Chemistry of Solutions, lecture 6 notes
Chemistry of Solutions, lecture 6 notes 202-NYB-05
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This 8 page Class Notes was uploaded by CatLover44 on Sunday September 25, 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 6 views. For similar materials see Chemistry of Solutions in Chemistry at Dawson Community College.
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Date Created: 09/25/16
Tuesday, September 13, 2016 Chemistry of Solutions Prof.: N. Schoonhoven Topics covered: solving equilibrium problems, and Le Châtelier's principle.. Solving Equilibrium Problems Example: assume we start with 3.00 mol o2(g)nd 6.00 mol of 2(g)n a 3.00-L flask, and K = 1.15x10 . What are the eqm (equilibrium) concentrations of the chemical species in this rxn (reaction)? Initial concentrations: [H2] = 3.00 mol / 3.00 L = 1.00 M [F2] = 6.00 mol / 3.00 L = 2.00 M [HF] = 0 We know the system shifts right because there aren't any products present initially. Calculate the value of x: H2 F2 2HF Initial M 1.00 2.00 0 Change 1 - x 2 - x 0 + 2x Equilibrium 0.03162 1.03162 1.9368 M 2 (115) = (2x) / (1 – x)(2 – x) 2 2 (115) (2 –3x + x ) = 4x 2 2 230 – 345x + 115x = 4x 2 111x – 345x + 230 = 0 Use the quadratic formula a = 111, b = -345, c = 230. Δ = b – 4ac = (-345) – 4(111)(230) = 16,905 √ Δ = ± 130.02 x1= (345 +130.02)/222 = 2.1397, discard x2= (345 – 130.02)/222 = 0.96838, keep this value Calculate the equilibrium concentrations: [H 2 = 1.00 – 0.96838 = 0.03162 M [F 2 = 2.00 – 0.96838 = 1.03162 M [HF] = 2(0.96838) = 1.9368 M Example: hydrogen iodide is prepared from hydrogen and iodine vapour, and K = 1.00x10 . 2 -1 -2 -3 Initially a container is charged with 5.00x10 atm of HI, 1.00x10 atm of H , and 5.00x12 atm of I2. Calculate the equilibrium pressures. Calculate the value of Q: Q = (0.5) / (0.01)(0.005) = 10,000 > K, so the system shifts left. 2 Calculate the value of x: H 2 I2 HI Initial M 0.005 0.01 0.5 Change 0.01 + x 0.005 + x 0.5 - 2x Equilibrium M 0.0405 0.0455 0.429 2 (100) = (0.5 - 2x) / (0.005 + x)(0.01+x) 2 2 (100) = (4x – 2x + 0.25) / (0.00005 + 0.015x + x ) (100) (0.00005 + 0.015x + x ) = 4x – 2x + 0.25 0.005 + 1.5x + 100x = 4x – 2x + 0.25 2 96x + 3.5x – 0.245 = 0 Using the quadratic equation, a = 96, b = 3.5, c = -0.245. The correct value for x is 3.55x10 , the other value for x is -7.19x10 , which is wrong because we can't have negative concentrations. The equilibrium concentrations are: [H2] = 0.0405 M [I2] = 0.0455 M [HI] = 0.429 M Don't forget to check your value for x. 3 Skipping the Quadratic Formula When Calculating the Value of x You can avoid using the quadratic formula to calculate x if the K is at least 1,000 times smaller than the initial concentrations of the species in a chemical rxn. For example, consider the rxn -5 2NOCl (g)= 2NO (g)+ Cl 2(g) with K = 1.6x10 and initial concentrations of 0.50 M and 0 M for NOCl, and NO and Cl respectively. K is much, much smaller than the initial concentrations, so 2 we can approximate the value of x. We know the system shifts right. (1.6x10 ) = (2x) (x) / (0.5 – 2x) 2 (1.6x10 ) ≈ (2x) (x) / (0.5) 2 1.6x10 ≈ (4x ) / (0.25) -5 1/3 3 1/3 ((1.6x10 )(0.25)) ≈ (4x ) x ≈ 0.01 M, which is such a small difference that it's negligible. So the initial concentrations are ≈ the equilibrium concentrations. Example: consider the following eqm: PbCl = Pb 2+ + 2Cl - . What will happen to [Pb ] if 2+ 2(s) (aq) (aq) some solid NaCl is added to the system? NaCl dissociates into Na and Cl , so the concentration of Cl - would increase, while the (aq) concentration of Pb would decrease to compensate for the increase of concentration of Cl (aq). Le Châtelier’s Principle If a change is imposed on a system at equilibrium, the position of the eqm will shift in a direction that tends to reduce that change. Le Châtelier's principle (LCP) allows us to make qualitative predictions about how a system responds to a disturbance of equilibrium. Example: in the Haber process, N 2(g)+ 3H 2(g)= 2NH 3(g) what happens if we add N 2(g) By adding more N 2(g) the concentration of N 2(g)in the system increases, so more 2NH 3(g)is produced. The system seeks to decrease the concentration of N , so it favours the forward rxn. 2(g) Example: in the rxn As O 4 6(s)+ 6C (s)s (g)+ 6CO ,(g)at happens if: a) CO (g)is added? 4 If CO (g)is added, the system will respond by favouring the rxn that reduces the concentration of CO , so the reverse rxn is favoured. (g) b) As O4 6(s)r 6C (s)removed? Removing a solid from this system has no effect on its eqm position. c) If As (g)is removed? If As(g)is removed, the system will shift in the direction that favours the formation of As(g). Eqm will shift to the right. Ways that Equilibrium may be Disturbed • Adding or removing a gaseous reactant or product is the same as changing the concentration of a chemical species. • Adding or removing an inert gas has no effect on a system’s eqm position. Partial pressures remain the same. • Changing the volume of the container changes the partial pressures or concentrations of a system, so eqm is disturbed. 5 In a gaseous reversible rxn, increasing the volume of a flask means decreasing its pressure. According to LCP, an increase in pressure causes a gas out system to favour the rxn that decreases the number of moles. Likewise, if volume is decreased, pressure is increased, and a gaseous system would favour the rxn that increases the number of moles. 6 7 Exercise: how will the following changes affect the eqm of this exothermic rxn? C(s)2H 2(g) CH 4(g) a) Adding more carbon to the mixture No change b) Adding CH4 to the mixture Eqm shifts left c) Raising the temp of the mixture Eqm shifts left d) Adding Ar to the mixture No change e) Increasing the volume of the rxn vessel Eqm shifts left to increase the number of moles 8
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