CH 102 1st Edition Lecture 13Outline of Current Lecture I. Example QuestionsII. Kc and Kp ConnectionIII. The Gas Phase Oxidation of Nitric OxideIV. Le Châtelier’s Principle: Disturbing and Restoring EquilibriumV. Disturbing Equilibrium: Removing ReactantsVI. The Effect of Concentration Changes on EquilibriumVII. Acids and BasesVIII. Strong AcidsIX. Strong BasesX. Weak AcidsXI. Weak BasesCurrent LectureI. Example Questiona. If the initial concentration of all species is 1.00 M, what is the equilibrium concentration of hydrogen iodide?i. 2.34 **b. Which of the following will increase the rate of a first order chemical reaction?i. Decreasing the concentration of the reactantii. Decreasing the temperatureThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.iii. Adding a catalyst ** II. Kc and Kp Connectiona. In calculating Kp, the partial pressures are always in atm.b. The values of Kp and Kc are not necessarily the same because of the difference in units.i. Kp = Kc when Δn = 0c. The relationship between them is as follows:i. Kp = Kc × (RT)Δnii. Where Δn is the difference between the number of moles of reactants and the number of moles of products.III. The Gas Phase Oxidation of Nitric Oxidea. 2 NO (g) + O2 (g) → 2 NO2 (g)b.IV. Le Châtelier’s Principle: Disturbing and Restoring Equilibriuma. Once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same.i. However, if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is restored.b. The new concentrations will be different, but the equilibrium constant will be thesame, unless you change the temperature.c. Le Châtelier’s principle guides us in predicting the effects various changes in conditions have on the position of equilibrium. i. It says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance. ii. Disturbances all involve making the system open.V. Disturbing Equilibrium: Adding Reactantsa. After equilibrium is established, a reactant is added, as long as the added reactant is included in the equilibrium constant expression.i. That is, as long as it is not a solid or liquid.b. How will this affect the rate of the forward reaction? i. Adding a reactant initially increases the rate of the forward reaction, but it has no initial effect on the rate of the reverse reaction. ii. The reaction proceeds to the right until equilibrium is restored. c. How will it affect the rate of the reverse reaction? i. At the new equilibrium position, you will have more of the products than before, less of the non-added reactants than before, and less of the added reactant. ii. But you will not have as little of the added reactant as you had before the addition.d. How will it affect the value of K?i. At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same.VI. Disturbing Equilibrium: Removing Reactantsa. After equilibrium is established, a reactant is removed, as long as the removed reactant is included in the equilibrium constant expression.i. That is, not a solid or liquidb. How will this affect the rate of the forward reaction? i. Removing a reactant initially decreases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction. 1. So the reaction is going faster in reverse.ii. The reaction proceeds to the left until equilibrium is restored. c. How will it affect the rate of the reverse reaction? i. At the new equilibrium position, you will have less of the products than before, more of the non-removed reactants than before, and more of the removed reactant.ii. But you will not have as much of the removed reactant as you had before the removal.d. How will it affect the value of K?i. At the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same.VII. The Effect of Concentration Changes on Equilibriuma. Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found that has the same K.b. Removing a product will increase the amounts of the other products and decrease the amounts of the reactants.VIII. Acids and Basesa. Acidsi. Are soluble in waterii. Lose one or more protons to make hydronium ions (H3O+)iii. Strong acids completely dissociate iv. Weak acids only partially dissociateb. Basesi. Are soluble in waterii. Generates one or more hydroxide ions (OH-)iii. Strong bases completely dissociate in wateriv. Weak bases only partially dissociate in waterIX. Strong Acidsa. Completely dissociate when dissolved in wateri. HX (aq) + H2O (aq) → H3O+ (aq) + X- (aq)b. For every mole of a strong acid dissolved in water, one mole of hydronium ion is generatedc. Partial list: i.X. Strong Basesa. Completely dissociate when dissolved in wateri. MOH (aq) → M+ (aq) + OH- (aq)b. For every mole of a strong base dissolved in water, one mole of hydroxide ions are generatedc. Partial List:i.XI. Weak Acidsa. Partially dissociate when dissolved in wateri. HX (aq) + H2O (aq) → H3O+ (aq) + X- (aq)b. For every mole of a weak acid dissolved in water, less than one mole of hydronium ion is generatedc. The concentration must be calculated from the value of KaXII. Weak Basesa. Partially dissociate when dissolved in wateri. MOH (aq) → M+ (aq) + OH- (aq)b. For every mole of a weak base dissolved in water, less than one mole of hydroxide ions are generatedc. The concentration must be calculated from the value of