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Condensed Matter 2 Physical Chemistry and Applications of Liquids and Solutions Course Aims The course aims to apply the principles of thermodynamics and kinetics to the study of the physical properties of gases, liquids and solids on both microscopic and macroscopic scales. There will be particular emphasis on the properties of liquids and solutions. This course provides a bridge between the elementary study of condensed matter phases and the advanced study of solid-state physics. Learning Objectives After completion of this course the student is expected to have a firm understanding the relationship between pressure, temperature, and volume on the behaviour of solids liquids and gasses. They should be able to apply these relationships to understand phenomena such as vapour pressure, surface tension, liquid crystals, etc. The student should understand the properties of liquids and solutions (both ideal and real) on microscopic and macroscopic scales, including electrolyte solutions, solvation, and equilibrium in solution, and appreciate the application of these principles to devices such as batteries and fuel cells. The student should have some understanding of intramolecular (chemical) and intermolecular bonding; including the bonding and structure of crystals. The student should be able to quantitatively solve problems involving all of the above. Course Content The Relative Stability of Gases Liquids and Solids (4 lectures): Including vapour pressure; surface tension; supercritical fluids; liquid crystals. Ideal and real solutions (4 lectures): Binary solution model; Gibbs-Duhem equation; suppression and elevation of freezing and boiling temperatures; osmotic pressure; dilute solutions; equilibrium in solution. Electrolyte Solutions (4 lectures): H, S, and G of ion formation; solvation; calculating activity coefficients by Debye-Huckel theory; equilibrium of electrolyte solutions. Electrochemical cells (4 lectures): Electrode potentials and cell potential; cell EMF and the equilibrium constant; electrochemical cells; batteries; fuel cells, atomic-scale electrochemistry and nano-machining; half-cell potentials. Introductory Solid State Physics (4 lectures): Including intermolecular and intramolecular bonding; and heat capacities of solids. Course Structure Lectures: Fridays 0900 and 1100 The course will consist of approximately 20 lectures. All lectures are compulsory and an attendance register will be taken. Failure to attend lectures at least 75 % of lectures may result in deregistration from the course. 8 Weekly Assignments (best 6 count) (Weeks, 4-12, save Reading Week). These sets are compulsory. Failure to submit at least 75 % of the assignments may result in deregistration. Deadline:1700 on the Wednesday preceding the problem class. The assignments will be available on the Module’s homepage (www.ph.qmul.ac.uk/~phy226) on the Monday of the week preceding the deadline. Problem Classes: Thursdays 1000 8 Weekly Problem Classes. (Weeks, 4-12, save Reading Week) All problem classes are compulsory and an attendance register will be taken. Failure to attend lectures at least 75 % of lectures may result in deregistration from the course. Course Assessment One Examination paper: 2 hours 15 minutes (80%). Eight assessed problem sets, of which the best six sets contribute to the mark (12%). Two in-class tests (8 % in total) to be run during the last lecture before reading week and the last lecture of term. Suggested Reading Physical Chemistry, P. W. Atkins. Properties of Liquids and Solutions, J.N. Murrell. Introduction to Solid-State Physics, C. Kittel. Equilibrium Things in Balance Mechanical Equilibrium Forces in balance. e.g., two equal masses, a rope and a pulley. From Newton II, the net force is zero. Note the system is reversible. Thermal Equilibrium Two objects in thermal contact. Heat flows from the warmer to the cooler. Until they reach the same temperature. (metal objects at room temperature initially feel cold – sensation diminished and vanishes at equilibrium) Chemical Equilibrium In a closed system, reactants turn into products. Eventually this process stops A chemical reaction is in equilibrium when there is no tendency for the quantities of reactants and products to change. Chemical Equilibrium +I2 → 2HI (synthesis) 2HI → H2 + I2 (dissociation) H2 The same chemical reaction system, but the roles are reversed. Both yield the same mixture when the change is completed (equilibrium mixture). Chemical Equilibrium composition of a chemical reaction system will tend to change in a direction that brings it closer to its equilibrium composition. Chemical Equilibrium The end is the same no mater where you start. Complete Reactions + B → C + D A reacts with B to form C and D. Implies that at the end of the reaction there is only C and D. Implies that C + D → A + B does not occur. The reaction is complete (the reactants turn completely into products). A Reversible Reactions If at equilibrium there are significant quantities of all reactants, the reaction is not complete, and the reaction is reversible. A + B = C + D In principle, all chemical reactions are reversible, but this reversibility may not be observable if the fraction of products in the equilibrium mixture is very small, or if the reverse reaction is kinetically inhibited (very slow.) Law of ‘Mass Action’ The law of mass action states that any chemical change is a competition between a forward and a reverse reaction. The rate of each of these processes is governed by the concentrations of the substances reacting; as the reaction proceeds. These rates approach each other, and at equilibrium they become identical. Law of ‘Mass Action’ Rate of forward reaction is: k f A B Rate of reverse reaction is: k r C D Thus, at equilibrium, a b c d k f A B kr C D a b c d How do we know a reaction is at equilibrium? Consider. 2H2 + O2 = 2H2O. We add the required amounts of H2 and O2 to a reaction vessel and leave them for a month. Is the mixture in equilibrium? How do we know a reaction is at equilibrium? [set of a spark in the vessel to find out]. The reaction now rapidly proceeds to equilibrium [provided the reaction vessel survives]. This mixture was not at equilibrium. Although the reaction is thermodynamically favoured it was kinetically inhibited. How do we know a reaction is at equilibrium? Test for equilibrium: Change P, or T, or an amount of one reactant (discussed later). If there is a change in the system we were at equilibrium. If no change is observed we were not at equilibrium (the reaction is kinetically inhibited). The Le Chatelier Principle The Le Chatelier Principle If a system at equilibrium is subjected to a change of pressure, temperature, or the number of moles of a component, there will be a tendency for a net reaction in the direction that reduces the effect of this change. The Le Chatelier Principle Example (amount of a reactant change) 2HI → H2 + I2 (in an arbitrary eq. mix) Now inject more H2: H2 now exceeds eq. concentration. No longer in equilibrium. A net reaction ensues until a new equilibrium is reached. The Le Chatelier Principle This happens in a direction that reduces the effect of the added H2. This can happen by some of the excess H2 reacting with some of the I2 to form more HI. Chemists would say "the equilibrium shifts to the left." The Le Chatelier Principle The Le Chatelier Principle Example (pressure change ) 2NH3 → 3H2 + N2 (in an arbitrary eq. mix, in a closed reaction vessel) Now increase the size of the vessel. Effect of increasing the size is that the pressure inside the vessel falls. How does the system react? The Le Chatelier Principle → 3H2 + N2 If only 2NH3 present, the pressure is 2 arbitrary units. If only 3H2 + N2 present, the pressure is 4 arbitrary units. Increasing volume decreases the pressure. Therefore system reacts to increase pressure 2NH3 The Le Chatelier Principle → 3H2 + N2 Pressure in increased by forming more products (and decomposing more NH3). Equilibrium shifts to the right. 2NH3 The Le Chatelier Principle Example (temperature change) heat + N2 + O2 → 2 NO (endothermic) Regard heat as a "reactant" or "product" in an endothermic or exothermic reaction respectively. Apply the Principle. The Le Chatelier Principle Suppose this reaction is at equilibrium at some temperature T1 . Increase temperature to T2. The system reacts in a way that absorbs heat. Equilibrium shifts to the right (product) [by analogy, equilibrium shifts to the left for exothermic reactions]. Equilibrium Quotient Earlier condition for the reaction aA + bB → cC + dD can be expressed as a quotient kf kr C D a b A B c d Qc Equilibrium Constant If expressed in terms of the equilibrium concentrations, this becomes: kf kr C D a b A B c d Kc Equilibrium Constant Kc is the value of Qc when the reaction is at equilibrium. The ratio Kc/Qc serves as an index how the composition of the reaction system compares to that of the equilibrium state. Thus, it indicates the direction in which any net reaction must proceed. Equilibrium Constant Qc/Kc > 1 Product concentration too high for equilibrium; net reaction proceeds to left. Equilibrium Constant Kc/Qc = 1 System is at equilibrium No net change will occur. < 1Product concentration too low for equilibrium; net reaction proceeds to right. Equilibrium Constant Kc/Qc < 1 Product concentration too low for equilibrium. Net reaction proceeds to right. Relating Q and K to Reactions Each tiny dot on the graph represents a possible combination of NO2 and N2O4 concentrations that produce a certain value of Qc for N2O4 → 2 NO2. Relating Q and K to Reactions Only those dots that fall on the red line correspond to equilibrium states of this system (those for which Qc = Kc ). Relating Q and K to Reactions If the system is initially in a non-equilibrium state, its composition will change in a direction that moves it to one on the line. Relating Q and K to Reactions Solid/vapour equilibria. E.g.,: Sublimation of iodine I2(s) = I2(g) The possible equilibrium states of the system are limited to those in which at least some solid is present (shaded) Relating Q and K to Reactions However, within this region, the quantity of iodine vapor (red line) is constant [as long as the temperature is unchanged]. Relating Q and K to Reactions The arrow shows the succession of states the system passes through when 0.28 mole of solid iodine is placed in a 1-L sealed container. Relating Q and K to Reactions The unit slope of this line reflects the fact that each mole of I2 removed from the solid ends up in the vapor. Relating Q and K to Reactions The decomposition of ammonium chloride NH4Cl(s) = NH3(g) + HCl(g) is another example of a solid-gas equilibrium. Relating Q and K to Reactions Arrow 1 traces the states the system passes through when solid NH4Cl is placed in a closed container. Relating Q and K to Reactions Arrow 2 represents the addition of ammonia to the equilibrium mixture; Relating Q and K to Reactions the system responds by following the path 3 back to a new equilibrium state. Relating Q and K to Reactions This state contains a smaller quantity of ammonia than was added (Consistent with the Le Châtelier principle). Does the reaction stop at equilibrium? At equilibrium Qc = Kc. There is no change in the concentrations of the reactants/products. The absence of any net change does not mean that nothing is happening. Why? Does the reaction stop at equilibrium? As any reaction is reversible, it can be expressed at the sum of the forward and the reverse reactions. A = B A → B rate = kf[A]. B → A rate = kr[B]. This implies B k f Kc A kr Does the reaction stop at equilibrium? This implies that the rates of the forward and backward reactions are equal: not that they are zero. I.e., at equilibrium As are turning into Bs as fast as Bs turn into As. Equilibrium is a dynamic process. Example The commercial production of hydrogen is carried out by treating natural gas (mainly methane) with steam at high temperatures and in the presence of a catalyst (“steam reforming of methane”): CH4 + H2O CH3OH + H2 Example Given the following boiling points: CH4 (methane) = –161°C H2O = 100°C CH3OH = 65° H2 = –253°C. predict the effects of an increase in the total pressure on this equilibrium at 50°, 75° and 120°C. Example Given the following boiling points: CH4 (methane) = –161°C H2O = 100°C CH3OH = 65° H2 = –253°C. predict the effects of an increase in the total pressure on this equilibrium at 50°, 75° and 120°C. Solution: Calculate the change in the moles of gas for each process: Example 50°C CH4 (g) + H2O (l) CH3OH (l) + H2(g) Solution: Calculate the change in the moles of gas for each process: One mole gas one mole of gas NO CHANGE Example 75°C CH4 (g) + H2O (l) CH3OH (g) + H2(g) Solution: Calculate the change in the moles of gas for each process: One mole gas two mole of gas To counter the increase in pressure, equilibrium shifts to the left. Example 120°C CH4 (g) + H2O (g) CH3OH (g) + H2(g) Solution: Calculate the change in the moles of gas for each process: two mole gas two mole of gas NO CHANGE How to find the equilibrium constant for a series of reactions Many chemical changes can be regarded as the sum or difference of two or more other reactions. If we know the equilibrium constants of the individual processes, we can easily calculate that for the overall reaction according to the following rule: How to find the equilibrium constant for a series of reactions The equilibrium constant for the sum of two or more reactions is the product of the equilibrium constants for each of the steps. Example Given the following equilibrium constants: CaCO3 (s) → Ca2+ (aq) + CO32– (aq) K1 = 10–6.3 HCO3– (aq) → H+(aq) + CO32–(aq) K2 = 10–10.3 Calculate the value of K for the reaction CaCO3 (s) + H+(aq) → Ca2+ (aq) + HCO3– (aq) Example The net reaction CaCO3 (s) + H+(aq) → Ca2+ (aq) + HCO3– (aq) is the sum of Reaction 1 CaCO3 (s) → Ca2+ (aq) + CO32– (aq) K1 = 10–6.3 And the reverse of Reaction 2 HCO3– (aq) ← H+(aq) + CO3 2– (aq) K-2 = 10-(-10.3) Example Therefore, K = K1 K-2 = 10(-8.4+10.3) = 10+1.9 The Harber process The Haber process for the synthesis of ammonia is based on the exothermic reaction N2(g) + 3 H2(g) = 2 NH3(g) Apply the le Châtelier principle in order to maximize the amount of product in the reaction mixture. The Harber process N2(g) + 3 H2(g) = 2 NH3(g) Would prefer N2(g) + 3 H2(g) = 2 NH3(l) The Harber process N2(g) + 3 H2(g) = 2 NH3(g) Would prefer N2(g) + 3 H2(g) = 2 NH3(l) it should be carried out at high pressure and low temperature. The Harber process N2(g) + 3 H2(g) = 2 NH3(g) But lower temperature slows the reaction. A choice has to be made. But Haber solved the first problem by developing a catalyst that would greatly speed up the reaction at lower temperatures. The Harber process N2(g) + 3 H2(g) = 2 NH3(g) A catalyst is a substance that lowers the energy barrier for a reaction. A catalyst plays no role in the net reaction. It is regenerated at the end of the reaction. Thus, they are only needed in trace quantities.