Unit 1 From Kinetics to Equilibrium

Transcription

1 Unit 1 From Kinetics to Equilibrium Kinetic energy S the energy of motion S the energy particles have because they are moving Potential energy S the energy of position S the energy that particles have because of their positions relative to one another Energy has a role to play in every chemical reaction Pp Kinetic Molecular Theory of matter S the tiny particles in all forms of matter are in constant motion S solid particles vibrate in fixed positions due to strong intermolecular (or interparticle) attractions S liquid and gas particles have more freedom of movement due to weaker intermolecular attractions. S when the temperature of a substance is increased, there is an increase in the motion of the particles, an increase in the frequency and intensity of collisions between particles, and a decrease in the density of the substance Describe two pieces of evidence that support KMT i. Pressure ii. Diffusion Evidence to support KMT 1. Pressure - moving bodies exert forces when they collide with other bodies - many simultaneous collisions of billions of gas particles on an object are what we call gas pressure! the kinetic molecular theory explains why gases, unlike solids and liquids, are compressible. If most of the volume of a gas is empty space, it should be possible to force the particles closer together ii. Diffusion! particles tend to move from areas of higher concentration to areas of lower concentration due to collision State the collision theory Pp

2 Collision Theory Pp S all reactions are the result of collisions between particles. For example, in order to burn butane gas, butane molecules have to collide with oxygen molecules. Relate the rate of reaction to the number of successful collisions between reacting particles The reaction rate is the number of atoms, ions, or molecules that react in a given time to form products.! reactions proceed at different rates, and collision theory provides an explanation why! most chemical reactions involve the transfer of atoms from one molecule to another, thus contact between the reactants is very important! for any reaction involving 2 or more reactants, the reacting particles must collide! the more often particles collide, the faster the reaction should go! chemical reactions also involve the making and breaking of chemical bonds, and therefore involve energy! the colliding particles must have enough energy for this process or no reaction will occur. The required energy is called activation energy. (E a )! in a sense, activation energy is a barrier that the reactants must cross to be converted to products! Collision theory and Concentration (470) - the rate of reaction increases if there are more collisions per unit of time - the more reactant particles (i.e., the greater the concentration) the faster the reaction rate - for gases, increasing the pressure has the same effect as increasing the concentration Collision theory and Surface Area (470) - collisions can occur only at the surface of a solid - increasing the surface area causes more surface to be exposed, increasing the rate of reaction Collision theory and the Nature of Reactants (471) - reactions that involve ionic compounds and simple ions are generally faster than those involving molecular compounds - reactions that involve breaking weaker bonds are generally faster than those involving breaking stronger bonds (e.g., ethane vs ethene; ethane is

3 faster...c=c bond in ethene is stronger) - reactions that involve breaking fewer bonds are generally faster than those involving breaking a greater number of bonds Collision theory and Temperature (471) - increased temperature increases the frequency of collisions - increased temperature increases the intensity of collisions This takes us back to activation energy (E a ). Activation energy is the minimum collision energy that is required for a successful reaction. Collision theory and Orientation - The particles must collide with the correct orientation in order for reactions to occur Particles must collide with sufficient energy and the correct orientation in order for reactions to occur. Otherwise the particles collide without reacting. Not all collisions, therefore, result in chemical change (example: butane + oxygen) In order to explain chemical reactions the kinetic molecular theory can be expanded to create a theory of chemical reactions. According to the KMT, the particles of a substance are in continuous random motion. This motion inevitably results in collisions among the particles. If different substances are present, all the different particles will collide randomly with each other. If the collision has a certain orientation and sufficient energy, the components of the particles will rearrange to form new particles. The rearrangement of the particles that occurs IS the chemical reaction. This is known as the collision-reaction theory. Evidence that reactions have taken place includes:! color change; the final product(s) may have a different color than the colors of the starting material(s) odor change;

4 the final products may have a different odor than the starting materials! state change; the final materials may include a substance in a different state than the originals, usually a gas or a solid (precipitate) volume change the volume of the product(s) may be greater or lesser than that of the reactant(s)! mass in chemical reactions the total mass of matter present before the change is ALWAYS the same as the total mass present after the change, no matter how different the new substance may appear (Law of Conservation of Mass) Harry Potter ph the product(s) may have different phs from the reactants, as in an acid/base neutralization reaction Homework: P 484, Q 1 and 2 Potential energy diagrams P i. Activation energy (p. 472) - minimum collision energy - depends on the kinetic energy of the colliding particles ii. Activated complex During a reaction, particles which are neither reactants nor products form momentarily. An activated complex is the arrangement of atoms at the peak of the activation energy barrier.! This group of atoms is on its way to becoming an ionic or molecular product.! very unstable! may go either way - to reactants or to product! lifetime of only second! also known as the transition state iii. ªH rxn (book has ªE)

5 - the enthalpy of reaction - the difference between the potential energy of the products and the potential energy of the reactants - breaking a bond is a process that requires energy; creating a bond releases energy iv. Reactants and products - original and final compounds in the reaction The increase in potential energy during a emical reaction can be represented by a tential Energy Diagram - a diagram which charts e potential energy of a reaction against the ogress of that reaction. c h Po t h p r For example: Let s look at this in more detail. A potential energy diagram illustrates several important stages in a chemical reaction:

6 1. The flat region labeled reactants (2H 2 + O 2 ) shows the potential energy content of the reactants relative to the products. 2. The rising part of the graph represents the increase in potential energy that occurs when reactants collide. S The kinetic energy of reactants is converted to potential energy as they collide because the force of the collision causes the chemical bonds to stretch and distort. S S The minimum energy the colliding particles must have in order to react is the activation energy. (E A ) activation energy is a sort of barrier that the reactants must cross in order to be converted into products. 3. The top of the graph is where the bonds of the colliding particles are stretched to the breaking point. S Here, particles which are neither reactants nor products form momentarily. S These particles form an activated complex - an unstable arrangement of atoms at the peak of the activation energy barrier. S The life of an activated complex is only about seconds and during that time the complex either forms new bonds to become products, or reforms old bonds and returns to being reactants. S At this stage the activated complex is in the transition state. 4. The falling part of the curve represents the energy released when bonds form between particles that make up the products. 5. The second flat region represents the potential energy of the products. S The difference between the potential energy of the reactants and products is called heat of reaction (ªH).[ªE] S the heat of reaction is the heat absorbed or released as reactants become products Collision theory explains why some exothermic reactions, like that of butane and oxygen, do not occur at room temperature. S The reaction of butane and oxygen is exothermic, but it has a high activation energy. S At room temperature, the collision of oxygen and butane molecules is not energetic enough to overcome the activation energy barrier, S no activated complex forms and S therefore no products will form. Look at the diagram above again:

7 What do you notice about the potential energy of the products (E p ) relative to the potential energy of the reactants (E r )? S it is lower than that of the reactants. S S this is an exothermic reaction since E p < E r, energy must have been released during the chemical reaction. Exothermic Reactions - the reactants have more potential energy than the products. The extra energy is released to the surroundings. A + B --> AB + Energy S the opposite is true for an endothermic reactions (E r < E p ) Example: Graph of endothermic reaction Endothermic Reactions - the reactants have less potential energy than do the products. Energy must be input in order to raise the particles up to the higher energy level. Energy + A + B --> AB

8 Reversible Reactions Some chemical reactions are reversible. That is, the products can be converted back into reactants. For example, water can be decomposed to produce hydrogen and oxygen, yet water is the product of a reaction between hydrogen and oxygen. Assume that the reaction above is reversible. How is the reverse reaction classified? It s exothermic (E r > E p ) The effect of a catalyst on the rate of reaction P 477, A catalyst works by lowering the activation energy of a reaction so that a larger fraction of the reactants have sufficient energy to react. It does this by providing an alternative mechanism for the reaction - the catalyzed reaction consists of a two (or more) step process e.g., A + B 6 AB A catalyst will increase the rate of this reaction by providing an alternative mechanism with lower activation energy. A possible mechanism would be: 1. A + catalyst 6 A-catalyst 2. A-catalyst + B 6 AB + catalyst

9 Overall reaction: A + B 6 AB A-catalyst is a reaction intermediate: produced in step 1 but consumed in step 2. The catalyst is regenerated in the reaction: it appears as a reactant in step 1 and a product in step 2. Both steps are faster than the original, uncatalyzed reaction. Homogeneous catalysts p482 - same phase as reactants Heterogeneous catalysts p different phase from reactants Homework: Question 9, page 484

10 Factors that affect the rate of reactions Chemical reactions occur at different rates. Compare a chemical explosion to a car rusting; both involve the oxidation of certain materials, but one takes place in seconds while the other may take many years. 1. Nature and reactivity of the reactants - the state of matter, type of bonding, valence orbital configuration, and many other features of a substance influence the speed of the chemical change The chemical and physical properties of substances are collectively called the nature of reactants. Examples: a. Solid lead nitrate and potassium iodide react very slowly. Aqueous solutions of these react instantly to produce a bright yellow precipitate.

11 * - Ions in solution react rapidly because the ions of opposite charge are attracted to each other and can bond upon colliding. * - Changing the state of matter can influence the rate of a reaction. b. A strong acid (one which ionizes completely) such as HCl (aq) will react with zinc metal faster than a weak acid (one that doesn t ionize completely) like acetic acid. S strong acids have higher hydrogen ion concentrations than weak acids of the same molar concentration * S the strength of an acid (the degree to which it ionizes) can influence the rate of a reaction. c. Cesium metal reacts so fast and so vigorously with water that the hydrogen gas produced explodes violently. However, lithium, which is also a group 1A metal, reacts relatively slowly with water. S * low electronegativity metals tend to react faster d. Chloride ions react faster with silver ions than chlorine gas reacts with solid silver. S In the reaction involving chlorine gas, the two covalently bonded chlorine atoms have to be separated before they can react with silver atoms, which in turn are held together by metallic bonds * S types and strengths of bonds influence reaction rates 2. Temperature S increasing the temperature of a chemical system increases the frequency of collisions between the reactant particles. S more collisions means that more particles will hit with greater frequency, intensity, and correct orientation and will therefore enable products to form S more colliding particles become energetic enough to slip over the

12 activation energy barrier to become products S Example: paper does not appear to be burning at room temperature (even though it is slowly oxidizing). Touching it with a match allows the paper and air molecules to collide with greater frequency and high energy. Some of the energy released when the products are formed enables more paper/air to get over the activation barrier. S removing energy from the system by decreasing the temperature will result in a decrease in the reaction rate because fewer particles will collide with the minimum energy and correct orientation * S even a small increase in temperature (like 10 o C) can double a reaction rate due to more particles having enough kinetic energy to overcome the activation energy barrier As a general rule: 8T, 8 intensity and frequency of collisions, 8 reaction rate 9T, 9 intensity and frequency of collisions, 9 reaction rate 3. Concentration S chemical reactants are more likely to collide and react if there are more of them occupying a volume of space at the same time S an increase in the concentration of a solution will increase the frequency of collisions (#/ sec) and thereby increase the chances for a chemical change to occur. This will result in an increase in the reaction rate S reducing the concentration of a solution will decrease the reaction rate As a general rule: 8 concentration, 8 frequency of collisions, 8 rate of reaction 9 concentration, 9 frequency of collisions, 9 rate of reaction S the same argument applies to increasing or decreasing the number of gas particles in a fixed volume of gas. S for gases S by decreasing the volume of a container, you force the particles closer together and increase the pressure on the gases in the container S this increases the collision frequency and thus the rate of reaction S by increasing the volume of a container, you lessen the chance for particle collision and lower the rate of reaction in the container

13 General rule (gases): 8 V, 9 P, 9 chance of collision, 9 rate of reaction 9 V, 8 P, 8 chance of collision, 8 rate of reaction Surface Area (effect of particle size) S total surface area of a reactant has an important effect on the reaction rate S the smaller the particle size, the larger the surface area for a given mass of particles (compare laundry tabs to detergent powder) S an increase in surface area increases the collision frequency and therefore the reaction rate S the best way to increase the surface area of solid particles is often to dissolve them in a liquid. This separates the particles S homogeneous mixtures of particles often react more quickly than heterogeneous mixtures (e.g., oxygen tanks vs air tanks) S it s often impossible to obtain homogeneous mixtures; heterogeneous mixtures of reactants result in heterogeneous reactions S General rule: 8 surface area, 8 number of collisions, 8 rate of reaction 5. Catalysts S a catalyst is a substance that increases the rate of reaction without being used up itself in that reaction S for any chemical reaction, there is a sequence of events that must occur if reactants are to become products S the series of steps involved in rearranging reactant particles to form products is called the reaction mechanism S a catalyst speeds up a reaction by providing a reaction mechanism that requires less activation energy (see diagram, page 481) S catalysts can cause the formation of activated complexes that have lower potential energy than those of uncatalyzed reactions S with a lower activation energy barrier, more collisions between the reactants will have the required energy to form products and, as a result, the rate of reaction will increase

14 S S catalysts do not appear as reactants or products in the equation for the reaction. They appear above the 6 line a substance that interferes with the action of a catalyst is called an inhibitor General rule: Catalysts increase reaction rates by providing alternate reaction mechanisms that have lower activation energy. There are many generalizations regarding the effect of nature of reactants on reaction rate. The type of bonding within reactants and products has a profound impact on reaction rate. For example: 1. Reactions involving ionic compounds are generally faster than reactions involving molecular compounds e.g. 2 CO (g) + O 2(g) 6 2 CO 2 (g) Very slow at 25 o C 5 Fe 2+ (aq) + MnO- 4 (aq) + 8 H + (aq) 6 5 Fe3+ (aq) + Mn2- (aq) + 4 H 2 O (l) Very fast at 25o C 2. Reactions involving breaking weaker bonds are generally faster than reactions involving stronger bonds. E.g. 2 C 2 H 6(g) + 7 O 2(g) 6 4 CO (g) + 6 H 2 O (l) V.v. fast at 25 o C C 2 H 4(g) + 4 O 2(g) 6 2 CO (g) + 4 H 2 O (l) V. fast at 25 o C 3. Reactions involving the breaking of fewer total number of bonds are generally faster than reactions involving the breaking of larger number of bonds. E.g. 2 C 5 H 12(g) + 11 O 2(g) 6 5 CO (g) + 12 H 2 O (l) fast at 25 o C 2 C 10 H 22(g) + 21 O 2(g) 6 20 CO (g) + 22 H 2 O (l) Slow at 25 o C Reaction mechanisms P A reaction mechanism is a series of steps that make up an overall reaction. Each

15 step, called an elementary reaction, involves a single event (e.g., a collision between atoms, molecules or ions) We have been studying elementary reactions. In elementary reactions: S reactants are converted to products in a single step S there is only one intervening activated complex e.g., NO + O 2 6 NO 3 elementary NO 3 + CO 6 NO 2 + CO 2 elementary NO + O 2 + CO 6 NO 2 + CO 2 overall Note that NO 3 is present in both elementary reactions but not in the overall reaction. It is produced in the first step and consumed in the second. Molecules (or atoms or ions) that are formed in elementary reactions and consumed in a subsequent elementary reaction are called reaction intermediates. Although they are not reactants or products in the overall reaction, they are essential for the reaction to take place. Most chemical reactions consist of a number of elementary reactions. A reaction mechanism includes all of the elementary reactions of a complex reaction. S the reaction progress of a complex reaction consists of a number of hills and valleys S the hills correspond to the energy levels of the activated complexes S the valleys correspond to the energy levels of the intermediate products Intermediates S a product of a reaction that immediately becomes a reactant of another reaction S have a significant lifetime compared with an activated complex S posses real ionic or molecular structures and some stability S are reactive enough to react further to eventually give the final product of the reaction

16 S do not appear in the chemical equation for a reaction Example: The decomposition of nitrous oxide (N 2 O) is believed to occur in two elementary steps: N 2 O 6 N 2 + O N 2 O + O 6 N 2 + O 2 2 N 2 O 6 2N 2 + O 2 Note the oxygen atoms are intermediates in this reaction. They disappear when the individual reactants are summed to give the final chemical equation. Thus the overall chemical equation for a complex reaction gives no information about the reaction mechanism. Definitions molecularity - the number of reactant particles involved in an elementary reaction - can be molecules, atoms or ions bimolecular - when two particles collide and react, the elementary reaction is said to be bimolecular unimolecular - when one molecule or ion reacts in an elementary reaction termolecular - when three particles collide in an elementary reaction - very rare ( ) Rate determining step The measurement of reaction rate is based on the rate of appearance of a product or disappearance of a reactant. It is usually expressed in terms of a change

17 in concentration of one of the participants per unit time. Experiments have shown that for most reactions the concentrations of all participants change most rapidly at the beginning of the reaction. That is, the concentration of the product shows the greatest rate of increase and the concentrations of the reactants the highest rate of decrease at this point. This means that the rate of reaction changes with time. Therefore a rate must be identified with a specific time. The reaction rate is usually proportional to the concentrations of the reactants. The usual dependence of the reaction rate on the concentration of the reactants can be simply explained by theorizing that if there are more particles of the reactant in the reaction area, then there is a greater chance that more reactions will occur (collision theory) The reason we say usually proportional to the concentration of the reactants is that some reactions do not occur directly between the reactants, but may go through intermediate steps to get to the final product. The series of steps by which the reacting particles rearrange themselves to form the products of a chemical reaction is called the reaction mechanism. For example: Step 1: A + B 6 I 1 (fast) Step 2: A + I 1 6 I 2 (slow) Step 3: C + I 2 6 D (fast) Net equilibrium 2A + B + C 6 D Step 2 would be the rate determining step. Elementary reactions in mechanisms all have different rates. Usually one elementary reaction - the rate determining step - is much slower (highest activation energy). It therefore determines the overall rate. E.g., Consider the reaction 2NO (g) + 2 H 2(g) 6 N 2(g) + 2 H 2 O (g)

18 Two step mechanism: 2NO (g) + H 2(g) 6 N 2 O + H O (g) 2 (g) slow (rate determining step) N 2 O + H 6 N + H O (g) 2(g) 2(g) 2 (g) fast A potential energy diagram for the reaction mechanism must take the rate determining step into account. As the potential energy diagram from the above reaction shows, there are 2 activation energy barriers - one for each proposed elementary reaction. As a result there are two transition states and one reaction intermediate. **Because the 1 st step is slower than the 2 nd, the activation energy barrier for the first step must be greater than that of the 2 nd. ** p 480 Reaction rate, fast or slow, involves many particles that must collide according to the balanced equation of a reaction mechanism step, and not the balanced overall equation for the reaction. E.G., For the following overall balanced reaction, a 3 step reaction mechanism is shown:

19 2 Cu I Cu + + I 2 + energy Fast Cu S 2 O 8 6 CuSO SO 4 Slow Cu + + CuSO Cu SO 4 Fast S 2 O I SO I 2 + energy Slow For the above: i. The overall reaction is exothermic since energy is produced in the first step, but is not mentioned in any other steps ii. The intermediate species are Cu + AND CuSO 4 since they are both produced in earlier steps and completely consumed in later steps iii. The catalyst is Cu 2+ since it is consumed in earlier steps and produced in equal amounts in later steps (i.e., a catalyst is added to the reactant vessel with the reactants but is re-produced.) iv. The step 2 is the rate determining step Note: 1. Each elementary step has its own activation energy barrier and activated complex. Each step could be visualized as 3 single step reactions. The step (i.e., the single reaction) with the highest activation energy is the rate determining step. In this example, step 2 is the rate determining step since it has the highest activation energy. 2. The reaction profile is consistent with the overall reaction being exothermic (as indicated in the example) Knowing the reaction mechanism provides the basis for predicting the effect of a concentration change of a reactant on the overall reaction rate. Catalysts, p 481 The catalyst provides an alternative pathway by which the reaction can proceed and in which the activation energy is lower. It is this which increases the rate at which the reaction comes to completion or equilibrium.

20 The catalyzed reaction will have a faster RDS with a lower E act than that of the noncatalyzed reaction. [students don t need to determine the actual catalyzed reaction mechanism It may have one or more steps. The largest energy barrier is smaller than the uncatalyzed mechanism. Dynamic Equilibrium Page Evidence obtained from many reactions contradicts the assumption that reactions are always quantitative (i.e., go to completion). In some reactions there is direct evidence for the presence of both reactants after the reaction appears to have stopped. This apparent anomaly can be explained consistently in terms of the collision reaction theory, by the idea that A reverse reaction has occurred. That is, the products can react to reform the original reactants. The final state of this chemical system can be explained as a competition between collisions of reactants to form products and collisions of products to re-form reactants. E.g., Na 2 SO 4(aq) + CaCl 2(aq) W CaSO 4(s) + 2 NaCl (aq) Eventually the rate of the forward reaction equals the rate of the reverse reaction and: Dynamic equilibrium occurs when opposing changes are occurring at the same time and

21 at the same rate This competition requires that the system be closed so that reactants and products cannot escape from the reaction container. When talking about dynamic equilibrium we are discussing a balance between forward and reverse reaction rates. At equilibrium, the rate of the forward reaction = the rate of the reverse reaction. P. 492 There are two chemical processes that reach equilibrium: 1. A reaction with reactants and products in the same phase (e.g., all gaseous or aqueous). In this case a homogenous equilibrium is reached 2. A reaction with reactants and products in different phases (e.g., aqueous ions combine to form semi-soluble product). In this case a heterogeneous equilibrium is reached. Any dynamic equilibrium can be described using a system of reversible rates. In the case of chemical equilibria they are forward and reverse chemical reactions. Establishment of a dynamic equilibrium involves 3 steps: 1. One reaction is initiated, or perturbed. The forward rate is initially very high 2. The reverse rate responds (products collide to form reactants) 3. Equilibrium is established. The forward and reverse reaction rates become equal Four conditions that apply to all equilibrium systems 1. Equilibrium is achieved in a reversible process when the rates of opposing changes are equal. This is indicated by X e.g., H 2(g) + Cl 2(g) X 2 HCl 2. The observable properties of the system at equilibrium are constant. At

22 equilibrium there is no overall change in the properties that depend on the total quantity of matter in the system (e.g., color, pressure, concentration and ph will not change) 8 there are changes, above, at the molecular level, but you won t see them at an observable level 3. Equilibrium can only be reached in a closed system. Nothing gets in or out, including energy. What this really means, in energy terms, is that the equilibrium can occur only in a system at constant temperature. 4. Equilibrium can be approached from either direction. For example, the proportions of H 2(g), Cl 2(g), and HCl in a closed container at constant temperature will be the same regardless of whether you started with H 2(g) and Cl 2(g) or HCl P 494 Let s consider how an equilibrium is established in a chemical system. Imagine that 2 moles of sulfur dioxide and one mole of oxygen are placed into an empty, sealable container. At first collisions between these reactants will result in the production of sulfur trioxide at a relatively fast rate: 2 SO 2(g) + O 2(g) 6 2 SO 3(g) However, the concentration of the reactants will eventually decrease because they are being used to make products. The forward rate of reaction will consequently decrease. As the sulfur trioxide molecules are being formed some of them will collide with each other to produce sulfur dioxide and oxygen. 2 SO 3(g) 6 2 SO 2(g) + O 2(g) This change will proceed slowly at first but will speed up as the SO 3 concentration increases. A point is reached when the rate of the forward reaction equals that of the reverse reaction. When this happens, an equilibrium is established.

23 Note: S S S 2 SO 2(g) + O 2(g) º 2 SO 3(g) the equal lengths of the arrows illustrates that the forward rate of reaction is equal to the reverse rate of reaction. the concentrations of the reactants do not have to equal those of the products only the rates of change have to be equal Problem: Write the chemical equation for the equilibrium system formed when hydrogen and iodine gases are sealed in a ml flask and explain what is happening at the molecular level in terms of reaction rates. H 2(g) + I 2(g) º 2 HI (g) le Chatelier s Principle Page An equilibrium is maintained as long as things like temperature, pressure and the amounts of matter present are kept constant. Le Chatelier observed that whenever an equilibrium was disturbed in some way, some property of the equilibrium changed and a new equilibrium became established. Le Chatelier s Principle When a chemical system at equilibrium is disturbed by a change in some property of that system, the system will react to partially counteract the change until a new equilibrium is established The application of Le Chatelier s Principle involves a 3 stage process: S an initial equilibrium state S a shifting non-equilibrium S a new equilibrium state Le Chatelier s Principle provides a method of predicting the response of a chemical system to an imposed change. There are three types of easily applied changes to a system:

24 6. Changes in concentration of species in the system ( p. 526) Many important equilibrium systems (especially those we ll be dealing with in this course) involve ions in aqueous solution. The common ion effect holds here. The addition of an ion to a solution in which the ion is already present causes a shift of equilibrium away from the added ion, as predicted by Le Chatelier. That is S the addition of more reactant or the removal of a product will increase the overall yield of product(s) by shifting the equilibrium to the right Consider the reaction: Fe 3+ (aq) + SCN- (aq) º FeSCN2+ (aq) If iron (III) ions are added to the system S more collisions will occur between Fe 3+ and thiocyanate ions, SCN - S S S S this will cause an increase in the rate of the forward reaction the concentration of SCN - will decrease, and the concentration of iron (III) thiocyanate ions, FeSCN 2+, will increase. This will, in turn cause the reverse reaction rate to increase until the rates equalize again In the new equilibrium the concentration of FeSCN 2+ will be higher and the concentrations of SCN - will be lower We may say that the equilibrium has shifted away from the side of the equation where the stress was imposed. We may also say that it has shifted right. Fe 3+ (aq) + SCN - (aq) º FeSCN 2+ (aq) Add 9 8 If iron (III) ions are removed from the system: S the opposite effect occurs S fewer collisions between SCN - and Fe 3+ will occur S the forward rate of reaction will slow down S with less FeSCN 2+ produced, the reverse rate will appear to be faster than the forward rate until it, too, slows down

25 S S the overall result will be a decrease in FeSCN 2+ concentration and an increase in the concentration of SCN - the equilibrium will have shifted towards the side where the concentration of a species was decreased (left) Fe 3+ (aq) + SCN - (aq) º FeSCN 2+ (aq) Remove 8 9 General rule: 8 concentration, equilibrium shifts away from increase 9 concentration, equilibrium shifts towards decrease Note: Changes in concentrations have the greatest effect on systems which involve gases and/or aqueous solutions (i.e., ions) 7. Change in the temperature of a system (page 526) The position of the equilibrium changes with temperature because the rates of the forward and reverse reactions are affected. We will deal with this again when we calculate equilibrium constants, but for now S energy in a chemical equilibrium is treated as though it were a reactant or a product. E.g., Reactants + energy º products reactants º products + energy S S energy can be added to or removed from a system by heating or cooling the container. According to Le Chatelier s principle, changing the temperature of a system at equilibrium should result in a shift to a new equilibrium if the system is cooled, the equilibrium shifts so that more heat is produced.. If heat is added, the equilibrium shifts in the direction in which energy is used. In either case, the equilibrium shifts to minimize the change for example, consider the system:

26 2 SO 2(g) + O 2(g) º 2 SO 3(g) kj the forward reaction is exothermic, because when we go to the right, energy is released the reverse reaction is endothermic because when we goto the left, we need energy to decompose the sulfur trioxide Since the reverse reaction requires energy, if we heat this reaction vessel, we will drive the equilibrium away from the energy term, to the left According to Le Chatelier s principle, changing the temperature of a system at equilibrium should result in a shift to a new equilibrium. S if the temperature is increased, the collisions between sulfur dioxide and oxygen molecules should be more frequent and intense. The same is true for the sulfur trioxide molecules S as a result, both the forward and reverse reaction rates will increase. S however, since the decomposition of sulfur trioxide is endothermic, the heat increase will favor the decomposition reaction so that it occurs at a faster rate than the forward reaction S when the reaction rates equalize, the concentration of sulfur trioxide will be lower while those of sulfur dioxide and oxygen will be higher. S the equilibrium will shift away from an increase in temperature, as predicted by Le Chatelier s Principle. 2 SO 2(g) + O 2(g) º 2 SO 3(g) kj add heat S S S S if the temperature is decreased, both the rate of forward and reverse reactions will decrease this is because collisions between particles will be fewer and less intense however, since the forward reaction is exothermic, it will be favored and will occur at a slightly faster rate than the reverse reaction as a result the concentrations of sulfur dioxide and oxygen will decrease while that of sulfur trioxide will increase

27 2 SO 2(g) + O 2(g) º 2 SO 3(g) kj remove heat General rule: 8 T, equilibrium shifts away from the energy term 9 T, equilibrium shifts towards the energy term Review this with note of the same system, written backwards, on page 526. Show that the last 2 rules for endo and exothermic change do apply to the general rule, stated above. 3. Change in the pressure (and volume) of a system (page 527) According to Le Chatelier s Principle, changing the pressure on a system at equilibrium should result in a shift to establish a new equilibrium. A change in pressure, however, will only affect an equilibrium with an unequal number of gaseous reactants and products. We must therefore consider: Boyle s Law S S S the concentration of a gas in a container is directly related to the pressure of the gas decreasing the volume by half doubles the concentration of every gas in the container, and may therefore cause a shift in the equilibrium doubling the volume halves the concentration of every gas in the container, and may therefore cause a shift in the equilibrium Avogadro s hypothesis S equal volumes of gases at the same temperature and pressure contain equal numbers of particles If you increase the pressure on a closed system at equilibrium (e.g., by decreasing the volume of the vessel) S the system should respond by decreasing the number of moles of gas in the system S this results in an equilibrium shift to the side of the equation with the least number of moles of gases (If there are fewer gas particles, they will require less space)

28 If you decrease the pressure on a closed system (e.g., by increasing the volume of the container) S there will be an equilibrium shift toward whichever side contains the greatest number of moles of gases General rule: 8 P, equilibrium shifts towards side with fewer moles of gas 9 P, equilibrium shifts towards side with more moles of gas A system with equal numbers of gas molecules on each side of the equation is not affected by a pressure change (no shift) Systems involving only liquids, aqueous and solids are not affected by pressure changes For example, consider this system: 3 H 2(g) + N 2(g) º 2 NH 3(g) Pressure increase S equilibrium will shift to the right because there are just two moles of gas compounds on the products side versus four moles on the reactants side S products occupy less space than the reactants and will reduce the magnitude of the pressure increase Pressure decrease S equilibrium will shift to the left because there are more moles of gas compounds on the reactants side than on product side S reactants occupy more space so a shift to that side increases the pressure within the system the products side of the equation Change in Volume S same effect as changing the pressure, except that since pressure and volume are inversely proportional, increasing volume is the same as decreasing pressure S note that changes in volume will only affect the gases in a system General Rule: 8 V of system, 9 P, shift to side with more moles of gas 9 V of system, 8 P, shift to side with fewer moles of gas Explain why the addition of a catalyst and varying surface area of a reactant or

29 product do not cause the equilibrium to shift, yet both factors do have an effect on the time it takes for a system to reach equilibrium. Effect of Adding a Catalyst (page 527) S according to Le Chatelier, adding a catalyst at equilibrium should result in a shift to establish a new equilibrium S however, catalysts lower the activation energies of both the forward and the reverse reaction equally S because of this, the equilibrium is not disturbed S the only effect that a catalyst has on the equilibrium is that it reduces the time it takes to reach equilibrium General Rule: Addition of a catalyst does not shift the position of an equilibrium Change in the Surface Area S has a similar effect to that of adding a catalyst S reduces the time required for a system to reach an equilibrium S no effect on the actual equilibrium concentrations

30 Summary: Variables Affecting Chemical Equilibria Variables Direction of Change Response of System General Rule Concentration increase Concentration decrease shifts to consume some of the added reactant or product shifts to replace some of the removed reactant or product 8concentration, equilibrium shifts away from increase 9concentration, equilibrium shifts towards decrease Temperature increase shifts to consume some of the added heat 8T, equilibrium shifts away from the energy term Temperature decrease shifts to replace some of the removed heat 9T, equilibrium shifts towards the energy term Pressure increase ( volume decrease) shifts toward the side with the smaller total amount of gas particles 8P, equilibrium shifts towards side with fewer moles of gas Pressure decrease (volume increases) shifts toward the side with the larger total amount of gas particles 9P, equilibrium shifts towards side with more moles of gas Catalyst none no change in position of equilibrium reduces time to reach equilibrium Change in surface area increase no change in position of equilibrium reduces time to reach equilibrium

31 Change in surface area decrease no change in position of equilibrium increases time to reach equilibrium

32 Homework: Due beginning of next class period Name: The production of freon-12, a CFC refrigerant, involves the following equilibrium reaction: CCl 4(l) + 2 HF (g) º CCl 2 F 2(g) + 2 HCl (g) 1. To improve the yield of the primary product, freon-12, more hydrogen fluoride is added to the initial equilibrium system. Describe what then occurs within the system. 2. The final step in the production of nitric acid is represented below: 3 NO 2(g) + H 2 O (l) º 2 HNO 3(aq) + NO (g) In this industrial process, NO (g) is removed from the chemical system by a reaction with oxygen gas. Describe what then occurs within the system. 3. In the endothermic salt-sulfuric acid process used in the production of hydrochloric acid, the system is heated. Describe what then occurs within the system. 2 NaCl (s) + H 2 SO 4(l) º 2 HCl (g) + Na 2 SO 4(s) 4. In producing sulfuric acid by the contact process (which is exothermic), the temperature of the system is kept low. Describe what then occurs within the system 2 SO 2(g) + O 2(g) º 2 SO 3(g) 5. Explain what would happen to the equilibrium in the following system if (a) the pressure were increased and (b) the pressure were decreased 2 SO 2(g) + O 2(g) º 2 SO 3(g) 6. Explain what would happen to the equilibrium in the following system if (a) the pressure were increased and (b) the pressure were decreased H 2(g) + I 2(g) º 2 HI (g)

33 Homework: Due beginning of next class period answers The production of freon-12, a CFC refrigerant, involves the following equilibrium reaction: CCl 4(l) + 2 HF (g) º CCl 2 F 2(g) + 2 HCl (g) 1. To improve the yield of the primary product, freon-12, more hydrogen fluoride is added to the initial equilibrium system. Describe what then occurs within the system. The additional amount of the reactant disturbs the equilibrium state and the system shifts to the right, consuming some of the added hydrogen fluoride by reaction with carbon tetrachloride. As a result, more freon-12 is produced and a new equilibrium state is attained The final step in the production of nitric acid is represented below: 3 NO 2(g) + H 2 O (l) º 2 HNO 3(aq) + NO (g) 2. In this industrial process, NO (g) is removed from the chemical system by a reaction with oxygen gas. Describe what then occurs within the system. The removal of the NO causes the system to shift to the right. Some nitrogen dioxide and water react, replacing some of the removed nitrogen monoxide. As the system shifts, more of the desired product, nitric acid, is produced. 3. In the endothermic salt-sulfuric acid process used in the production of hydrochloric acid, the system is heated. Describe what then occurs within the system. 2 NaCl (s) + H 2 SO 4(l) º 2 HCl (g) + Na 2 SO 4(s) The energy that is added causes the system to shift to the right, increasing the % yield of hydrogen chloride gas and absorbing some of the added energy 4. In producing sulfuric acid by the contact process (which is exothermic), the temperature of the system is kept low. Describe what then occurs within the system 2 SO 2(g) + O 2(g) º 2 SO 3(g) Removing energy causes the system to shift to the right. This yields more sulfur trioxide while partially replacing the energy that was removed

34 5. Explain what would happen to the equilibrium in the following system if (a) the pressure were increased and (b) the pressure were decreased 2 SO 2(g) + O 2(g) º 2 SO 3(g) a. There were be a shift in the equilibrium to the right, which decreases the number of gas molecules (2 moles as opposed to 3) b. There were be a shift in the equilibrium to the left, which increases the number of gas molecules (3 moles as opposed to 2) 6. Explain what would happen to the equilibrium in the following system if (a) the pressure were increased and (b) the pressure were decreased a. Nothing b. Nothing H 2(g) + I 2(g) º 2 HI (g)

35 Law of Chemical Equilibrium (494) At equilibrium there is a constant ratio between the concentrations of the products and reactants in any change. Chemists use a mathematical relationship to provide a constant value for a chemical system over a range of concentrations. This constant value is a mathematical expression called the equilibrium constant, K, for the reaction system. The subscript c is often used to show that the equilibrium constant is expressed in terms of molar concentration. E.g., K c. Other subscripts are possible. We ll see those later. The equilibrium constant, K, is a mathematical expression that represents the ratio of product concentration to reactant concentration with each concentration raised to the power of its coefficient. These numbers relate the amounts of reactants to products at equilibrium. That is, for any reaction: aa + bb º cc + dd a = mol of reactant A b = mol of reactant B c = mol of product C d = mol of product D Because the reaction is at equilibrium, there is no net change in the amounts of A, B, C, or D at any given moment. The expression for the equilibrium constant, K (or, the equilibrium law) is: K = [C] c [D] d [A] a [B] b [ ] - indicate that the amounts of substances are in moles per litre K - can also be written K eq

36 ** By convention, chemists always write the concentrations of the products in the numerator (on top) and the concentrations of the reactants in the denominator.

37 Note: S only quantities that have a concentration ([ ]) that can appreciably change can be used to calculate K. The concentrations of solids and liquids do not appreciably change (i.e., they are already constants), therefore these are not included into the equilibrium constant S only concentrations of gases and solutions (i.e., ions... aq) are included in the expression K S each chemical system has its own constant. Thus, (9) S a specific value of K is only accurate when calculated at a specific temperature S the value of K will change with changes in temperature For a given system at equilibrium, the value of the equilibrium constant depends only on temperature. - changing the temperature changes the rate of forward and reverse reactions by differing amounts - this is because the forward and reverse reactions have different activation energies - a reacting mixture at one temperature has an equilibrium constant whose value changes if the mixture is allowed to reach equilibrium at a different temperature This contrasts with adding or removing of a reactant or product. The rate of the reaction will at first change, but the equilibrium will be re-established with the same equilibrium constant. (This is because if you increase the [] of the reactants, you decrease the [] of products proportionally.) The value of the equilibrium constant is also affected by the way the equilibrium is stated. For example, the K of an equilibrium between dinitrogen tetraoxide and nitrogen monoxide is very different from the K of an equilibrium between nitrogen monoxide and dinitrogen tetraoxide. E.g., N 2 O 4(g) º 2 NO 2(g) vs 2 NO 2(g) º N 2 O 4(g) K = [NO 2 ] 2 K = [N 2 O 4 ] [N 2 O 4 ] [NO 2 ] 2 To write an equilibrium constant expression: 1. Write the formulas for aqueous or gaseous products in the numerator of the expression, and use their coefficients as exponents 2. Write the chemical formula of the aqueous or gaseous reactants in the denominator and use their coefficients as exponents.

38 Example 1: Write the equilibrium constant for the decomposition of trinitrogen hexaoxide into nitrogen dioxide. Both compounds are in the gas phase. Balanced equation: N 3 O 6(g) º 3 NO 2(g) Expression for K: K = [NO 2 ] 3 [N 3 O 6 ] Example 2: Write the equilibrium constant for this system: Zn (s) + 2 HCl (aq) º ZnCl 2(aq) + H 2(g) solid: Write the expression for K as before, excluding zinc from the expression because it s a Expression for K: K = [ZnCl 2 ][H 2 ] [HCl] 2

39 Chemistry 3202 K worksheet. 1. Write the expression for the equilibrium constant for each of the following formulae: a. 2 NO (g) + O 2(g) º 2 NO 2(g) b. 2 HBr (g) º H 2(g) + Br 2(g) c. 2 SO 3(g) º 2 SO 2(g) + O 2(g) d. CO 2(g) + H 2(g) º CO (g) + H 2 O (l) 2. Write the balanced chemical equilibrium equation, then the expression for the equilibrium constant for each of the following reactions. Assume all species are gases unless otherwise indicated. a. Hydrogen and carbon disulfide react to form methane and dihydrogen sulfide b. Phosphorus pentachloride decomposes into phosphorus trichloride and chlorine gas c. Nitrogen oxide reacts with atmospheric oxygen to form nitrogen dioxide d. Carbon monoxide and liquid water react to form molecular hydrogen and carbon dioxide