Presentation on theme: "Reaction Rates and Equilibrium"— Presentation transcript:
1 Reaction Rates and Equilibrium Chapter 17aReaction Rates and Equilibrium
2 Dark Energy and Dark Matter-Not in Textbook How Chemical Reactions OccurConditions That Affect Reaction RatesThe Equilibrium ConditionChemical Equilibrium: A Dynamic Condition
3 Dark EnergyIn a 1998 study led by Adam Riess, and a group of scientists from the Mount Stromlo Observatory, which is part of the Australian National University, Harvard University and Johns Hopkins University and Space Telescope Science Institute found, by observing supernovas in distant galaxies that the universe is expanding faster and faster. This violates Newton’s second law of motion that says acceleration is due a force. What force?????
4 Dark EnergyBasically, dark energy is what they attribute to the accelerated expansion of the universe which means it isn’t dark as to light, but dark as to they know nothing about it. Attempts to explain or measure this energy have largely failed.
5 Dark MatterDark matter was discovered by observing that the arms of spiral move at the same speed in violation of Kepler’s Law of planetary motion. It was explained by some unknown (dark) matter that distorted known gravitational effects. It was later found that the gravity of super massive black holes at the center of galaxies was not enough to hold the galaxies together. Gravity was not enough to hold the small local clusters of galaxies together as well as the super clusters of galaxies. Long filamental lines of matter, evident of some sort of attractive forces, has also been found between the super clusters of galaxies. Dark Matter has been used to explain all of these
6 New Model of the CosmosThe new model of the Cosmos puts about 4.6% of the universe being made up of atoms and molecules like what we think we know something about, about 23% is made up by dark matter and the rest (72%) is composed of dark energy.They say the Universe was different 13.7 billion years ago?????
7 Rates of Chemical Reactions 4 C3H5N3O9 6 N H2O CO2 + O kJ Nitroglycerine Exothermic
8 An Example of Reaction Rates 35/97 people died in 1937Fast Reaction vs. Slow Reaction
9 Molecules must collide in order for a reaction to occur. Collision ModelMolecules must collide in order for a reaction to occur.Rate depends on concentrations of reactants and temperature.
10 Concentration – increases rate because more molecules lead to more collisions. Temperature – increases rate.Why?
11 How to Tame Allergic Reactions How can you slow down a histamine attack?Histamine attacks are greater when you are hot. Cooling down affected areas can reduce allergy symptoms.
12 Minimum energy required for a reaction to occur. Activation EnergyMinimum energy required for a reaction to occur.Wood must have Ea to light and burn!
15 Chemical Reactions must go over an energy hill like a mountain (Swiss Alps).
16 A substance that speeds up a reaction without being consumed. CatalystA substance that speeds up a reaction without being consumed.Enzyme – catalyst in a biological system
17 A substance that speeds up a reaction without being consumed. CatalystA substance that speeds up a reaction without being consumed.Enzyme – catalyst in a biological system
18 A substance that speeds up a reaction without being consumed. CatalystA substance that speeds up a reaction without being consumed.Chlorofluoro Carbons (CFC’s) are acting as catalysts to decompose the ozone (O3) layer. The ozone layer is formed from cosmic radiation and protects us from UV light.
19 Depletion is measured by T.O.M.S. “Total Ozone Mapping Spectrometer”The below dark shaded are shows the amount of depletion around the Antarctica
24 Macroscopically static Microscopically dynamic Chemical EquilibriumOn the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.Macroscopically static Microscopically dynamic
39 Heterogeneous Equilibria Heterogeneous equilibria – involve more than one phase:2KClO3(s) KCl(s) + 3O2(g)2H2O(l) H2(g) + O2(g)
40 The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.The concentrations of pure liquids and solids are constant.2KClO3(s) KCl(s) + 3O2(g)H2O(l) H+(aq) + OH-(aq) K=[H+][OH-]
42 Equilibrium shifts to counter a disturbance. Hills and Valleys! 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.Equilibrium shifts to counter a disturbance. Hills and Valleys!O3 (g) + Cl (g) O2 (g) + OCl(g)
44 Effect of a Change in Volume The system is initially at equilibrium.The piston is pushed in, decreasing the volume and increasing the pressure. The system shifts in the direction that consumes CO2 molecules, lowering the pressure again.
52 The Extent of a Reaction A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right.Reaction goes essentially to completion.
53 The Extent of a Reaction A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left.Reaction does not occur to any significant extent.
54 The value of K for a system can be calculated from a known set of equilibrium concentrations. Unknown equilibrium concentrations can be calculated if the value of K and the remaining equilibrium concentrations are known.
56 At a given temperature, K = 50. for the reaction: H2(g) + I2(g) 2HI(g) Concept CheckAt a given temperature, K = 50. for the reaction:H2(g) + I2(g) HI(g)Calculate the equilibrium concentration of H2 given:[I2] = 1.5 × 10–2 M and [HI] = 5.0 × 10–1 M1.5 × 10–2 M3.0 × 10–2 M5.0 × 10–1 M3.3 × 10–1 MK = (HI)2/(H2)(I2)50 = (5.0 × 10–1)2/(H2)(1.5 × 10–2)(H2) = 3.3 × 10–1 MThe correct answer is d.K = (HI)2/(H2)(I2)50 = (5.0 × 10–1)2/(H2)(1.5 × 10–2)(H2) = 3.3 × 10–1 M
57 The equilibrium conditions also applies to a saturated solution containing excess solid, MX(s). Ksp = [M+][X] = solubility product constantThe value of the Ksp can be calculated from the measured solubility of MX(s).
59 Concept CheckIf a saturated solution of PbCl2 is prepared by dissolving some of the salt in distilled water and the concentration of Pb2+ is determined to be 1.6 × 10–2 M, what is the value of Ksp?2.6 × 10–42.0 × 10– PbCl Pb Cl-3.2 × 10– x x x1.6 × 10–5Ksp = [Pb2+] [Cl–]2 = (x)(2x)2=(1.6 × 10–2) (3.2 × 10–2)2 = 1.6 × 10–5The correct answer is d.The equilibrium for the dissolution of the salt is written as: PbCl2(s) <==> Pb2+(aq) + 2Cl–(aq)We are given a value for the concentration of lead(II) ion. The concentration of chloride ion must be twice that of the lead. Next we put these values into the Ksp expression and solve:Ksp = [Pb2+] [Cl-]2 = (1.6 × 10-2) (3.2 × 10-2)2 = 1.6 × 10-5
60 Calculate the solubility of silver chloride in water. Concept CheckCalculate the solubility of silver chloride in water.Ksp = 1.6 × 10–101.3 × 10–5 M1.6 × 10–10 M AgCl(s) Ag+ + Cl-3.2 × 10–10 M x x x8.0 × 10–11 MKsp = [Ag+][Cl–]1.6 × 10–10 = (x)(x) = x2x = 1.3 × 10–5 MThe correct answer is a.The equilibrium for the dissolution of silver chloride is written as: AgCl(s) <==> Ag+(aq) + Cl–(aq)Ksp = [Ag+][Cl-]We are given the Ksp value so that we can solve for the solubility of silver chloride:1.6 × = (x)(x) = x2x = 1.3 × 10-5 M
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