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Lecture 7 (edited), Chemistry of Solutions Notes

by: CatLover44

Lecture 7 (edited), Chemistry of Solutions Notes 202-NYB-05

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These edited notes cover equilibrium rium (different ways a system at equilibrium can be disturbed), Le Châtelier's Principle, and an intro to/review of acids and bases.
Chemistry of Solutions
Nadia Schoonhoven
Class Notes
Equilibrium, Le Chatelier's Principle, pressure, concentration, temperature, acids and bases
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This 13 page Class Notes was uploaded by CatLover44 on Saturday October 8, 2016. The Class Notes belongs to 202-NYB-05 at Dawson Community College taught by Nadia Schoonhoven in Fall 2016. Since its upload, it has received 3 views. For similar materials see Chemistry of Solutions in Chemistry at Dawson Community College.


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Date Created: 10/08/16
Chemistry of Solutions Lecture no. 7 Thursday, September 15, 2016 Topics Covered: Le Châtelier’s principle, the nature of acids and bases, acid strength, the pH scale, and calculating the pH of strong acid solutions. Le Châtelier's Principle If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. A chemical reaction is not a sentient being, if a substance is added the rates of the forward and reverse reactions are disturbed. The net reaction occurs in one direction, until the system reaches equilibrium again, At the new equilibrium position, the reactants and products have different concentrations (compared to their initial concentrations). 1. Changing concentrations: disturbing a system’s equilibrium position by adding a substance causes the reaction to shift in the direction that absorbs some of the added substance. If a substance is removed from an equilibrium system, the reaction will shift in the direction that produces more of the lost substance. Note: adding noble gases or solids to a gaseous equilibrium system does not disturb its equilibrium position! Adding a noble/inert gas only changes the total pressure of the system, not the partial pressures of each species. The same rule applies if you add a substance that won't react with any of the chemical species in the reversible reaction. 2 2. Changing Pressures (gaseous systems): the pressure of a gaseous system is changed if certain gaseous substances are added or removed. For example adding more reactant and removing some product would disturb a system’s equilibrium position. Increasing or decreasing the volume of the container holding an equilibrium system also disturbs its equilibrium position because the partial pressures are changed. A decrease in volume causes the system’s partial pressures to increase (increased rate of effective collisions), which has the same effect as increasing the concentrations of all the gaseous concentrations. The biggest disturbance occurs where the stoichiometric coefficient is greatest. An increase in pressure causes the reaction to favour the direction that reduces the number of moles until a new equilibrium is reached. Example: what happens if we add 1.00 M of N to th2 reaction N 2(g)+ H 2(g) 2NH 3(g) 3 Adding N t2 the system increases the concentration of N , which2causes the system to favour the forward reaction – production of more product. Equilibrium shifts right. 2 3 -2 K p (0.202) / (1.399)(1.197) = 5.96x10 Q = (0.202) / (1.399)(1.197) = 1.70x10 -2 Example: in the reaction As O 4 6(s) 6C (s)⟺ As (g) 6CO , (g)t happens if: - CO is added? The partial pressure of CO (g)will increase, causing the system to favour the reverse reaction (a decrease in pressure). There are more gaseous molecules on the right side of the equation than on the left (ratio of 0:2 for products). The reverse reaction will be favoured because this is the reaction that decreases the number of moles of CO . (g) - As 4 6(s)or 6C (s)is removed? This has no effect on the system’s equilibrium position. - As (g)is removed? The system responds by producing more As (g)to reduce the effect of the disturbance. The equilibrium shifts to the right. Example: predict the shift in equilibrium position when the volume is reduced for each reaction. 1) P4(s) + 6Cl2(g) ⟺ 4PCl3(l) 6 moles of gaseous reactant, 0 moles of gaseous product. (6:0 ratio for reactants). Decreasing volume causes an increase in the number of moles, causing the system to favour the reaction that reduces the number of moles. The equilibrium shifts right. 2) PCl3(g) + Cl2(g) ⟺ PCl5(g) 4 2 moles of gaseous reactant, 1 mole of gaseous product (2:1 ratio for reactants). Decreasing volume causes an increase in the number of moles, causing the system to favour the reaction that reduces the number of moles. The equilibrium shifts right. Same explanation as the previous example. 3) PCl3(g) + 3NH3(g) ⟺ P(NH2)3(g) + 3HCl(g) 4 moles of gaseous reactant and product, 1:1 ratio. Changing the volume will not alter the equilibrium of this system. Example: complete the following table. Some Gas Ratio Decrease Volume/ Increase Volume/ Reactions Increase Pressure Decrease Pressure SO 3(g)⟺ SO 2(g) 1 : 3/2 Eqm shifts left Eqm shifts right 1/2O 2(g) 2H 2(g) O 2(g)⟺ 2 : 2 Eqm isn't disturbed Eqm isn't disturbed 2H 2 (g) SO 2(g)+ NO 2(g)⟺ 1 : 1 Eqm isn't disturbed Eqm isn't disturbed SO 3(g)+ NO(g) 5 3. Changing the temperature: if energy, in the form of heat, is added to a reversible reaction, the system shifts to absorb the added energy. Ask yourself if the forward reaction is endo- or exothermic. Example: what happens if the temperature is increased in the following reactions? 1) N + O ⟺ 2NO , ΔH = 181 kJ 2(g) 2(g) (g) N 2(g) O 2(g)+ 181 kJ ⟺ + 2NO (g) The reaction is endothermic, so increasing temperature will cause it to heat itself up. Equilibrium shifts right. 2) 2SO 2(g)+ O 2(g) SO 3(g) ΔH = -198 kJ 2SO 2(g) O 2(g)⟺ SO 3(g)– 198kJ The forward reaction is exothermic, so increasing the temperature will cause it to cool itself down. The reverse reaction is favoured, so equilibrium shifts left. 6 Example : consider the reversible reaction, 58 kJ +2N 4(g)⟺ 2NO .(g) Predict which way the equilibrium will shift following each disturbance. Change Shift Addition of N 2 4(s) Right Addition of NO 2(g) Left Removal of N O Left 2 4(g) Removal of NO 2(g) Right Addition of He (g) No shift Decrease volume Left Increase volume Right Increase temperature Right Decrease temperature Left Exercise: how would the following disturbances affect the reversible reaction, C (s) 2H 2(g)⟺ CH 4(g)+ energy? a) Adding more carbon to the system? No effect on the system. 7 b) Adding CH 4(g)to the system? The equilibrium will shift left. c) Raising the temperature? The equilibrium shifts left. d) Adding Argon to the system? No effect, Argon is an inert gas. e) Increase the volume of the reaction vessel? The equilibrium shifts left to increase the number of moles. -5 Exercise: At 1000 K, I 2(g)molecules dissociate with K = 3.c6x10 according to the following reaction, I2(g)⟺ 2I .(g) we introduce 1.00 mole of Iodine into a 2.00 L flask, what will [I ] and [2] be at equilibrium, and what percentage of the iodine molecules will have dissociated at equilibrium? Solution: 8 The Nature of Acids and Bases 1. Arrhenius definition of acids and bases: - Acids produce hydrogen ions in aqueous solutions; - Bases produce hydroxide ions in aqueous solutions - The Arrhenius definitions of acids and bases were very limited (too specific). 2. Brønsted-Lowry definition of acids and bases: 9 - Acids are proton donators; - Bases are proton acceptors; - Focus on this definition (it's more broad, and includes bases that don't initially contain OH in their chemical formulas, such has NH ). 3 Acid Dissolution in Water Consider the dissolution of HCl in H O.2Water acts as a Brønsted-Lowry base because it accepts a proton. Water is excluded in K aad K becbuse its concentration doesn't vary significantly during acidic or basic reactions. Above are two ways you can express Ka. As long as the reaction is balanced, it doesn't matter whether you write H O 3r H . + - when an acid loses a proton, the compound it becomes (its corresponding compound on the product side of the reaction) is the conjugate base. 10 - When a base becomes protonated, the compound it becomes is the conjugate acid. - The acid is the compound with the proton (the proton is attached to A on the reactant side, so HA is the acid). + + - A- and H O 2re competing for the proton (H O or H ). 3he stronger base gets protonated. Exercise: what's the conjugate base of formic acid, HCO H? 2 HCO H 2s a weak acid, so it loses a proton (by the Arrhenius definition). - + HCO H 2 HCO + H 2 The conjugate base is HCO . 2- Example: write the dissociation equations for the following acids: HCl, HOC H , Al(H O) 6 5 2 63+. 1) HCl is a strong acid. HCl → H+ + Cl- 2) HOC H i6 a5weak acid. + - HOC H 6 H5+ OC H 6 5 3+ 3) Al(H O2 6 is a hydrated metal, so H O is2the source of protons. 3+ - Al(H 2) 6 ⟺ H+ + Al(H O) O2 5 11 Non-aqueous Solutions The Brønsted-Lowry definition isn't limited to aqueous solutions (Arrhenius definition is limited to aqueous solutions). Note: NH 3an act as a base when mixed with an acid, as long as protons are being transferred. Acid Strength - Defined by the equilibrium position of an acid’s dissociation reaction. - For a strong acid, equilibrium lies far to the right, and is dissociated completely at equilibrium. - The conjugate base of a strong acid is weak, and the conjugate acid of a strong base is a weak acid. 12 - Memorize the strong acids! Note: most organic compounds (with carbon backbones) are weak acids, and most hydro halide acids are strong acids, except HF; HCl, HI, and HBr are strong acids. 13


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