Lesson 7.1.1a Reversible Reactions PDF

Lesson 7.1.1a - Reversible Reactions Specification Points Command words are highlighted. Do you know what they mean? 1. State that no chemical reaction proceeds only in one direction. 2. Define equilibrium as a dynamic system in which concentration of both reactants and products do not change with time. 3. Explain that at equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. 4. Recognize that equilibrium is achieved only in a closed system. 5. Use equilibrium arrow (⇋ to indicate reversibility in a chemical reaction. 6. Illustrate that when the forward and the reverse reaction occur at the same rate, concentrations no longer change and the system, has reached equilibrium. The equilibrium constant, K, is a number based on a particular ratio, at a given T, of product(s) and reactant(s) concentrations 7. Elucidate the value of KC for a generic reaction in the form of S6 - Thermodynamics (Recap) The system = the area of interest (the area of the reaction) The surroundings = everything outside of the system (in theory the entire universe!) Open System: energy and matter can be exchanged between the system and the surroundings. Most reactions take place in an open system. (Examples: a test tube, a conical flask). Closed system: energy can be exchanged between the system and the surroundings but NO matter! (Examples: a sealed test tube, a closed-tight reaction flask). Isolated system: cannot exchange matter or energy between the system and the surroundings. Is there such thing as a truly irreversible reaction? Considering a closed system Strictly speaking, all reactions are equilibrium reactions. Hence, no reaction proceeds fully in one direction. However, in many reactions, the equilibrium mixture nearly fully consists of products. In this situation, the reaction has virtually gone to completion, hence the symbol → instead of ⇌ is used. Example: Combustion is normally considered to be irreversible, but technically there is a very small statistical possibility for the reverse reaction. This reverse reaction will not be observed macroscopically because the product is so thermodynamically stable. Deeper understanding →This is an example of the kinetics of a reaction (reaction rate), being affected by thermodynamics of the system (in this case, the stability of products vs the stability of the reactants) Macroscopic = visible to the naked eye. Dynamic equilibrium → Physical Systems In a reversible reaction, there is a forward reaction and reverse reaction, where both reactions are having a clear impact on the overall reaction mixture. In certain situations, a dynamic equilibrium can be reached (in a closed system) such that the forward and reverse reaction rates are the same. If the forward and reverse reaction rates are the same, there will be no net change in concentration for reactants or products. Mathematically, at equilibrium: rate FORWARD = rate REVERSE Hence: Δ [Reactants] = 0 moldm-3 , Δ [Products] = 0 moldm-3 Volatile liquid: a liquid that readily (easily) vaporizes to transition to the gas phase. Consider the situation where bromine (Br2) is placed in a sealed container at room temperature, hence we have established a closed system necessary for dynamic equilibrium to possible. What is happening to the particles in this system? As bromine is a volatile liquid and has a boiling point close to room temperature (B.P = 58.8oC): Several molecules of Br2 (l) will have enough energy such that some intermolecular forces are broken to enable the change from the liquid state to the gas state (evaporation). The container is sealed (closed system), so the bromine vapour cannot escape. Overtime, the concentration of Br2 (g) will increase. As the number of Br2 (g) particles increases, there is an increased chance of collision between the bromine vapour molecules and the surface of the liquid. When the bromine vapour molecules collide with the liquid surface, these particles lose energy and become a liquid again (condensation). The rate of condensation increases as the concentration of bromine vapour increases. At dynamic equilibrium: Rate of Condensation = Rate of Evaporation. Δ [Br2 (g)] = 0 AND Δ [Br2 (l)] = 0 KEY: The dynamic equilibrium is only reached in this system because a closed system was established. If the system was open, then the Br2 (g) would have escaped before condensation was possible. Dynamic equilibrium → Chemical Systems We will investigate the dissociation reaction of hydrogen iodide gas (HI (g) ). What is happening to the particles in this system? Assuming that we are working with a closed system (hydrogen iodide gas begins in a sealed container: The reaction can be monitored by observing the colour change in the reaction flask given that I2 (g) is the only coloured chemical species. At the beginning of the reaction, there will be an increase in the purple colour, due to the iodine gas being produced. After some time, the colour will not become any more pronounced. KEY IDEA: rate of dissociation HI (g) will be fastest at the beginning of the reaction, [HI (g) ] is highest at t = 0. Even if the colour change stops increasing, the reaction is still in progress and has not stopped. Over time, the reverse reaction rate will start to compete against the forward reaction rate. As the [HI (g) ] decreases, due to the dissociation reaction, [ H2 (g) ] and [ I2 (g) ] will increase. This causes an increase in the rate of the reverse reaction. KEY IDEA: At beginning of the reaction: Rate REVERSE = 0 moldm-3s-1 Quick Question: Rationalise why at the beginning of the reaction, rate REVERSE = 0 moldm-3s-1 After a certain time, the rate of dissociation will have the same rate as the reverse reaction – this is dynamic equilibrium. At dynamic equilibrium: Both the forward and reverse reaction are still happening at dynamic equilibrium, however, there will be no overall change in the concentrations of reactants or products. Δ [HI (g) ] = 0 moldm-3 Δ [ I2 (g) ] = 0 moldm-3 Δ [ H2 (g) ] = 0 moldm-3 The Equilibrium Constant and Equilibrium Constant Expression The size of the equilibrium constant, K, provides us with information about the composition of an equilibrium mixture. Equilibrium mixture = the mixture formed when a chemical system is in a state of dynamic equilibrium. The equilibrium constant is a thermodynamic measurement that is temperature dependent! So a change in temperature will cause the value of K to increase OR decrease. Factors like a catalyst do not affect the value of the equilibrium constant ➔ We will investigate the roles of catalysts in equilibria in 7.1.1e How can the equilibrium constant be defined? The equilibrium constant expression links together: the equilibrium constant, K, [Reactants] and [Products]. Because concentrations are used we say that the equilibrium constant is Kc, where the “c” represents “concentration”. KEY: The stoichiometry of the equation MUST be considered, so your chemical equation needs to be fully balanced, or this won’t work! Consider the following generic equation: The concept of activities and the relation to Kc values Activities are used to account for how there are differences in behaviours between ideal and real solutions. Activities measure effective concentrations and become very important for concentrated solutions. For most solutions, we can model solution activities as the same as concentrations. Activity of a substance = a measure of the presence of a chemical relative to its relative state. Phase Reference State Activity Aqueous 1 moldm-3 Concentration Concentration in moldm-3 / 1 moldm-3 Gaseous 1 atm partial pressure Partial Pressure in atm / 1 atm Liquid Pure Liquid 1 Solid Pure Solid 1 Extra information: pure liquids and solids are modelled as having an activity of 1 because they do not change significantly during reactions, as long as there is enough to reach equilibrium. If we are being really specific: In theory, we should write in the expression for Kc that every concentration stated is an equilibrium concentration. For example, [NH4+] eq instead of just [NH4+] but this does not need to be written as it is assumed (the Kc expression always contains equilibrium concentrations). KEY IDEAS: Activities of solutions and gasses can be approximated as being equal to concentration for the purpose of this course; this assumption only stops working in systems with very high concentration of solutions. In this example, H20 is not included in the equilibrium expression, because activity of liquids = 1. KC is a unitless constant. At dynamic equilibrium: Rate FORWARD = Rate REVERSE By comparing the rate of the forward and reverse reaction, we can define the equilibrium constant in another way using the rate constant, k. k FORWARD / k REVERSE = K Just like before, K is indicating the extent of a reaction (the ratio of products compared to reactants (at equilibrium) as indicated by the rate of forward and reverse reactions. Independent Practice – Equilibrium Constant Expressions Construct an expression for the equilibrium constant (Kc) for the following chemical equilibria, assuming that dynamic equilibrium has been reached in each example. NOTE: Make sure you use square brackets in the equilibrium constant expressions, so it is clear that you are talking about concentrations of each chemical species! Mak e sure to consider the state of the reactant/product, is the chemical species going to be present in the equilibrium constant expression? NOTE: As equilibrium can be approached from either direction, the equilibrium constant expression depends on the form in which the chemical reaction is written. independent Practice – Equilibrium Constant Expressions → Workings Independent Practice – Equilibrium Constant Expressions (Solutions) Independent Practice – Mixed Questions Lesson 7.1.1b - Equilibrium Constants Specification Points Command words are highlighted. Do you know what they mean? 1. State that the magnitude of the equilibrium constant indicates the extent of a reaction at equilibrium and is temperature dependent. 2. Recognize that Kc has no units and that the only thing that changes the value of Kc for a particular reaction, is the temperature. 3. Deduce that the equilibrium constant for a reaction, that is the sum of two or more reactions, is equal to the product of the individual equilibrium constants. 4. Recognize the law of mass action for reactions involving pure solids and liquids (heterogeneous equilibrium). 5. Recognize that pure liquids and pure solids, have an activity of 1 and, therefore, do not feature in the equilibrium equation for KC What does the magnitude of K tell us about the extent of the reaction? RECAP From the last lesson we saw that KC is a thermodynamic property, which is used to describe a reaction which is in a state of dynamic equilibrium. KC is temperature dependent, the value will change as a function of temperature (as temperature changes) Equilibrium concentrations of chemical species are inputted into the KC equilibrium equation. Whenever we are talking about equilibria, we are talking about closed systems where matter is contained within the system. If matter was lost from the system (open system), then no dynamic equilibrium would be reached. The magnitude of K (the equilibrium constant) provides information about the extent of the reaction, in other words to what extent has the reaction gone to completion. Remember: there are different types of equilibrium constant, last lesson we saw KC but we also explore KA in more detail in the course (KA is used to investigate acid-base closed-system equilibria) Going to completion (chemically) = when there is a very high conversion of reactants into products. This information about K (shown above) is equally valid for different sub-types of equilibrium constant, so it works for KC and KA values, in both cases we are talking about equilibria! The value of K tells you about to what extent the reaction has gone to completion when dynamic equilibrium is reached, the value of K does NOT tell you about how quickly it takes for dynamic equilibrium to be reached! KC units and Activities (Reminder) KEY IDEAS (DON’T FORGET! 😊) From last lesson we saw that KC is unitless because the equation for KC is derived originally from the activities of all the chemical species in the chemical equilibrium. In fact, we can extent this rule to be even more broad by saying that all equilibrium constants are unitless because all equilibrium constants are derived originally from the activities of chemical species. Don’t forget, pure liquids and pure solids do not feature in the equation for KC (if you need a reminder of the theory behind this, go back to 7.1.1a) Extent of Reaction - Independent Practice KC values are determined for the following chemical systems that are in a state of dynamic equilibrium. For each chemical system, comment on the position of equilibrium with reference to the provided KC value Extent of Reaction – Application-Style Question Combining Equilibrium Constants Consider the following two-step chemical reaction which results in the production of NO2. NO2 is responsible for giving urban smog its distinctive brown colour. The reaction takes place by a two-step pathway, with the intermediate NO (g) initially being formed before the second reaction takes place to convert the intermediate into the final product, NO2 (g). Both stages of the reaction pathway take place at a constant temperature of 373 K. For the following reactions we will use some abbreviations: Let KC for Reaction 1 = K1, Let Kc for Reaction 2 = K2 Let KC for Reaction 3 = K3 The reaction reaches dynamic equilibrium, and the equilibrium constants for Reaction 1 and Reaction 2 are determined. The results are shown below: K1 = 2.0 x 10-25 K2 = 6.4 x 109 1.) By summing reactions 1 and 2, describe the overall reaction between N2 and O2, label this overall reaction as reaction 3. 2.) Construct three equilibrium constant expressions to describe reaction 1, reaction 2 and reaction 3 (K1, K2 and K3) 3.) Deduce the numerical value of the equilibrium constant Kc (REACTION 3). Question 1 → Worked Example (Teacher-Led) Question 2 → Independent Work Question 2 → Independent Work (Solutions) Question 3 → Deduction Combining Equilibrium Constants - Mini Application Question If there is an overall reaction that can be described by summing together two or more reactions together, the equilibrium constant for the overall reaction is equal to the product of the equilibrium constants for the individual reactions. Considering a three-step reaction where: K1 = equilibrium constant (reaction 1) K2 = equilibrium constant (reaction 2) K3 = equilibrium constant (reaction 3) K4 = equilibrium constant (reaction 4) K5 = equilibrium constant (overall reaction) In this four-step reaction, when reactions 1,2, 3 and 4 are summed together, they are equal to the overall reaction. Quick Question: Construct an equation with K5 as the subject to show the connection between K5 and the other equilibrium cosntants Combining Equilibrium Constants - Independent Practice Combining Equilibrium Constants - Independent Practice (Solutions) Given: two balanced equilibrium equations, values of K𝐾, and an equilibrium equation for the overall reaction. Asked for: equilibrium constant for the overall reaction. Strategy: Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of K𝐾 for that equation. Calculate K𝐾 for the overall equation by multiplying the equilibrium constants for the individual equations. Solution: The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2: The Law of Mass Action - Quick Explainer For a system at equilibrium, the law of mass action relates the equilibrium constant, K, to the ratio of the equilibrium concentrations of the products and reactants raised to their powers as described in the equilibrium constant expression. As we have seen before, pure liquids and pure solids have an activity of 1 and do NOT feature in the equilibrium constant expression. Why is this the correct answer? Lesson 7.1.1bi - Inverse K Relationships Specification Points Command words are highlighted. Do you know what they mean? 1. State that the magnitude of the equilibrium constant indicates the extent of a reaction at equilibrium and is temperature dependent. 2. Recognize that Kc has no units and that the only thing that changes the value of Kc for a particular reaction, is the temperature. 3. Deduce that the equilibrium constant for a reaction, that is the sum of two or more reactions, is equal to the product of the individual equilibrium constants. 4. Recognize the law of mass action for reactions involving pure solids and liquids (heterogeneous equilibrium). 5. Recognize that pure liquids and pure solids, have an activity of 1 and, therefore, do not feature in the equilibrium equation for KC Approaching Equilibrium from the reverse direction Because equilibrium can be approached from either direction in a chemical reaction, the equilibrium constant expression and thus the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. ➔ Consider that the reaction (shown below) represents the forward rate of reaction in a generic chemical reaction. ➔ We can rewrite the equation to obtain the reverse equation and obtain the corresponding equilibrium constant for the reverse equation (K’). ➔ The two equilibrium constants are now inverses of each other. K’ = 1/K Where K’ = the equilibrium constant for the reverse reaction. Where K = the equilibrium constant for the forward reaction. Lesson 7.1.1c - Direction of a Reaction Specification Points Command words are highlighted. Do you know what they mean? 1. Elucidate that the reaction quotient of a reaction, Q, is the ratio of the concentrations of products and reactants at one moment in time. 2. Construct an expression to describe the reaction quotient Q. 3. Derive the following relationships between Q and KC: If Q = KC, reaction is at equilibrium If Q < Kc, reaction proceeds in favour of products If Q > Kc, reaction proceeds in favour of reactants. Introduction → The Reaction Quotient The reaction quotient (Q) measures the relative concentrations of products and reactants present during a reaction at a particular point in time. The most important reason why reaction quotients are used is to determine the direction in which a reaction will proceed. Why can’t we just use Kc? Kc is used to describe the relative concentrations of products and reactants ONLY in the state of dynamic equilibrium. Q can be used to describe the relative concentrations in the system at ANY time chosen. KEY IDEAS: Just like with Kc we model activities of all the chemical species as being approximately the same as the concentrations of the chemical species in the system. Just like with Kc, pure liquids and pure solids are NOT included in the expression to describe Q both pure solids and pure liquids have an activity of 1. The value of Q is not a fixed value as it can be measured at any time but the value of K is constant at a given temperature. Creating an expression to describe Q Striving to dynamic equilibrium All chemical systems that are reversible are striving towards equilibrium because this is the thermodynamically preferred situation. Consider this theoretical reversible system A+B⇌C The value of Q is determined at t = x and compared to the value of Kc; the temperature remains constant. Kc = 20 Q = 10 What does the system do in this situation? Kc = 20 Q = 10 Q < KC in this case (at t= x): but the system is always striving toward equilibrium concentrations for reactants and products. Because Q < Kc: : this tells us that [C]x is less than [C]eq. The system will strive to increase [C] because at t = x, [C] is not as high as [C]eq. The system will therefore respond by shift to the right and producing more products! Comparing and understanding → Q vs K values (Worked Example) A comparison of 𝑄 and K values indicates which way the reaction shifts, and which side of the reaction is favored: We are going to consider the chemical system shown below: The value of Q is determined at t = x and compared to the value of Kc; the temperature remains constant. Q = 30 KC = 10 Using this information, would we predict for the system to shift to produce more products or shift to produce more reactants? Q > KC in this case (at t= x): but the system is always striving toward equilibrium concentrations for reactants and products. Because Q > Kc: : this tells us that [CO2][H2] > [CO2]eq [H2]eq The system will strive to decrease [CO2] and [H2] so that equilibrium concentrations are reached. The system will therefore respond to this change and shift to produce more reactants! Q vs K values → Summary If Q = KC, reaction is at equilibrium If Q < Kc, reaction proceeds in favour of products If Q > Kc, reaction proceeds in favour of reactants. NOTE: You have to be able to explain these three ideas, it is not sufficient to memorize it only because you can be provided with questions that involve rationalizing the relationships. Independent Practice - The Reaction Quotient The equilibrium constant (KC) for the reaction below is 5.1 x 10-2 at 298 K. COI2 (g) ⇌ CO (g) + I2 (g) Deduce whether the following reaction mixture concentrations represent a reaction at equilibrium or not. For the reactions that are not at equilibrium indicate the direction that the reaction will shift in in order to reach equilibrium. HINT: Start by constructing an expression to describe the reaction quotient for this reaction. Independent Practice - The Reaction Quotient (Workings) Independent Practice - The Reaction Quotient (Solutions) Reaction 1: in this mixture Q >> K, so Q has to decrease to reach K, this means the reaction must shift to the left, favouring reactants. Reaction 2: in this mixture Q = K, hence the system is in a state of dynamic equilibrium and there will be no change to the concentration of reactants or products. Reaction 3 in this mixture Q < K, so Q has to increase to reach K, this means the reaction must shift to the left, favouring products. Lesson 7.1.1d - Reaction Composition at Equilibrium Specification Points Command words are highlighted. Do you know what they mean? 1. Produce an I.C.E. (I: initial, C: change, E: equilibrium) table and calculate concentration of products/reactants or the K for a reaction at equilibrium. What is an I.C.E table? An ICE (Initial, Change, Equilibrium) table is a method used (matrix approach) that simplifies the calculations in reversible equilibrium reactions. I stands for initial concentration. This row contains the initial concentrations of products and reactants. C stands for the change in concentration. This is the concentration change required for the reaction to reach equilibrium. It is the difference between the equilibrium and initial rows. In this row, there are positive (+) or negative (-) signs used; this is because this row represents an increase or decrease (or no change) in concentration. E is for the concentration when the reaction is at equilibrium. This is the summation of the initial and change rows. Once this row is completed, its contents can be substituted into the equilibrium constant equation to solve for KC Implementation of an I.C.E. Table - Worked Example KEY INFO: It is essential before doing any work with an I.C.E table that the reaction scheme is fully balance, otherwise the table will provide incorrect results! A sample consisting of 0.500 mol of X is placed into a system with a volume of 0.750 liters. At dynamic equilibrium, the amount of X is known to be 0.350 mol. Use an I.C.E table to determine the KC for this chemical system. What does the question tell us? Come Moles Klane Initial amount of X = 0.500 mol Equilibrium amount of X = 0.350 mol The question provides information about X in terms of moles, this means eventually we will need to do a conversion moles → concentration in order to calculate KC The questions says that the system starts with X. Hence: Amount of Y = 0 mol Amount of Z = 0 mol Chemical 2X 3Y 4Z Species Initial Amount / 0.500 0.000 0.000 mol Change in -0.150 ? ? Amount / mol Equilibrium 0.350 ? ? Amount /mol Chemical 2X 3Y 4Z Species Initial Amount / 0.500 0.000 0.000 mol Change in -0.150 +0.225 +0.300 Amount / mol Equilibrium 0.350 0.225 0.300 Amount /mol Key Information The biggest challenge with multi-step calculations involving Kc values and equilibrium concentrations is that there is a massive range of potential questions and your Maths skills have to be secure! You are strongly advised to have completed all the lessons in the “Maths for Chemistry” Unit. It is also important to understand whether your answer makes “Chemistry sense”. For example: if you are dealing with concentrations or moles of a substance, the value cannot be negative, meaning that one of the solutions from the quadratic formula can be discarded! Implementation of an I.C.E. Table - Independent Practice 1 Q: A chemical system is established between a generic acid (HA) and its conjugate base (A-). Determine the equilibrium concentration of A- for this dissociation reaction. You are advised to construct an I.C.E. table, however this is NOT obligatory. [ HA (aq) ] INITIAL = 0.260 M and Ka = 2.4 x 10-3 INFO: The Ka constant is used to describe the equilibrium concentrations in an acid-base chemical system. The equilibrium constant works in the same way as Kc and the equilibrium constant expression is formed in the same way. You will learn about Ka in detail during Unit 7.2. REMEMBER: Pay attention to the states of the products and reactants, do they need to be included in the equilibrium expression for Ka? HINT: Constructing an equilibrium expression for Ka will help 😉 Implementation of an I.C.E. Table - Independent Practice 1 (Workings) CATCH O KA CHA 2 1 10 3 A HO 0.2 or Implementation of an I.C.E. Table - Independent Practice 1 (Workings) Implementation of an I.C.E. Table - Independent Practice 1 (Workings) Implementation of an I.C.E. Table - Independent Practice 1 (Solutions) Implementation of an I.C.E. Table - Independent Practice 2 There are several different types of chemical system which can form chemical equilibria. It is also possible to form a chemical equilibrium between stereochemical isomers, an example is shown below where the position of the high-priority groups is either together (Z isomer) or separate (E isomer). If you need a reminder about E/Z labelling, go back to S6.6 (Organic Chemistry). (Z)-But-2-enenitrile (aq) (E)-But-2-enenitrile (aq) Q: At 303°C, a 2.0 M sample of (Z)-But-2-enenitrile at is allowed to come to equilibrium according to the balanced chemical equation above. The equilibrium constant, Kc, for this reaction at 303°C is 1.40. What are the concentrations of (Z)-But-2-enenitrile and (E)-But-2-enenitrile after equilibrium is established? You may use the following abbreviations in this question only: (Z)-But-2-enenitrile (aq) = ZBUT2 (aq) (E)-But-2-enenitrile (aq) = EBUT2 (aq) Implementation of an I.C.E. Table - Independent Practice 2 (Workings) Kc 1.4 2 male ZBut21nl EBut2caq I Initialconcentration 2 0 Change in concentration 1.12 1.12 Find concentration 0.83 1.17 EBut2 Kc 1.4 zBut2 1 4 14 1 4 2 x x 28 1 4x x 2 8 x 1 44 2 8 2 4x 8 2.1667 2 Implementation of an I.C.E. Table - Independent Practice 2 (Workings) Implementation of an I.C.E. Table - Independent Practice 2 (Workings) Implementation of an I.C.E. Table - Independent Practice 2 (Solutions) Additional Questions - Independent Practice 1 NH cags H2O NH NH initial 0.350M NHTeq 0.325M Ke 1.8 10 6 Ke 11 NHI NH4 initial come µ 0.350 0.0 change 0.025 0.025 inane n hindane in 0.325 0.025 Kc 8 0.0714 Additional Questions - Independent Practice (Solutions) Lesson 7.1.1e - Le Châtelier Principle Specification Points Command words are highlighted. Do you know what they mean? 1. State that when a system at equilibrium is disturbed, the system will shift its equilibrium position so as to counteract the effect of the disturbance. 2. Predict the effect of the following changes on a reaction at equilibrium: Concentration of reactant(s) or product(s) Temperature changes (endothermic and exothermic reactions) Addition of a catalyst (no effect on equilibrium) Le Châtelier Principle - Increasing Concentration Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. Consider the situation above: a system is in a state of dynamic equilibrium. At t = x, [ A] is increased. A value of Qc is determined at t= x. At t = x → Qc < KC (Go back to Lesson 7.1.1d if you don’t understand this conclusion) In this case, the equilibrium position will move to the right so that the concentration of A decreases again by reacting it with B to form more C and D. This will continue until [A] returns to [A]eq. In a practical sense, this is a useful way of converting the maximum possible amount of B into C and D; this is advantageous if, for example, B is a relatively expensive material whereas A is cheap and plentiful. Le Châtelier Principle - Decreasing Concentration Consider the situation above: a system is in a state of dynamic equilibrium. At t = x, [A] is decreased. A value of Qc is determined at t= x. At t = x → Qc > KC (Go back to Lesson 7.1.1d if you don’t understand this conclusion) According to the Le Châtelier principle, the position of equilibrium will move so that the [A] increases. More C and D will react to replace the A that has been removed. In this case, the equilibrium position will move to the left so that the concentration of A increases This will continue until [A] returns to [A]eq. This is essentially what happens if one of the products is removed as soon as it is formed. If, C was removed in this way, the position of equilibrium would move to the right to replace it. If it is continually removed, the equilibrium position shifts further and further to the right, effectively creating a one-way, irreversible reaction. Le Châtelier Principle - Increasing Temperature Key Point: the equilibrium constant Kc is a thermodynamic measure that is temperature- dependent! So, changing the temperature in the reaction will change the value of Kc and hence change the equilibrium concentrations for reactants and products. You have to know the enthalpy change for the reaction to determine how temperature will affect the equilibrium. Consider the situation above: a system is in a state of dynamic equilibrium. Quick Think: Is the forward reaction exothermic or endothermic? Quick Think: Is the reverse reaction exothermic or endothermic? In this reaction, 250 kJ of energy is released into the surroundings (indicated by the negative sign) when 1 mole of A reacts completely with 2 moles of B. For reversible reactions: the enthalpy value is always given as if the reaction was one-way in the forward direction. The back reaction (the conversion of C and D into A and B) would be endothermic, absorbing the same amount of heat. Le Châtelier Principle - Increasing Temperature If the temperature is increased, then the position of equilibrium will move so that the temperature is reduced again. Example: imagine the system is at equilibrium at 300°C, and the temperature is increased 500°C. To cool down, it needs to absorb the extra heat added. In this case, the reverse reaction is endothermic and will, therefore absorb the extra heat energy. The position of equilibrium therefore moves to the left and favours the production of reactants. Increasing the temperature in this equilibrium is, therefore, a poor way of maximizing the amount of products made. Le Châtelier Principle - Decreasing Temperature If the temperature is decreased, the position of equilibrium will move in such a way that the temperature increases again. Example: suppose the system is in equilibrium at 500°C and the temperature is reduced to 400°C. The reaction will tend to heat itself up again to return to the original temperature by favouring the exothermic reaction. The position of equilibrium will move to the right and favour the production of products. Catalysts - Effect on Position of Equilibrium Adding a catalyst makes absolutely no difference to the position of equilibrium, and Le Châtelier's principle does not apply. This is because a catalyst speeds up the forward and back reaction to the same extent and adding a catalyst does not affect the relative rates of the two reactions, it cannot affect the position of equilibrium. Catalysts are very useful to increase the rate at which a reaction which dynamic equilibrium. Without a catalyst, some reactions would take years to reach equilibrium!! The energy profile diagram (above) shows how the activation of both forward and reverse reactions are decreased by the same amount in a reversible reaction Le Châtelier Principle - Summary Le Châtelier Principle - Summary Questions 1 Le Châtelier Principle - Summary Questions 1 → Solutions Le Châtelier Principle - Summary Questions 2 NOTE: For question 15, please ignore subsection a and b Le Châtelier Principle - Summary Questions 2 → Solutions YOU DID IT! :D Lesson 7.2.1 – Brønsted-Lowry Acids and Bases Specification Points Command words are highlighted. 1. Define an acid as a proton donor and a base as a proton acceptor: 2. Identify an acid-base reaction as one in which a proton is transferred from an acid to a base. 3. Recognise that acids with more than one ionizable proton are polyprotic acids. 4. Write the acid/base pair as: 5. Identify an amphoteric substance and define it as a species which can act as both a Brønsted-Lowry acid and base. 6. Illustrate amphoteric behaviour with two equations taking water, hydrogen carbonate, or hydrogen phosphate as example. 7. Write the dissociation of water into ions. 8. Define the self-ionization constant (Kw) and explain the temperature-dependence of the Kw, given that it is an equilibrium constant. Brønsted Acids and Bases → Definitions In 1923, chemists Johannes Brønsted and Martin Lowry independently developed definitions of acids and bases based on compounds abilities to either donate or accept protons (H+ ions) Bronstead Acid = a chemical species that can act as a proton (H+) donor. Bronstead Base = a chemical species that act as a proton (H+) acceptor The reaction shown below is an example of an acid-base reaction ➔ “HA” notation can be used to describe a generic acid. ➔ Consider the theoretical reaction between the acid (HA) and water. ➔ Water is acting as a Bronstead base because it accepts a proton (H+) from HA Labelling: Identify the conjugate base of HA (aq) Independent Practice → Identification Bronstead Acid/Base Independent Practice → Identification Bronstead Acid/Base (Solutions) Acid/Base Reactions → Considering forward and reverse reaction Proton donation doesn’t happen in isolation. If there is proton donation, there has to be a chemical species that acts as the proton acceptor. Consider the acid-base reaction between a generic acid (HA) and a base (B) ➔ In the forward reaction: we see HA acting as an acid (proton donor) and B acts as a base (proton acceptor) BUT ➔ In the reverse reaction: we see A- acting as a base (proton acceptor) and BH+ acts as a base (proton donor) ➔ Acids react to form bases and bases react to form acids. ➔ The acid-base pairs that are related are known as conjugate acid-base pairs. Independent Practice → Conjugate acid/base pairs identification Polyatomic ions and Basicity Polyatomic ions normally act as Bronstead Bases and will act a proton to form the conjugate acid. What is an amphiprotic species? Some chemical species can act as both Bronstead acid AND a Bronstead base (depending on the exact situation). Example → Water is an example amphiprotic species ➔ When reacting with a carboxylic acid, water can act as a base. ➔ However, when reacting with ammonia, water can act like an acid. ➔ Amphiprotic species will: have a lone pair of electrons can release H+ ions. Application Question → Amphiprotic Behaviour Application Question → Amphiprotic Behaviour a.) Construct equations to show how the hydrogen phosphate ion, [HPO4]2-, can act as Bronsted Acid. b.) Construct equations to show how the hydrogen phosphate ion, [HPO4]2-, can act as Bronsted Acid. Dissociation of water ➔ The dissociation of water relies on the amphoteric nature of water, being able to either act as a proton acceptor (base) OR a proton donor (acid). ➔ About one water molecule in half a billion dissociates into an OH- ion by losing a proton to another water molecule. The molecule that receives a proton becomes a H3O+ ion ➔ The reaction is reversible; the conjugate acid (H3O+) and the conjugate base (OH-) react to re-form the two water molecules. ➔ The dissociation of water is an equilibrium reaction in which one water molecule donates its proton to another water molecule. ➔ Because water causing an ionization of itself (one water molecule ionizes another), we describe this as the self-ionization of water. ➔ There is a specific equilibrium constant to define the self-ionization of water (Kw) ➔ Kw = the self-ionization constant The Water Dissociation Constant (Kw) The dissociation of water is an equilibrium reaction. ➔ At equilibrium: the rate of the forward reaction is equal to the rate of the reverse reaction and the concentration of the reactants and products do not change. Q: Construct an expression to describe the value of Kw. The Water Dissociation Constant (Kw) Remember: Pure liquids do not feature in an expression describing an equilibrium constant! The expression to describe Kw is as follows: Q: At T = 298K, the value of Kw is measured as being 1 x 10-14 Using this data, comment on the extent of self-ionization at 298K. The Water Dissociation Constant (Kw) As you saw from the previous section, the extent of self-ionization is very limited. The position of equilibrium very slightly favours the products, but the values [H3O]+ and [OH]- will be very small. ➔ The product of [H3O]+ and [OH]- is equal to Kw , as derived earlier. ➔ At T = 298K, Kw = 1 x 10-14 Q: Calculate the concentration of [H3O]+ and [OH]- present in pure water at 298K. HINT: Think about the reaction stoichiometry, what is the ratio between [H3O]+ and [OH]- ions? The Water Dissociation Constant (Kw) ➔ If one water molecule self-ionizes, this will lead to one molecule of [H3O]+ and one molecule of OH-. ➔ Hence, [H3O]+ = [OH]- (take some time to think about this, if needed!) ➔ Hence, at 298K [H3O]+ = [OH] - = 10-7 Temperature-Dependance ➔ Kw is an equilibrium constant, just like Ka or Kc. ➔ Kw is temperature dependent and will change as temperature changes! Q: In a sample of pure water, the value of Kw at 350K is measured as being 1.0 x 10-14 a.) Comment on the extent of self-ionization for water, at 350K, compared to the self- ionization at 298K. b.) Calculate the value of [H+], at 350K, for a sample of pure water; using the data provided. Lesson 7.2.2 – pH Definitions Specification Points Command words are highlighted. 1. Use the definition of pH of an aqueous solution as 2. Use the definition of pOH of an aqueous solution as: 3. Explain that pH + pOH = 14 Definitions → The Concept of pH ➔ Chemical species that can act as acids liberate H3O+ ions (hydronium ions) into solution. ➔ H+ ions don’t exist in aqueous situations because the more stable hydronium ion forms. ➔ When HCl gas dissolves in water, there is a release of hydronium ions. Label: Acid, Base, Conjugate Acid, Conjugate Base ➔ Expressing the acidity of a solution by expressing [H3O+(aq)] is difficult because the concentrations are very small. ➔ Danish scientist Søren Sørensen (1868 - 1939) proposed an easier system for indicating [H3O+(aq)] called the pH scale. ➔ The letters pH stand for the power of the hydrogen ion. ➔ pH is temperature dependent because changing the temperature affects the position of equilibrium and therefore changes [H3O+(aq)] and [OH-(aq)] pH Equation → Rearrangements The equation can be rearranged to make [H3O+(aq)] as the subject. Equation 1 is provided in the formula booklet. You are advised to know key equation 2, or be comfortable deriving it. Key Equation 1: pH = -log [H3O+] Key Equation 2: 10-pH = [H3O+] ➔ Note that a logarithmic scale is a convenient way of bringing numbers of differing magnitudes on to the same scale. ➔ The pH scale does not have an upper or lower limit, and even though many solutions fall between the 0 and 14 range, it is possible to have values below 0 and above 14. Independent Practice → Le Chatelier Principle and Self-Ionization 1. Construct an expression to describe the self-ionization constant (Kw) 2. Calculate the value of [OH-(aq)] at T = 50oC 3. Usng the data provided, and with reference to Le Chatelier’s principle: Rationalize whether the formation of self-ionization of water is an endothermic or exothermic process. 4. Pure water, has a pH = 7. Determine the value of [H3O+(aq)] 5. In a sample of pure water: at T = 400K, the self-ionization constant (Kw) has a value of 2.92 x 10-12. Calculate the pH of water at T = 400K Definitions → The Concept of pOH (Aqueous Solutions) Looking back at the self-ionization of water, the forward reaction releases H3O+ ions AND OH- ions. The concentration of OH- ions can be expressed logarithmically by using the pOH concept. Derivation → Prove that pH + pOH = 14 at T = 298K, given that Kw = 1.00 x 10-14 Hint: Construct an expression for Kw for the self-ionization of water. Independent Practice → Le Chatelier Principle and Self-Ionization Determine the concentration of OH- (aq) ions when the pH of a solution is 4.42. Lesson 7.2.3a - Strength of Acids and Bases Specification Points Command words are highlighted. 1. Define strength as a measure of the extent to which an acid can donate a H+, or the extent to which a base can accept a H+. Use forward arrow for strong acid/base and equilibrium arrow for weak acid/base. 2. Elucidate the definition of ionisation constants (Ka and Kb) of weak acids and bases. 3. Use the mathematical relationship between Ka, Kb and Kw. Discuss the strength of an acid/base, in relation to the strength of its conjugate base/acid. 4. Rank the strength of acids and bases using their Ka and Kb value (or pKa and pKb). 5. Rank the strength of acids and bases, comparing reaction rates of acid/base solutions of the same concentration. What does “acid strength” mean? Acid strength = a measure of the extent to which an acid can donate a H+. A strong acid will almost dissociate fully in aqueous solutions and donates a H+ ion easily. Strong Acids → HCl example HCl is a strong acid, the position of equilibrium is so far to the right that we can just model this reaction as being an irreversible reaction. State symbols: add in the lesson. Weak Acids → CH3COOH (Ethanoic Acid) example Ethanoic acid is a weak acid and does not fully dissociate, we have to model the dissociation of ethanoic acid as being an equilibrium reaction. State symbols: add in the lesson. Weak vs Strong Bases → Examples Strong Bases → G1 Hydroxides Example ➔ Strong bases will almost dissociate fully in aqueous solutions. ➔ G1 metal hydroxides are commonly used strong bases → KOH, LiOH, NaOH. ➔ The position of equilibrium is so far to the right; we can model as an irreversible reaction. Weak Bases → ➔ Strong bases will partially dissociate in aqueous solutions. ➔ Ammonia and amines are commonly used weak bases. ➔ The position of the equilibrium is not fully to either side = an equilibrium reaction. Worked Example → Interpreting the value of Kb Base strength = a measure of the extent to which a base accepts a proton (H+ ) Acid strength = a measure of the extent to which an acid can donate a H+. ➔ You are provided with the theoretical reaction between a base (B) and H2O. As you have seen previously, H2O is an amphoteric species (do you remember the word?!) ➔ In this situation, H2O is acting as an acid. ➔ Kb is an equilibrium constant which measures how easily a base is protonated. ➔ Just like all equilibrium constants, Kb is also temperature dependant and unitless. Δ = -70 kJ/mol 1.) Construct an expression to describe the base dissociation constant (Kb); the concept is exactly the same as other equilibrium constants that we have seen. 2.) At a given temperature (Tx), the value of Kb for this reaction is determined to be 1 x 108. a.) Comment on the position of equilibrium given this value of Kb b.) State whether B is acting as a strong base or a weak base. c.) Justify your response from Q2b. 3.) The temperature of the reaction mixture is increased. Comment on how the position of equilibrium changes after this temperature change. Worked Example → Interpreting the value of Ka ➔ You are provided with the reaction between a benzoic acid (C6H5COOH) and H2O. H2O is acting as the base in this equation. 1. Label the conjugate acid and conjugate base for this reaction. 2. Construct an equilibrium expression to describe the value of the acid dissociation constant (Ka). 3. Write the skeletal formula for benzoic acid and highlight the hydrogen atom that will be deprotonated. 4. Justify the usage of an equilibrium arrow for this reaction. Derivation → Relationship between Ka and Kb ➔ There is a simple relationship between the magnitude of Ka for an acid and Kb for its conjugate base. ➔ Consider the dissociation of the weak acid, hydrocyanic acid (HCN), in water to produce an acidic solution and the reaction of CN− with water to produce a basic solution: Equation 1 Equation 2 The equilibrium dissociation constant (Ka) → Constructing an overall equation (KREACTION) ➔ As we saw in equilibrium KREACTION is calculated by multiplying the individual equilibrium constants of each reaction. ➔ Perhaps you noticed that the overall equation in this reaction is a variation of the self- ionization of water. Derivation → Relationship between Ka and Kb Lesson 7.2.3b - Strength of Acids and Bases Specification Points Command words are highlighted. 1. Rank the strength of acids and bases using their Ka and Kb value (or pKa and pKb). 2. Rank the strength of acids and bases, comparing reaction rates of acid/base solutions of the same concentration. 3. Explain that polyprotic acids have different Ka for each dissociation step and outline that for all polyprotic acids: Ka1 >> Ka2 >> Ka3 4. Calculate the pH of solutions of strong acids and strong bases. 5. Calculate the pH of aqueous solutions of weak monoprotic acids and bases working out [H3O+(aq)] from the expression for Ka, and [OH-(aq)] from the expression for Kb. 6. Compare pH and pKa to find the predominant species at different pH values (predominance diagrams). 7. Determine the position of equilibrium in an acid-base reaction by looking at Ka or pKa values of the species involved in the reaction (already discussed in previous lessons). 8. Calculate the equilibrium constant of an acid-base reaction using pKa values, (i.e. the pKa value of the acid and the pKa value of the conjugate acid of the base in the reaction). Bringing together key knowledge so far…. RECAP! ➔ The pH scale is a logarithmic way of representing the [H3O+]. ➔ The pH scale is used in aqueous situations. ➔ Indicators can be provide a colourful representation of the pH of solution ➔ The pH and pOH scales are interrelated! ➔ This leads to the following key equations from the Chemistry formula booklet. Key Equations 1 and 2 → pH = - log [H30+] AND pOH = -log [OH-] Key Equation 3 → pKw = pH + pOH RECAP → pOH relationships at 298K Ranking Strengths of Acids and Bases → pKa vs pKb Acid Strengths ➔ In the previous lesson we saw how the acid dissociation constant (Ka) is related to the strength of the acid. We investigated the example below with the weak acid ethanoic acid in the previous lesson. ➔ Increasing Ka value means that the equilibrium for the acid dissociation is positioned increasingly to the right meaning an increase in [H3O+]. ➔ Remember pKa and Ka are inversely related; increasing Ka will cause a decrease of the pKa value. pKa = - log (Ka) Base Strengths ➔ In the previous lesson we saw how the base ionization constant (Kb) describes how easily a base is protonated. An example is shown below with the weak base, ethylamine. As Kb increases, this indicates an increase in the concentration of the conjugate acid (the protonated ethylamine). ➔ Remember pKb and Kb are inversely related; increasing Kb will cause a decrease of the pKa value. pKb = - log (Kb Independent Practice All data has been collected at T = 298K 1. Rank the following weak acids in order of strength; start with the least strong acid and end with the strongest acid. 2. Rank the following weak bases in order of strength; start with the least strong acid and end with the strongest acid. 3. Construct a balanced equation to represent the protonation of aniline by water. 4. Construct a balanced equation to show the deprotonation of hydrofluoric acid Strong Acids vs Weak Acids → pKa All data has been collected at T = 298K You can see from the table below that the pKa of strong acids is considerably lower than those of weak acids. Strong Bases vs Weak Bases → pKb Comparing strength of acid by monitoring reaction rate Experiment 1: 10 mL of a 1 molL-1 NaOH solution reacts with 10 mL of a 1 molL-1 strong acid (HCl), the following results are generated of reaction rate against time. Reaction rate against time → y axis = reaction rate, x axis = time In-lesson: finish axis-labels and graph title. Experiment 2: 10mL of a 1 molL-1 NaOH solution reacts with 10 mL of a 1 molL-1 solution of the weak acid (CH3COOH). Independent Practice → Acid Strength and Reaction Rate Formula from S5: Concentration = Moles / Volume ➔ To fairly compare the relative strength of the two acids, the same concentration of solution must be used for both acids being compared. 1. Calculate the number of moles of HCl that have reacted in Experiment 1. 2. Describe how the speed of the reaction differs for the two reactions 3. Construct an equation to show the reaction between NaOH (aq) and HCl (aq). 4. Construct an equation to show the reaction between the weak acid CH3COOH and NaOH. 5. Rationalise why the rate of reaction is faster in one of the experiments; refer to the reaction equations. ➔ The same logic used in this question can be applied with bases. Reaction rates can be used to compare the strength of acids of the same concentration. Polyprotic Acids and Acid Strength The acids we have seen so far are able to release one proton (H+) per molecule; these acids are known as monoprotic. Whether an acid is monoprotic or not as nothing to do acid strength! Examples of monoprotic acids: HCl, HNO3, CH3COOH Polyprotic acids: they can lose several protons (H+) per molecule. They can be further categorized into diprotic acids and triprotic acids, those which can donate two and three protons, respectively. Examples of polyprotic acids: H2S, H2SO4 Polyprotic Acids and Acid Strength Consider the example of H2S with two ionizable hydrogen atoms. Hence, we can consider two reactions. Reaction 1: Deprotonation of H2S by water Ka1 = 9.1×10-8 Reaction 2: Deprotonation of HS- by water Ka2 = 1.3×10−13 1. Comment on how the Ka value changes with each successive deprotonation. 2. Construct an equation to represent the overall reaction taking place. 3. Calculate the value of Ka for the overall reaction. Polyprotic Acids and Acid Strength ➔ For each successive deprotonation of a polyprotic acid the Ka for the acid will decrease. ➔ Hence, pKA is increasing for each successive deprotonation. ➔ Both trends can be seen the in the data below. Data collected at 298K. pH of solutions → Strong Acids Key relationship (provided in the formula booklet): pKw = pH + pOH At T = 298K, pKw = 14 Worked Example → Strong Acid A solution of hydrochloric acid, HCl (aq), has a concentration of 1 x10-3 molL-1 1. Construct an equation to show the dissociation of HCl (aq) and the formation of hydronium H30+(aq) ions. You can assume a full ionization takes place. 2. Calculate the pH of the hydrochloric acid solution. pH of solutions → Strong Bases (Questions) Key relationship (provided in the formula booklet): pKw = pH + pOH At T = 298K, pKw = 14 Worked Example → Strong Base The strong base Ca(OH)2 can be dissolved in water to form a solution. 1. Construct an equation to show complete ionization of Ca(OH)2. 2. Calculate the pH of a solution containing 1.2345×10−4 M of Ca(OH)2. pH of solutions → Strong Bases (Solutions) Key relationship (provided in the formula booklet): pKw = pH + pOH At T = 298K, pKw = 14 1M = 1 molL-1 = 1 mol/L 1. Construct an equation to show complete ionization of Ca(OH)2. 2. Calculate the pH of a solution containing 1.2345×10−4 M of Ca(OH)2. *Add state symbols into equation. Additional Exercises (Showbie) → pH of solution (Strong Acid/Strong Base Qs and Answers) Worked Example → pH of solutions → Weak Acids Calculate the pH of a 2.00 molL-1 solution of nitrous acid (HNO2). The Ka for nitrous acid is 4.5×10−4 To start: write an equation to show the deprotonation of nitrous acid by water. Independent Practice → pH of solutions → Weak Acids You will investigate the following acid-base pair: CH3COOH / CH3COO- CH3COOH is dissolved in water and a solution forms of concentration = 2 x 10-3 molL-1 pH of the solution formed when ethanoic acid dissolves in water is 2.3. This experiment takes place at 298K. 1. Construct an equation to show the deprotonation of ethanoic acid by water. 2. Calculate the pKa for ethanoic acid at 298K. Worked Example → pH of solutions → Weak Bases Independent Question → Weak Bases and pH Predominant Species and Predominance Diagrams Predominance diagrams are a graphical way to visualize how the distribution of species changes with pH. Predominance Rules ➔ When pH < pKa, the protonated form HA will be the predominant species, the concentration of H30+ is high. ➔ When pH < pKa, the deprotonated form A- (the conjugate base) will be the predominant species in the solution. ➔ When pH = pKa, there is no predominant species in the solution. Worked Example → Predominance Diagrams Consider the ionization of ethanoic acid (Label in Class) ➔ pKa of ethanoic acid at 298 K = 4.76 ➔ A solution of ethanoic acid (CH3COOH) in water is made with a pH = 3.5 Deduce the structural formula of the conjugate base of ethanoic acid. Construct a predominance diagram State whether ethanoic acid or its conjugate base will be predominant at pH = 3.5 pH Determining equilibrium constant from Ka values Lesson 7.2.4 - Composition and Chemical Properties Specification Points Command words are highlighted. 1. Write the balanced chemical equation for the reaction of acids with reactive metals, metal oxides, metal hydroxides, hydrogen carbonates, and carbonates. 2. Define a buffer solution as a solution that resists large changes in pH upon addition of small quantities of strong acid or strong base. 3. Write the equation for the reaction of a buffer solution with an acid or a base. 4. Describe three methods to prepare a buffer solution. 5. Prepare a buffer solution at a given pH given a list of acids/bases and their dissociation constants. 6. Calculate the pH of a buffer solution made from a monoprotic acid and its conjugate base (acidic buffer) or from a monoprotic base and its conjugate acid (basic buffer). 7. Rationalise that the pH of a buffer solution depends on two factors: 8. State that a buffer solution is efficient in the pH range is pKa ± 1. Chemical Equations → Acids with reactive metals (G1,G2) ➔ Acids react with reactive metals, such as Group 1 and Group 2 metals, to release hydrogen gas. ➔ An example is shown below: the reaction of sodium with hydrochloric acid. Chemical Equations → Acids with metal oxides ➔ Acids react with metal oxides to produce a dissolved ionic compound and water. ➔ An example is shown below: the reaction of magnesium oxide with hydrochloric acid. Chemical Equations → Acids with metal hydroxides ➔ Acids react with metal hydroxides to produce a dissolved ionic compound and water. ➔ An example is shown below: the reaction of calcium hydroxide with hydrochloric acid. Chemical Equations → Acids with hydrogen carbonates ➔ Acids react with hydrogen carbonates to produce a dissolved ionic compound, water and carbon dioxide. ➔ An example is shown below: the reaction of calcium hydroxide with hydrochloric acid. Chemical Equations → Acids with carbonates ➔ Acids react with carbonate to produce a dissolved ionic compound, carbon dioxide and water. ➔ An example is shown below: the reaction of sodium carbonate with hydrochloric acid. ➔ Drawing chemical equations are common 1-mark BAC questions (just be aware!) ➔ Drawing the chemical equation is normally the first step. ➔ Often calculations can be performed using the chemical equation as a guide. What is a buffer solution? Buffer solution → a solution that resists large pH change on addition of small quantities of strong acid or strong base. ➔ It is useful to see how a non-buffered solution behaves so we can compare later in the course. ➔ Below is an example for a pure water sample (pH = 7). The diagram shows how even a very small volume of strong acid or strong base can have a very large impact on the pH of the solution! Composition of a buffer solution Buffer solutions can be made by: ➔ Mixing a weak acid with a solution of the conjugate base ionic salt. ➔ Mixing a weak base with a solution of the conjugate acid ionic salt. We are going to investigate a commonly used buffer system (ethanoic acid / sodium ethanoate) Ethanoic acid is a weak acid → partial ionization in water Sodium ethanoate is a salt that complete ionizes in solution Buffer Solutions → Reactions with small acid volumes ➔ Ethanoate ions (CH3COO-) react with the H3O+ to prevent a decrease in the pH. ➔ The equilibrium position shifts to the left as H3O+ ions react with CH3COO- ions to form more CH3COOH until equilibrium is re-established. ➔ As there is a large reserve supply of CH3COO- the concentration of CH3COO- in solution doesn’t change much as it reacts with the added H+ ions. ➔ As there is a large reserve supply of CH3COOH the concentration of CH3COOH in solution doesn’t change much as CH3COOH is formed from the reaction of CH3COO- with H3O+. ➔ As a result, the pH remains broadly constant. Buffer Solutions → Reactions with small base volumes ➔ Ethanoic acid is deprotonated by the hydroxide ions. ➔ Hydroxide ions are removed from the solution directly. ➔ Hence, pH remains broadly unchanged. Preparing an acidic buffer solution → Method 1 (Weak Acid + Conjugate Base) This method involves mixing a weak acid with one of its salts containing its conjugate base, the For example: ➔ Mix ethanoic acid (CH3COOH) with sodium ethanoate (CH3COO-Na+) ➔ This creates a buffer solution that resists pH change on addition of small volumes of acid/base. Preparing a basic buffer solution → Method 2 (Weak Base + Conjugate) This method involves mixing a weak base with one of its salts containing its conjugate acid. For example: ➔ Mix ammonia (NH3) with ammonium chloride (NH4Cl). ➔ In solution, ammonia is protonated by water, leading to the formation of (OH-) ions. ➔ This method of buffer solution leads to an alkaline solution being formed. Buffer solution → Method 3 (Partial Neutralization of a Weak Acid) This method involves reacting a weak acid with a strong base to partially neutralize the acid. For example: ➔ Add a small mass of sodium hydroxide (NaOH) to ethanoic acid (CH3COOH). ➔ The reaction mixture has sodium ethanoate (CH3COONa), as well as unreacted CH3COOH – a buffer is formed. Buffer solution → Method 4 (Partial Neutralization of a Weak Base) This method involves reacting a weak base with a strong acid to partially neutralize the acid. For example: ➔ Add a small volume of hydrochloric acid (HCl) to ammonia (NH3) ➔ The reaction mixture has ammonium chloride (NH4Cl), as well as unreacted NH3 – a buffer is formed. pH of a buffer solution -- Henderson-Hasselbalch Equations ➔ The pH of a buffer solution will depend on the interactions between its components. ➔ Consider the example of an acidic buffer made up of the generic weak acid (HA) and its salt (MA). The equilibria that exist in this acidic buffer will be: pH of a buffer solution -- Henderson-Hasselbalch Equations Conclusions → Henderson-Hasselbalch Equations ➔ For basic buffer solutions: pOH is calculated from the equation, however pH can be indirectly calculated via use of an equation that you should be very familiar with by now! pKw = pH + pOH ➔ The pH of a buffer solution is dependent on two key factors, as indicated by the Henderson-Hasselbalch equations: ➔ Changing the value of Ka or Kb leads to major changes in the pH measurements. Changing the Ka or Kb, so that pH can be modified; this is known as “coarse tuning”. ➔ Slowly changing the ratio of [conjugate base] : [acid] will allow for smaller change to the pH of the buffer; this is known as “fine tuning”. pH of a buffer solution → Worked Example Calculate the pH of a buffer solution at 298 K, prepared by mixing 25 mL of 0.1 M ethanoic acid (CH3COOH), with 25 mL of 0.1 M sodium ethanoate (Na+CH3COO-) Ka of CH3COOH = 1.8 x 10-5 at 298K pH of a buffer solution → Worked Example 2 1. Calculate the concentration of potassium butanoate in the 1.00 dm3 buffer solution using the Henderson-Hasselbalch equation. 2. Calculate the mass of potassium butanoate needed for the 1.00 dm3 buffer solution. pH of a buffer solution → Worked Example 2 Buffer Efficiency → pH vs pKa Statement: A buffer solution is most effective when the pH of the solution lies in the range of pKa ± 1. Justification: The key idea lies in the Henderson-Hasselbalch Equations ➔ The efficiency of a buffer is related to its ability to resist changes in pH upon the addition of small amounts of acid or base. ➔ The ability to resist pH change is greatest when the [HA] and [A-] are similar, this way the buffer has considerably capacity to neutralize H3O+ or OH-. ➔ Beyond this range, either the acid or the base predominates so heavily that the buffer loses its ability to efficiently neutralize added acid/bases that could be added to the buffer solution. Independent Practice → Calculating pH of a buffer solution This question investigates the basic buffer solution composed of the base/conjugate acid buffer system of: NH3 / NH4+ At 298K, 25 mL of 0.5 M ammonia (NH3), is mixed with 25mL of 1 M ammonium chloride solution (NH4+Cl-). The value of Kw = 1 x 10-14 pKw = pH + pOH 1. Deduce the value of the ratio [conjugate acid] / [base] for this buffer system. 2. Calculate the value of pOH for this buffer system 3. Determine the pH of the buffer system. Independent Practice → Determining ratio of acid/conjugate base The normal pH of blood is between 7.35 and 7.45, which is considered a slightly alkaline pH. Any deviation from these values is very serious, and puts health at risk, including the risk of death. Normally, the pH of the blood is kept practically constant, in part by the action of the buffer system H2 CO3 /HCO–3 corresponding to the equilibrium below. CO2 (g) + H2 O (l) ⇌ H2 CO3 (aq) H2 CO3 (aq) + H2 O (l) ⇌ HCO–3 (aq) + H3 O+ (aq) i. Write chemical equations that show how hydrogen carbonate ions can control the pH within the range considered. [HCO3− ] ii. Calculate the ratio [H in the blood so that the pH is 7.40, knowing that the value of 2 CO3 ] Ka(H2CO3) = 4.30 x 10–7 Independent Practice → Determining ratio of acid/conjugate base (Solutions) When acid is added: HCO–3 (aq) + H3 O+ (aq) ⇌ H2 CO3 (aq) + H2 O (l) When base is added: H2 CO3 (aq) + OH- (aq) ⇌ HCO–3 (aq) + H2 O (l) Lesson 7.2.5a - Titrations Specification Points Command words are highlighted. 1. State that in any titration the known concentration of a solution is used to determine the unknown concentration of another. 2. Draw a diagram of a typical titration apparatus. 3. Define the equivalence point. 4. Write the equation for a titration reaction. 5. Perform a titration experiment → (Completed in A-Mark Experiment) 6. Sketch the profile of a titration curve for the following titration types: strong acid/strong base, weak acid/strong base, strong acid/weak base 7. Calculate the concentration of an acid or a base using the titration curve. 8. Plot a titration curve, pH = f(Volume added) → (Completed in A-Mark Experiment) 9. Predict and explain the value of the pH at the equivalence point (acidic, neutral or basic). 10. Calculate the pH value at the equivalence point. Acid-Base Titration → What is the purpose? Titration = a technique used to determine the concentration of an unknown solution by adding a certain volume of another solution where the concentration is known. Titration is widely used in the food production and cosmetic industries! ➔ A volume of reactant is measured using a measuring pipette and placed into a flask. ➔ Controlled volumes of the other reactant are slowly added from the burette into the flask. ➔ The solution in the burette is known as the titrant. ➔ The solution in the flask is known as the analyte. ➔ Volumes from the burette are added into flask until the equivalence point is reached (definition on the next page). In this example acid has been placed in the burette; this is not always the case. A basic solution could be added to the burette instead! Acid-Base Titrations → Equivalence Point Equivalence point (acid-base titration) = the point at which the moles of reactant added from the burette is exactly equal to the mole of reactant in the flask. At this point, the reaction has just completed, there is no excess of either reactant. The equivalence point is very important for determining amounts that have reacted! When a base is added to acid, a neutralization reaction takes place. But the change in pH recorded as base is added is not linear. ➔ A pH meter can be used to record the pH changes during a titration. ➔ Using a pH meter, volumes of acid or base can be added, and the pH can be monitored. ➔ The equivalence point can be determined by drawing a graph. Strong Acid - Strong-Base → Neutralization of HCl (aq) and NaOH (aq) This is the classic system used to demonstrate an acid-base titration. 1. Construct a balanced equation to show the reaction between NaOH (aq) and HCl (aq) including state symbols. Strong Acid - Strong-Base → Neutralization of HCl (aq) and NaOH (aq) Consider a situation where: ➔ 0.1M NaOH (aq) is the titrant and the concentration of HCl (aq) is unknown. ➔ A known volume of the HCl (aq) solution is added to a flask. ➔ Volumes of 0.1M NaOH (aq) solution are added slowly from the burette. ➔ A pH probe is used to monitor the pH change after every addition of 0.1M NaOH (aq) ➔ At the equivalence point all of the (H3O+) ions have been neutralized. When this strong acid-strong base neutralization takes place, the following curve is generated. Independent Practice → Neutralisation Questions 1. State the pH of the solution formed when a strong acid is exactly neutralized by a strong base. 2. Construct an equation to represent the neutralization reaction between aqueous dilute nitric acid and potassium hydroxide solution. 3. In a titration, the equivalence point is reached when 10.00 cm3 of 0.01 M NaOH (aq) has been added to 12.00 cm3 of H2SO4 (aq) a) Calculate the moles of NaOH (aq) that have reacted. b) Deduce the number of moles of H2SO4 (aq) that have reacted. c) Calculate the concentration of H2SO4 (aq) that have reacted. Concentration = Moles / Volume (Review S5 – Calculations Review if needed!) Titrations → Strong Acid - Weak Base Consider a situation where: ➔ NH3 (aq) is the titrant and the concentration of HCl (aq) is unknown. ➔ A known volume of the HCl (aq) solution is added to a flask. ➔ Initially, there are only H3O+ ions present in solution from the complete dissociation of the strong acid (HCl) - initial pH about 1-2. ➔ As the volume of weak alkali (NH3) added increases, the pH of the analyte solution slightly increases too as more and more H3O+ ions react with the NH3. ➔ The change in pH is not that much until the volume added gets close to the equivalence point. ➔ The equivalence point is the point at which all H3O+ ions have been neutralised by the NH3 however the equivalence point is not pH = 7 in this case. The equivalence point here is approximately pH = 5.5. ➔ This is because all H3O+ have reacted with NH3 to form NH4+ which is a relatively strong acid, causing the solution to be acidic ➔ As more of the NH3 is added, the pH increases to above 7 but below that of a strong base as NH3 is a weak base. Titrations → Strong Acid - Weak Base Titrations → Weak Acid - Strong Base The weak acid-strong base titration leads to a point labelled as the “half-equivalence point”, the point where half of the acid has been neutralized by the base. We will investigate the half equivalence point further in the next lesson. Independent Practice → Determining acid concentration from a titration curve System = Strong Base – Weak Acid) In this question we will determine the concentration of the weak acid, propanoic acid by analysing the equivalence point via titration. Note: The equivalence point is not always pH = 7, however it is an important point that tells you the point where the acid has been fully reacted by the base. 25 mL of an unknown concentration of C2H5COOH (aq) is added into a reaction flask. Small volumes of 0.2 molL-1 KOH (aq) solution are added slowly to the reaction flask via burette. Volume of 0.2 molL-1 KOH (aq) added / mL Independent Practice → Determining acid concentration from a titration curve 1. From the graph estimate the value of pH at the equivalence point for this reaction. 2. Define the term “equivalence volume”. 3. Determine the moles of KOH (aq) that reacted. 4. Calculate the concentration of C2H5COOH (aq). Lesson 7.4.1a - Intermediates and Electronic Effects Specification Points Command words are highlighted. Do you know what they mean? 1. Use curly arrow notation to show movement of electrons in chemical reactions. 2. Explain stability of intermediates formed in heterolysis of covalent bonds in terms of the nature of the atoms, or group of atoms, attached to the carbon atom. 3. Discuss how the electronic properties of atoms, or groups of atoms, help to predict how organic compounds react. 4. Describe the two factors that influence the electronic properties of atoms, or groups of atoms: inductive and mesomeric effects. 5. Explain the inductive effect in terms of electronegative differences that between atoms or group of atoms with consequent push and pull of electrons in the 𝜎 bonds of molecules (e.g. halogens, alkyls, hydroxyl groups, etc.). 6. Explain the mesomeric effect in terms of delocalization of electron density through 𝜋 bonds. 7. Elucidate how the size of substituents influences the stability and reactivity of ions and neutral molecules. What is an intermediate? A multi-step reaction will involve a set of individual processes where at least one intermediate will be formed. Intermediate = a temporary chemical species which is formed during a chemical reaction. Intermediates are not very stable, so there is a thermodynamic driving force for the reaction to continue to the more stable product. Introduction to the carbocation Cation = a positively charged ion. Cations and anions can be unstable for the simple reason that charge separation costs energy There are a few cases in which these cations are quite stable, for example: alkali cations such as Na+ because sodium is electropositive, meaning that the atom will readily react to lose an electron. We are investigating positive charges centered on carbon atoms, known as the carbocation. Stability of the carbocation Carbocations are generally unstable. Carbon is in the upper right part of the periodic table, so it is not particularly electropositive like sodium. A positive charge on carbon frequently makes a molecule reactive. Nevertheless, this intermediate is frequently encountered during organic reactions so we have to study ways that this intermediate can be stabilized. Below: some examples of some cations with the positive charge centered on an organic element. Some carbocations are more stable than others….. Carbocations can be stabilized by increasing the degree of substitution around the central carbocation. Left to right: A primary carbocation, a secondary carbocation, a tertiary carbocation. “Why does stability of the carbocation increase with increasing substitution?” The methyl groups are actually providing electron density to the positively charged central carbon. When this electron density is provided, the positive charge is slightly neutralized = MORE stable carbocation. How does this happen? → Next page The Inductive Effect Polarization → Inductive effect = a permanent state of polarization. The electron density in a σ bond between two unlike atoms is not equally shared. The electron density is distributed more toward the more electronegative of the two atoms. KEY: The degree of polarization decreases with distance. The inductive-effect is distance- dependent. Electron-withdrawing effect (-I effect) Here is a chain of carbon atoms without any hydrogen atoms, in this case the electronegative X is removing withdrawing electron density from the carbon chain. The carbon directly bonded to X is considerably polarized, the other carbons in the chain will all be very slightly polarized because the degree of polarization decreases with distance. Because X is withdrawing electron density, this is described as a (-I effect). Electron-releasing effect (+I effect) In this case, carbon is more electronegative than Y, hence an electron donating effect is seen. The carbon directly bonded to Y is considerably polarized, the other carbons in the chain will all be very slightly polarized because the degree of polarization decreases with distance. Inductive Effects → Quick Review There are two categories of inductive effects: the electron-withdrawing (-I effect) and the electron-donating (+I effect). In the image above, X is electron-withdrawing and Y is electron-donating. Inductive Effects → Carbocation Stability KEY: Alkyl groups are electron releasing groups, when bonded they show a +I effect. The tertiary substituted carbocation is the most stable out of the three because there are the most +I effects. Inductive Effects → Carbocation Stability and Alkyl Chain Length THINKING: Which carbocation do you think will be better stabilized? Inductive Effects → Carbocation Stability and Alkyl Chain Length The carbocation on the left (with the ethyl substituent) will be more stabilized. The +I group increases with alkyl length (up to a limit!) because there is more electron density that can be pushed to stabilize the carbocation. Extra information (if you are interested interested): the induction effect decreases with distance, so a pentyl group will show a very similar +I effect as a butyl group. Stereochemical Representation → Carbocations It is useful to think about the 3D arrangements of carbocations, especially when we consider typical reactions that they participate in. We saw from S6.6 (Organic Chemistry) that carbon with 4 single bonds forms a tetrahedral structure (109.5o bond angles). However, the carbocation (3 single bonds) forms a completely flat structure with an empty 2p orbital. Next lesson we will see about a type of substitution reaction (SN1) where a reaction can either take place from the top face or bottom face of the carbocation → Use some MOLYMODS to see this concept! Typical +I groups and typical - I groups Electron withdrawing groups (-I groups) include: the halogens (Cl,Br,F,I), hydroxyl groups (OH), nitro groups (NO2), carboxy groups (COOH) and ester groups (COOR). There are loads of others, though! Electron donating groups (+I groups) include: alkyl groups. Independent Practice → Carbocation Stability Mesomeric effects → +M and -M groups Mesomeric effects are caused when pi electrons are moved or removed from a conjugated (connected by p-orbitals) system. +M groups → provide electron density in the form of pi electrons. -M groups → remove electron density from a conjugated pi system. ➔ Maximum carbocation stability is achieved if there is a dual electron donating effect, in other words if there are +M effects and +I effects which both provide electron density to the unstable carbocation. Mesomeric effects → Carbocation Stability KEY: To produce the +M effect the group must have either a LONE PAIR of electrons (non- bonded), or a negative charge centered on the atom. Mesomeric effects → Carbocation Stability Typical +M groups and typical - M groups Examples of groups that show a +M effect include: O-, -NH2, O-CH3. Examples of groups that show a +M effect include: COO-, CHO. Independent Practice → Carbocation Stability Applying your knowledge of inductive and mesomeric effects to decide the order of carbocation stability. Draw any relevant resonance structures to demonstrate any mesomeric effects. Justify your choices fully. Independent Practice → Carbocation Stability Applying your knowledge of inductive and mesomeric effects to decide the order of carbocation stability. Draw any relevant resonance structures to demonstrate any mesomeric effects. Justify your choices fully. Independent Practice → Carbocation Stability Applying your knowledge of inductive and mesomeric effects to decide the order of carbocation stability. Draw any relevant resonance structures to demonstrate any mesomeric effects. Justify your choices fully. Independent Practice → Cation Identification Carbon → 4 total bonds = Neutral atom Nitrogen → 3 total bonds = Neutral Atom Oxygen → 2 total bonds = Neutral Atom Lesson 7.4.b - Nucleophiles and Substitution Reactions Specification Points Command words are highlighted. Do you know what they mean? 1. State that nucleophiles are negatively charged ions or neutral molecules that donate a pair of electrons to form a covalent bond. 2. State that electrophiles are positively charged ions or neutral molecules that accept a pair of electrons to form a covalent bond. 3. Predict and draw the mechanisms for a substitution reaction (SN1 or SN2) using curly arrows. 4. Explain the factors influencing whether the SN1 or SN2 mechanism is followed: steric hindrance and stability of carbocation. 5. Write the balanced chemical equation for a substitution reaction. Nucleophiles vs Electrophiles Nucleophile: negatively charged ions or neutral molecules that donate a pair of electrons to form a covalent bond. Electrophile: positively charged ions or neutral molecules that accept a pair of electrons to form a covalent bond. We just did a whole lesson about a very common electrophile used in organic synthesis – the electrophilic carbon in a carbocation. Nucleophilic Substitution → SN2 There are two main ways that nucleophilic substitution reaction can take place (SN1 or SN2). The two substitution methods proceed by different pathways! We will investigate a group of compounds called the haloalkanes (a type of substituted alkane) Under certain conditions, halogenoalkanes undergo nucleophilic substitution by an SN2 mechanism. Background information ‘S’ stands for ‘substitution’ ‘N’ stands for ‘nucleophilic’ ‘2’ means that the rate of the reaction (which is determined by the slowest step of the reaction) depends on the concentration of both the halogenoalkane and the nucleophile ions ➔ The SN2 mechanism is a one-step reaction. ➔ The nucleophile donates a pair of electrons to the δ+ carbon atom to form a new bond ➔ At the same time, the C-X bond is breaking, and the halogen (X) takes both electrons in the bond (heterolytic fission) The halogen leaves the halogenoalkane as an X- ion. We can describe the X- as being a leaving group. Nucleophilic Substitution → SN2 Mecanism (Haloalkane) This example shows the nucleophilic substitution of bromoethane by hydroxide ions to form ethanol. Nucleophile = the HO- group Electrophilic centre = the delta δ+ carbon atom. The C-Br bond is polarized because of the electronegativity of Br. For an SN2 reaction, the rate of the reaction depends on [Haloalkane] and [Nucleophile] Nucleophilic Substitution → SN1 Mecanism (Haloalkane) Haloalkenes substituted with several alkyl groups follow a different substitution mechanism. ➔ Later in the lesson, we will investigate why some compounds proceed via SN2 and some proceed via SN1 The example shows the nucleophilic substitution of 2-bromo-2-methylpropane by hydroxide ions to form 2-methyl-2-propanol mesh C Br handisbsken i.) Describe the type of stabilization force acting on the carbocation. I 3 ii.) On the diagram, label the inductive effects that are present for the carbocation stabilization. Rate of reaction depends on stability of carbocation Nucleophilic Substitution → Return of the Carbocation! In SN1 processes involving carbon centres, the reaction proceeds with formation of a carbocation intermediate. NOTE: Make sure you have understood and are aware of the factors that stabilize and destabilize carbocations! ➔ In the SN1 mechanism, a tertiary carbocation is formed ➔ This is not the case for SN2 mechanisms as a primary carbocation would have been formed which is much less stable than tertiary carbocations. ➔ This has to do with the positive inductive effect of the alkyl groups attached to the carbon which is bonded to the halogen atom Determination → Which substitution pathway will be followed? Factor 1 → Steric Hinderance around the electrophilic centre Steric hinderance = the slowing of chemical reactions because of the interference of bulky side groups. ➔ The image (above) represents a sterically hindered carbon center. ➔ An HO- nucleophile begins an approach to the delta positive carbon (attracted by electrostatic attraction). ➔ It is very difficult for the nucleophile to donate electrons to the electrophilic carbon center because the alkyl groups hinder (prevent) the nucleophilic attack. Factor 2 → Stability of the carbocation (SN1 substitution) As you saw from the mechanism, SN1 substitution requires the formation of a carbocation intermediate. ➔ The more stable the carbocation intermediate, the more likely the SN1 substitution path will be followed. Independent Practice → Determining the substitution method Determine which substitution reaction pathway will be followed for the compound shown below. Justify your answer. H I A Cfco H Nstwellstablers Idea 1 Carbocation notfound Sn2 pathasy Independent Practice → Determining the substitution method (Solutions) Summary Questions → Nucleophiles and Substitution dementia Summary Questions → Nucleophiles and Substitution 5m 5m Theyhave an intermediate which makes thecarbocation more state Summary Questions → Nucleophiles and Substitution Summary Questions → Nucleophiles and Substitution (Solutions) Example Chemical Equation → General Subsistution Lesson 7.4.c - Elimination Reactions Specification Points Command words are highlighted. Do you know what they mean? 1. Predict and draw the mechanisms of an elimination reaction (E1 or E2) using curly arrows, (limited to the dehydration of alcohols). 2. Explain the factors influencing whether the E1 or E2 mechanism is followed: steric hindrance and stability of carbocation. 3. Write the balanced chemical equation of elimination reactions. Elimination Reactions (Dehydration of Alcohols) We are going to investigate a style of reaction which involves the removal of a water molecule from an alcohol (compound involving a hydroxyl - OH functional group). There are two different modes of elimination, in the same way that there are two modes of substitution → E1 and E2. To work out which substitution mode is followed, we have to investigate the extent of substitution for the alcohol E2 reactions for dehydration of alcohols Concerted = bond making and bond breaking take place in the same step. ➔ Primary alcohols dehydrate in a concerted process. ➔ No cation is formed! ➔ The reaction proceeds under acidic conditions. ➔ The acid is regenerated at the end of the reaction. Mechanism → Acid-assisted dehydration of pentan-1-ol E1 reactions for dehydration of alcohols Leaving group (seen in SN1 and E1 reactions): an atom or group of atoms that are able to break away from a molecule with a lone pair, breaking the bond between it and the molecule. ➔ E1 is not a concerted reaction; there are multiple steps ➔ An acid is required to protonate the hydroxyl (-OH group). The protonation of the hydroxyl make it a better leaving group – there is a driving force for there to be no positive charge on the electronegative oxygen atom. ➔ The loss of the H2O leaving group is the SLOW rate determining step. ➔ There can sometimes be more than one product from the dehydration of alcohols (as shown in the example). Mechanism → Acid-assisted dehydration of butan-2-ol Determination → Which elimination pathway will be followed? For the elimination pathway for alcohols → the factors that need to be considered are similar to the ones that we considered when we investigated substitution reactions. ➔ Stability of the carbocation ➔ The solvent used in the reaction (this is outside the syllabus but good to know) Factor 2 → Stability of the carbocation (SN1 substitution) Consider the primary alcohol shown below. Sample Question → Rationalise why this primary alcohol will dehydrate following an E2 pathway. ➔ To determine this, we can first model what would happen if an E1 pathway is followed and then work out what is preventing the E1 pathway. The Theoretical E1 pathway ➔ Primary alcohols like this example would form a theoretically unstable carbocation with minimal inductive effect stabilization. ➔ The lack of stability for the carbocation encourages the alternative elimination pathway. ➔ Primary alcohols will hence eliminate via the E2 mechanism. Conclusion → Which elimination pathway will be followed? The stability of the carbocation formed must be considered. ➔ Tertiary alcohols will form a heavily stabilized carbocation (E1 pathway) ➔ Secondary alcohols will form well-stabilized carbocations (E1 pathway) ➔ Primary alcohols will form poorly stabilized carbocations (E2 pathway) Extension → Stabilization of a carbocation via interaction with the solvent This is outside the scope of the course, but it’s good to know about. ➔ Remember these reactions are NOT happening in a vacuum. ➔ Certain solvents can stabilize the carbocation formed in the E1 or SN1 pathways. ➔ Polar protic solvents (a solvent which has a H-atom bound to an oxygen or nitrogen atom). Lesson 7.4.1d - Alcohols Specification Points Command words are highlighted. Do you know what they mean? 1. Define and recognize primary, secondary and tertiary alcohols. 2. Recall nomenclature for C1 to C10 compounds, and physical properties (S6.5 – Organic) 3. Explain that alcohols undergo elimination and substitution reactions. 4. Write balanced chemical equations for: ➔ Dehydration (inter-molecular) to form ether (130°C – H2SO4) ➔ Dehydration (intra-molecular) to form alkene (170°C - Al2O3) Primary, Secondary and Teritary Alcohols → Definitions Primary alcohols = alcohols in which the carbon atom bonded to the -OH group is attached to one other carbon atom (or alkyl group). Secondary alcohols = alcohols in which the carbon atom bonded to the -OH group is attached to two other carbon atoms (or alkyl groups) Tertiary alcohols = alcohols in which the carbon atom bonded to the -OH group is attached to three other carbon atoms (or alkyl groups) S6 Review → Nomenclature Primary Alcohols Alcohols are a family of molecules that contain the hydroxyl functional group, -OH ➔ Their general formula is CnH2n+1OH. ➔ The nomenclature of alcohols follows the pattern alkan + ol. ➔ If there are two - OH groups present the molecule is called a diol. Q: Draw the displayed and skeletal formulae for: HI I I a 0H not a.) butanol b.) butane-1,4-diol b ot Ottoman You are provided with the information that the bromine functional group leads to “bromo” nomenclature. Example 1-bromobutane Q: Provide the IUPAC name for: heptane 3 fictional 3 2bismpentose 2 Properties of Alcohols → Hydrogen Bonding (S6 Review) ➔ Hydrogen bonding occurs between molecules where you have a hydrogen atom attached to one of the very electronegative elements → fluorine, oxygen or nitrogen. ➔ In an alcohol, there are hydroxyl groups with OH bonds present in the structure. ➔ Hydrogen bonds set up between the slightly positive hydrogen atoms (δ+ H) and lone pairs on oxygens in other molecules. Properties of Alcohols → Hydrogen Bonding (S6 Review) ➔ In alkanes, with only C-H bonds, the only intermolecular forces are London Dispersion forces, which are very weak (S6 Intermolecular Forces). ➔ Boiling points of alkanes are considerably lower than the corresponding alcohol with the same number of carbon atoms. ➔ Hydrogen bonds are much stronger than these and therefore it takes more energy to separate alcohol molecules than it does to separate alkane molecules. Properties of Alcohols → Solubility (S6 Review) Miscible → forming a homogenous mixture when mixed. If an alcohol is miscible, it means that the alcohol dissolves fully to form a single homogenous mixture. ➔ The small alcohols are completely soluble in water ➔ If they are mixed with water, there will be one solution → the alcohol is miscible in water. ➔ However, solubility falls as the length of the hydrocarbon chain in the alcohol increases ➔ The hydrocarbon chains are forcing their way between water molecules and so breaking hydrogen bonds between those water molecules ➔ The OH end of the alcohol molecules can form new hydrogen bonds with water molecules, but the hydrocarbon cannot form hydrogen bonds = decrease in solubility as carbon chain increases! Substitution Reactions → Alcohols When treated with HBr, a nucleophilic substation reaction can take place which leads to the formation of an alkyl bromide. 1. Determine and justify which substitution reaction pathway will be followed for reactant containing a hydroxyl group. 5h7 Substitution Reactions → Alcohols (Mechanism) When treated with HBr, a nucleophilic substation reaction can take place which leads to the formation of an alkyl bromide. ➔ HBr(aq) will dissociate fully in solution to release H+ ions and Br- ions into solution. ➔ HBr(aq) is an example of a strong acid because the species dissociates fully in solution. ➔ This is an example of acid dissociation which we will investigate in detail in S7.2 – Acids and Bases. Pistastion Slow Independent Practice

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