will determine the direction the reaction will proceed. Q < Keq Reaction shifts to gRight Q > Keq Reaction shifts to fLeft

Reaction Quotient: Direction of Reaction Consider, T =532 °C, Kc = 0.19 what direction will the reaction proceed if the initial concentration is change to- N2 (g) + 3H2 (g)  2NH3 (g) i) 0.30 M 0.20 M 0.10 M Q = 4.2 > Kc=0.19 0.19 Kc 4.2 Q Direction of Reaction: Proceeds to the left

LeChatelier Principle: A Review Teeter•Totter At Equilibrium Stress applied Self Adjust Re-establish Equilibrium

Equilibrium: Stress / Relief on Reactant Altering Chemical Concentrations Stress on Reactant, Rxn shift right Relief on Reactant, Rxn shift left

Equilibrium: Stress / Relief on Product Altering Chemical Concentrations Stress on Product, Rxn shift left Relief on Product, Rxn shift right

Exothermic Heats of Solution Exothermic Process. Energy is a product Heating a solution in which the Hsoln is exothermic Energy is a product, Hsoln ( - ), results in a shift of the reaction to the left, or more solute precipitating out of solution.

Endothermic Heats of Solution Endothermic Process. Energy is a reactant Heating a solution in which the Hsoln is endothermic, Energy is a reactant, Hsoln ( + ), results in a Shift of the reaction to the right, or more solute dissolving in solution.

Solubilities of Solids Vs. Temperature Solubility of several ionic solid as a function of temperature. Most salts have positive heats of solution. When the salt solution is heated, more solute dissolves. Some salts have negative enthalpy of solution, (exothermic process) i.e., Ce2(SO4)3. When these salt solutions are heated, the solute becomes less soluble. Other example: Mg(OH)2 , Ce2(SO4)3 and Starch

2ii) Temperature & Solubility: Gases Temperature - (Gas) Consider the extent in which O2 or CO2 dissolves in water. What are the conditions which will increase the solubility of gas in water. [Solute]  [Solute] As the temperature increase, both solute and solvent will be moving faster, the gas solute however will now have enough energy to leave the liquid interface Is this an exothermic or endothermic process? + E gas Solution

Gas solute; Exothermic Hsoln As the temperature increase, both solute and solvent will be move faster. The gas solute however will now have enough energy to leave the liquid interface because IMF can be overcome.

Disaster: (1700 dead) from Gas Solubility In the African nation of Cameroon in 1986 a huge bubble of CO2 gas escaped from Lake Nyos and moved down a river valley at 20 m/s (about 45 mph). Because CO2 is denser than air, it hugged the ground and displaced the air in its path. More than 1700 people suffocated. The CO2 came from springs of carbonated groundwater at the bottom of the lake. Because the lake is so deep, the CO2 mixed little with the upper layers of water, and the bottom layer became supersaturated with CO2. When this delicate situation was changed, perhaps because of an earth-quake or landslide, the CO2 came out of the lake water just like it does when a can of soda is opened. Lake Nyos in Cameroon, the site of a natural disaster. In 1986 a huge bubble of CO2 escaped from the lake and asphyxiated more than 1700 people. http://hubpages.com/hub/The-Exodus-Decoded

The effect of a Change in Temperature in terms of Reaction Quotient and LeChatelier In an endothermic process, energy is a reactant and an increase in temperature results in a shift of the reaction to the right. When the temperature is decreased the reaction shifts to the left. In an exothermic process, energy is a product and an increase in temperature results in a shift of the reaction to the left. When the temperature is decreased the reaction shifts to the right. The question is raised, why does the reaction adjust itself, when according to the Mass Action Expression, the concentrations of chemicals are not altered with temperature changes? What causes the Mass Action Expression not to equal Keq(new) ( keq(old)). Consider an Endothermic reaction: E + R D P: Keq = [Prod] / [React] . If the temperature is raised, the reaction shifts to the right, (Q < Keq ) A shift to the right means that [Prod] = hwill raised) and [React] = i (will lowers). This will only occur if the [Prod] / [React] (or Q) is now less than Keq @ new the temperature. In other words when the temperature is change (increase T), the equilibrium constant changes, the current [P] / [R] ratio is still equal to the old Keq which is now Q (the reaction quotient) and the reaction shifts to re-establish equilibrium that is to attain the new Keq.

Temperature Effect (increase) & Reaction Quotient Endothermic Rxn: Increase in Temperature Exothermic Rxn: Increase in Temperature K eq(old Temp) = Q Direction Reaction Direction Reaction K eq(old Temp) = Q Keq @ New Temperature When the temperature is raised for an endothermic reaction, the Keq constant changes. Because the current concentrations yields a reaction quotient less than Keq (new) the reaction must shift to the right. When the temperature is raised for an exothermic reaction, the Keq constant changes. Because the current concentrations yields a reaction quotient greater than Keq (new) the reaction must shift to the left.)

Temperature Effect (decrease) & Reaction Quotient Exothermic Rxn: Decrease in Temperature Endothermic Rxn: Decrease in Temperature K eq(old Temp) = Q Direction Reaction Direction Reaction K eq(old Temp) = Q Keq @ New Temperature When the temperature is lowered for an exothermic reaction, the Keq constant changes. Because the current concentrations yields a reaction quotient lower than Keq (new) the reaction must shift to the right. When the temperature is lowered for an endothermic reaction, the Keq constant changes. Because the current concentrations yields a reaction quotient greater than Keq (new) the reaction must shift to the left.

Temperature Effect on [CoCl4]2- D [Co(H2O)] Consider the reaction: DH < 0 [CoCl4]2- + 6H2O D [Co(H2O)] + 4Cl- blue pink In this experiment when the solution was placed in cold water, the solution turned vivid pink. The pink color indicates a shift to the right for the reaction shown above. This will only occur if the new Keq at the instant the temperature is altered, is now higher than the old Keq, (which is now called Q). Therefore the reactant concentration decreases, the product concentration increases as the reaction adjust itself so that the Mass action equals Keq: Q  Keq K eq(old Temp) = Q Direction Reaction K eq (new Temp)

3i) Pressure on Solubility: Solids / Liquid Pressure - (Solid and Liquid) The solubility of solids and liquids are hardly affected by pressure. Solids and liquids are already very close to each other. An increase in pressure will not affect solubility.

3 ii) Pressure on Solubility: Gas Pressure - (Gas) Solubility of gas is greatly affected by pressure Henry’s Law: PA = KHXA Gas solutes are very sensitive to pressure. Because gas particles are separated by void space, an increase in pressure will increase the solubility of the gas. Divers must be careful when diving to great depths because the potential of dissolved N2 gas in the blood will lead to the bends.

Pressure Affect: Teeter Totter Analogy [Solute]  [Solute] Pressure Sensitive gas Solution In utilizing LeChatelier Principle to determine the direction of solubility for a gaseous solute with variation in pressure, the first thing that must be establish is which side is more sensitive to pressure. In our case the gas is more sensitive than the solution.

The effect of a Change in Volume of reaction vessel or better known as a Pressure change in terms of Reaction Quotient and LeChatelier For a chemical reaction, LeChatelier Principle can be verified in terms of the Reaction Quotient. Consider the reaction: N2 + 3H2 D 2NH3 Let [H2] = 0.1207 M, [N2] = 0.0402 M and [NH3] = 0.00272 M for a 1-L vessel at equilibrium. In which direction will the reaction shift if the reaction vessel is decreased by half such that V = Vo/2. The concentrations will now double for all specie at the instant the volume is decreased. Molarity = mol / L [H2] = 0.2414 M, [N2] = .0804 M and [NH3] = 0.00544 M . Plugging in to the mass action expression and solving for Q, In the reaction quotient equation Q < Keq which means that in order to regain equilibrium, the product must increase and the reactant decrease, a shift to the right in the overall reaction. Direction Reaction Q= .0262 Keq = 0.105

Summary of Pressure, Temperature Affect on Solubility H (s, l or g) Temp Direction Solubility (+) Endothermic   Prod  increase (+) Endothermic   React  decrease (-) Exothermic   React  decrease (-) Exothermic   Prod  increase Pressure Direction Solubility Gas solute   Prod  increase Gas solute   React  decrease

Example Summary Consider the following system at equilibrium: 6H2O (g) + 6CO2 (g) D 2 C6H12O6 (s) + 6O2 (g) Complete the following table. Indicate changes in moles and concentrations by entering I, D, N, or ? in the table. (I = increase, D = decrease, N = no change, ? = insufficient information to determine) B