Chemistry AS Textbook

NES/Chemistry/AS Change in Enthalpy Equation ∆H is the enthalpy change for a reaction Hproducts is the total enthalpy of all the products Hreagents is the total enthalpy of all the reagents  Example 2: Calculate the enthalpy change for the following reaction, using average bond energies: C2H6 + Cl2  C2H5Cl + HCl If you draw out the molecules and use the data booklet, you will see that: bonds broken C-H = 410 Cl-Cl= 242 bonds formed C-Cl = 340 H-Cl = 431 enthalpy change, ∆H = (410 + 242) - (340 + 431) enthalpy change, ∆H = -119kJmol-1 This reaction has a negative enthalpy change, so the reaction is exothermic.  Example 3: Write the symbol equation for the enthalpy change for the formation of propene. Using the definition of enthalpy change of formation we get: 3C (s, graphite) + 3H2 (g)  C3H6 (g) Note - carbon's standard condition is the solid graphite allotrope. 100

NES/Chemistry/AS  Example 4: Write the symbol equation for the enthalpy change for the combustion of butane. C4H10 (g) + O2 (g)  4CO2 (g) + 5H2O (l)  Example 5: Write the symbol equation for the reaction of sulphuric acid and sodium hydroxide. H2SO4 (aq) + NaOH (aq)  Na2SO4 (aq) + H2O (l) Note - the equation must be balanced per 1 mole H2O because of the definition.  Example 6: Write the symbol equation for the enthalpy change for the atomisation of hydrogen. H2 (g)  H (g)  Example 7: Write the symbol equation for the enthalpy change for the hydration of copper(II) sulphate. CuSO4 (s) + 5H2O (l)  CuSO4. 5H2O (aq)  Example 8: Write the symbol equation for the enthalpy change for sodium chloride dissolving in water. NaCl (s) + aq  NaCl (aq) Note - (aq) is used to represent water in the equation. 101

NES/Chemistry/AS Measuring Enthalpy Changes It is not possible to measure enthalpy directly. We can measure the temperature change of the surroundings and from this we can calculate the enthalpy change of the chemical reaction. The thermometer is placed in the reaction mixture, which will have at least one aqueous reagent. The thermometer is actually in contact with more water than reagent, so we are really just measuring the temperature change of the water (surroundings). If we assume that all of the heat energy is transferred from the chemical to the surroundings, then the change in temperature is proportional to the enthalpy change of the reaction. 102

NES/Chemistry/AS Calculating Enthalpy Change The following equations are used to calculate the enthalpy change for a reaction: First: E is the change in energy due to the reaction, J m is the mass of water that is being heated, g c is the specific heat capacity, J/g/0C ∆T is the change in temperature of the water, 0C  When finding the mass of water, we can use the volume of water in cm3 to be equal to the mass of water in g as the density of water is 1gcm-3.  We are assuming the density of the solution is the same as the density of water.  We are also assuming the mass of the excess solid has no effect on the energy change and do not add it to the mass of water. Then: ∆H is the change in enthalpy, kJmol-1 mol is the number of moles of the limiting reagent 103

NES/Chemistry/AS  Example 9: A reaction was carried out to measure the enthalpy change of a displacement reaction. 2.00g of magnesium was placed in 50cm3 of copper(II) sulphate and the temperature rise was found to be 200C. Calculate the change in enthalpy during the reaction: Mg (s) + CuSO4 (aq)  MgSO4 (aq) + Cu (s) First: m is the mass of the water, we are going to assume that the mass of the copper(II) sulphate solution and the mass of the magnesium is going to be equal to the volume of the copper(II) sulphate solution. m = 50g c is the specific heat capacity of the reaction mixture, we are going to assume that this is the same as the specific heat capacity of water. c = 4.18J/g/0C ∆T is the change in temperature during the reaction. We are going to assume that there was no heat loss from the insulated plastic cup. ∆T = +200C Then: -E = -4.18 kJ - we have to change units to kJ for the second equation mol = = 0.0823mol Mg ∆H = -50.7kJmol-1 104

NES/Chemistry/AS 5.2 Hess’ Law We cannot measure the energy change in some reactions, so an alternative method has to be used to calculate the enthalpy change. If the starting reagents are the same and the products are the same, then a series of different reactions can be carried out and the enthalpy changes for each reaction can be added together to find the total enthalpy change. This is usually drawn out as a Hess cycle. Reagents Products Alternative Route Whichever route is taken, the overall enthalpy change is the same. Setting up a Hess cycle can be quite tricky, you must know your enthalpy definitions. 105

NES/Chemistry/AS  Example 10: Using enthalpy changes of formation to calculate the enthalpy change for the reaction: 2NaHCO3  Na2CO3 + CO2 + H2O Given the following enthalpy changes of formation: ∆HƟf (NaHCO3) = -950.8 kJmol-1 ∆HƟf (Na2CO3) = -1130.7 kJmol-1 . ∆HƟf (CO2) = -393.5 kJmol-1 ∆HƟf (H2O) = -285.8 kJmol-1 The Hess cycle: Na2CO3 (s) + CO2 (g) + H2O (l) 2NaHCO3 (s) 2Na (s) + H2 (g) + 2C (s,graphite) + 3O2 (g) Note - the arrows are pointing up from the elements because of the definition of enthalpy change of formation. Note - the alternative route is balanced with the reagents and the products. ∆HƟR = -(2 x ∆HƟf (NaHCO3)) + (∆HƟf (Na2CO3)) + (∆HƟf (CO2)) + (∆HƟf (H2O)) ∆HƟR = -(-1901.6) + (-1130.7) + (-393.5) + (-285.8) ∆HƟR = +91.6 kJmol-1 Note - the value for ∆HƟf (NaHCO3) has a -sign outside of the brackets as the alternative route is going in the opposite direction, so the sign for the enthalpy change of formation must be the opposite. Note - the signs for the enthalpy changes of formation of the products stays the same as the alternative route is in the same direction as the arrows. 106

NES/Chemistry/AS Setting up a Hess cycle is considerably more difficult than doing the actual calculation. It will take practice.  Example 11: Calculate the enthalpy change for the formation of ethane: 2C (s,graphite) + 3H2 (g)  C2H6 (g) Given the following enthalpy changes of combustion: ∆HƟc (C(s,graphite)) = -393.5 kJmol-1 ∆HƟc (H2 (g)) = -285.8 kJmol-1 ∆HƟc (C2H6 (g)) = -1559.7 kJmol-1 The Hess cycle: C2H6 (g) 2C (s,graphite) + 3H2 (g) 2CO2 (g) + 3H2O (l) Note - the arrows are pointing down from the equation because of the definition of enthalpy change of combustion. ∆HƟf (C2H6) = (2 x ∆HƟc (C)) + (3 x ∆HƟc (H2)) - (∆HƟc (C2H6)) ∆HƟf (C2H6) = (2 x -393.5) + (3 x -285.8) - (-1559.7) ∆HƟf (C2H6) = -84.7 kJmol-1 107

NES/Chemistry/AS Energy Level Diagrams We can also draw an energy level diagram for any Hess cycle.  Example 12: Using the values from example 11, draw an energy level diagram 2C (s,graphite) + 3H2 (g) -84.7 kJmol-1 C2H6 (g) -1644.4 kJmol-1 -1559.7 kJmol-1 2CO2 (g) + 3H2O (l) The totals for each path have been put on the arrows already. We need to split the cycle into 3 parts, reagents, products and alternatives, and place these on the graph. 2C + 3H2 -84.7 kJmol-1 C2H6 Enthalpy, kJmol-1 -1644.4 kJmol-1 -1559.7 kJmol-1 2CO2 + 3H2O 108

NES/Chemistry/AS Topic 6 - Electrochemistry This topic illustrates the relationship between electricity and chemical changes. 6.1 Redox processes: electron transfer and changes in oxidation number (oxidation state) a) calculate oxidation numbers of elements in compounds and ions b) describe and explain redox processes in terms of electron transfer and changes in oxidation number c) use changes in oxidation numbers to help balance chemical equations 6.2 --A2 Only-- 6.3 --A2 Only-- 6.4 --A2 Only-- Oxidation is loss of electrons, gain of oxygen, or loss of hydrogen Reduction is gain of electrons, loss of oxygen, or gain of hydrogen A redox reaction is when an oxidation reaction and a reduction reaction take place together Reducing Agents causes another species to be reduced - reducing agents themselves are oxidised during the reaction Oxidising Agents cause another species to be oxidised - oxidising agents themselves are reduced during the reaction A disproportionation reaction is when one species is oxidised and reduced (in the same reaction) to form products with two different oxidation numbers 109

NES/Chemistry/AS 6.1 Redox Oxidation numbers can be determined for an atom, or ion in a compound using the following rules: Elements:  Elements have an oxidation number = 0, e.g. Zn  Elemental molecules have an oxidation number = 0, e.g. Cl2 or P4 Ions in compounds:  Group I = +1  Group II = +2  Fluoride = -1  Hydrogen = +1  Hydride (only when bonded to metals) = -1  Oxide = -2  Peroxide = -1 In compounds:  the sum of the oxidation numbers in a compound is zero  the sum of oxidation numbers in an ion is equal to the charge on that ion  the more electronegative element is considered negative For an element (not Transition Metal, or Noble Gas):  The maximum positive oxidation number = Group number  The maximum negative oxidation number = 8 - Group number 110

NES/Chemistry/AS Naming Complexes and Elements with Variable Valency Elements with a variable valency have their charge in brackets after their name.  Example 1: Iron(III) oxide, Fe2O3 Iron(II) oxide, FeO Complexes also have their charge in brackets, the charge is for the non-oxygen element and is positive.  Example 2: Sodium nitrate(III), NaNO2 Sodium nitrate(V), NaNO3 Vanadium(V) nitrate(V), V(NO3)5 Copper(I) sulfate(VI), Cu2SO4 Calculating Oxidation Numbers - Examples  Example 3: Calculate the oxidation numbers for the elements in MgCl2 The overall charge on the compound is zero The charge on Mg = +2 as it is in Group II So the charge on each Cl must be -1  Example 4: Calculate the oxidation numbers for the elements in NO3- The overall charge on the ion is -1 Oxygen is more electronegative, so the charge = -2 So the charge on N = +5  Example 5: Calculate the oxidation numbers for the elements in NaClO The overall charge on the compound is zero The charge on Na = +1 as it is in Group I Oxygen is the most electronegative, so the charge = -2 So the charge on Cl = +1  Example 6: Calculate the oxidation numbers for the elements in H2O2 The overall charge on the compound is zero Hydrogen is less electronegative, so the charge = +1 So the charge on each peroxide = -1 111

NES/Chemistry/AS Changing Oxidation Numbers Oxidation numbers can be calculated in reagents and products in a chemical reaction to see how they have changed. This helps to work out which species has been oxidised and which species has been reduced.  Example 7: List the changes in oxidation numbers for the reaction: CuSO4 + Mg  MgSO4 + Cu Cu has been reduced Cu2+ + 2e-  Cu Mg has been oxidised Mg  Mg2+ + 2e- SO42- has not changed (spectator ion) Spectator ions are ions that are the same (formula, charge and state) left and right. As they do not take part in the reaction, they are ignored in an ionic equation. Reducing Agents and Oxidising Agents Any species that has been reduced is an oxidising agent. Any species that has been oxidised is a reducing agent. In example 7, the Cu2+ ion is a oxidising agent and the Mg atom is a reducing agent. Think of it this way - the copper(II) ion has taken electrons from the magnesium atom. 112

NES/Chemistry/AS Balancing Redox Equations Redox equations can be balanced using the following sequence: 1. Balance the species that is changing 2. Balance the different charges on the species that is changing by adding electrons, e- 3. Balance the total charges left and right in the equation by adding hydrogen ions, H+ 4. Balance the number of water molecules  Example 8: Balance MnO4-  Mn2+ Step 1 Balance the species that is changing Step 2 Step 3 MnO4-  Mn2+ There is already 1 Mn on each side Balance the number of electrons, e- MnO4- + 5e-  Mn2+ 5e- are needed as the charge on the Mn changes from 7+ to 2+ Balance the number of hydrogen ions, H+ MnO4- + 5e- + 8H+  Mn2+ 8H+ are needed to balance the total charges left (6-) and right (2+) Step 4 Balance the number of water molecules MnO4- + 5e- +8H+  Mn2+ + 4H2O 4H2O are required to balance H and O 113

NES/Chemistry/AS  Example 9: Balance Cr2O72-  Cr3+ Step 1 Balance the species that is changing Step 2 Cr2O72-  2Cr3+ Step 3 Balance the number of electrons, e- Cr2O72- + 6e-  2Cr3+ Balance the number of hydrogen ions, H+ Cr2O72- + 6e- + 14H+  2Cr3+ Step 4 Balance the number of water molecules Cr2O72- + 6e- + 14H+  2Cr3+ + 7H2O Sometimes you will get two ionic half-equations to balance. Each half equation is balance separately, then the two equations are combined together.  Example 10: Balance Fe2+ + NO3-  Fe3+ + N2 Step 1 Separate the two half equations Fe2+  Fe3+ NO3-  N2 Step 2 Balance each half equation Fe2+  Fe3+ + e- 2NO3- +10e- +12H+  N2 + 6H2O Step 3 Make both equations have the same number of electrons - multiply the first equation by 5 10Fe2+  10Fe3+ + 10e- 2NO3- + 10e- + 12H+  N2 + 6H2O Step 4 Combine the equations - add anything on the same side, and cancel anything that appears on opposite sides 10Fe2+ + 2NO3- + 12H+  10Fe3+ + N2 + 6H2O 114

NES/Chemistry/AS Topic 7 - Equilibria This topic illustrates that many chemical reactions are reversible and involve an equilibrium process. The consideration of the many factors that can affect an equilibrium is an important aspect of physical chemistry. 7.1 Chemical equilibria: reversible reactions; dynamic equilibrium a) explain, in terms of rates of the forward and reverse reactions, what is meant by a reversible reaction and dynamic equilibrium b) state Le Chatelier’s principle and apply it to deduce qualitatively (from appropriate information) the effects of changes in temperature, concentration or pressure on a system at equilibrium c) state whether changes in temperature, concentration or pressure or the presence of a catalyst affect the value of the equilibrium constant for a reaction d) deduce expressions for equilibrium constants in terms of concentrations, Kc , and partial pressures, Kp (treatment of the relationship between Kp and Kc is not required) e) calculate the values of equilibrium constants in terms of concentrations or partial pressures from appropriate data f) calculate the quantities present at equilibrium, given appropriate data (such calculations will not require the solving of quadratic equations) g) describe and explain the conditions used in the Haber process and the Contact process, as examples of the importance of an understanding of chemical equilibrium in the chemical industry 7.2 Ionic equilibria a) show understanding of, and use, the Brønsted-Lowry theory of acids and bases, including the use of the acid-I base-I, acid-II base-II concept b) explain qualitatively the differences in behaviour between strong and weak acids and bases and the pH values of their aqueous solutions in terms of the extent of dissociation c) - k) & 7.3 --A2 Only-- 115

NES/Chemistry/AS Reversible Reaction is a reaction that can proceed either forwards, or backwards depending on the conditions. Dynamic Equilibrium is when the forward reaction and the backward reaction happen at the same rate. Le Chatelier's Principle states if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. Equilibrium Constant is a numerical value that represents where the equilibrium position is. Less than 1 = equilibrium position lies to the left (more reagent) 1 = equilibrium position lies in the middle (same reagent and product) More than 1 = equilibrium position lies to the right (more product) Partial Pressure is the pressure due to one gas in a mixture of gases. Brønsted-Lowry Theory of Acids and Bases is what we use to identify acids (proton donors) and bases (proton acceptors). 116

NES/Chemistry/AS 7.1 Chemical Equilibria A lot of this content is the same as IGCSE. Those of you who are familiar with the IGCSE book will recognise the content up to 'equilibrium constants'. Reversible Reactions Reagents react together to form products. A+BC+D In some reactions, when the products are made, they can react together to form the reagents. These are called reversible reactions. C+DA+B Instead of writing two separate equations, they are combined into one equation. A+B⇌C+D The ⇌ symbol means a reversible reaction.  Example 1: Hydrated copper(II) sulfate ⇌CuSO4.5H2O(aq) CuSO4(s) + 5H2O(l) Reading the equation forwards: when the hydrated copper(II) sulfate is heated, it decomposes into anhydrous copper(II) sulfate and steam. The process is endothermic and requires heating, which is why the water is steam. Reading the equation backwards: when water is added to anhydrous copper(II) sulfate it forms hydrated copper(II) sulfate. The process is exothermic. 117

NES/Chemistry/AS Equilibrium This is when a reaction is reversible and the forward and backward reactions happen at the same time. All the reagents and products will be present in the equilibrium mixture. At dynamic equilibrium:  Rate of forward reaction = rate of backward reaction  The concentrations of reagents and products are constant A+B⇌C+D Concentration at start maximum minimum (zero) As the reaction happens decreases increases Rate of reaction at start maximum minimum (zero) As the reaction happens decreases increases At some point, the rate of the forward reaction must be equal to the rate of the backward reaction. This is called dynamic equilibrium. Once the rates of reaction are equal, the concentrations will not change any more and become constant. rate of reaction rate of forward reaction decreases with time dynamic equilibrium rate of backward reaction increases with time time (s) 118

NES/Chemistry/AS Even though the forward and backward reactions are still occurring, there is no visible change observed because the rate of the two reactions in opposite directions are equal. Dynamic equilibrium can only be achieved in a closed system (materials cannot enter or leave the system) like a sealed gas jar. Position of Equilibrium The amount of reagent and product in an equilibrium reaction can vary and does not have to be 50% reagent and 50% product. If a reaction has more than 50% reagent, we say that the equilibrium position lies to the left and sometimes the symbol is used. If a reaction has more than 50% product, we say that the equilibrium position lies to the right and sometimes the symbol is used.  Example 2: The Haber Process (for making ammonia) N2(g) + 3H2(g) ⇌ 2NH3(g) The equilibrium position lies to the left. In fact the %Yield (how much product there is) is 15%. This means that there is 85% reagent.  Example 3: The Contact Process (for making sulfuric acid) 2SO2(g) + O2(g) ⇌ 2SO3(g) The equilibrium position lies to the right. In fact the %Yield is 90%. This means there is 90% product and 10% reagent. 119

NES/Chemistry/AS Changing Equilibrium Position The equilibrium position can be moved:  To the left - making more reagent at dynamic equilibrium  To the right - making more product at dynamic equilibrium We can alter the equilibrium position by altering the: 1. Concentration of reagents or products 2. Temperature of the system 3. Pressure (if there are gases as reagents, or products) 4. Adding/removing reagent/product 5. Adding acid/alkali to a reaction with H+/OH- ions Le Chatelier's Principle If you increase the concentration of reagents or products, or increase the temperature, or pressure in a sealed reaction vessel; the equilibrium position shifts in such a way as to try to cancel what you are doing. In other words, the equilibrium position will shift in such a way as to do the opposite of what changes away from room temperature and pressure we do. 120

NES/Chemistry/AS 1. Changing Concentration If we increase the concentration of a reagent the equilibrium position will shift to the right as the system tries to use up the extra reagent that we added. If we reduce the concentration of a reagent then the opposite will happen. If we increase the concentration of a product the equilibrium position will shift to the left as the system tries to use up the extra product that we added. If we reduce the concentration of a product then the opposite will happen. What are we Equilibrium Change in Concentration Concentration doing to the Position will Equilibrium of reagents of products shift to try to: will will system? Position increasing decrease moves from left to right concentration concentration of decrease increase moves from of reagents reagents right to left decreasing increase moves from right to left concentration concentration of increase decrease moves from of reagents reagents left to right increasing decrease concentration concentration of increase decrease of products products decreasing increase concentration concentration of decrease increase of products products 121

NES/Chemistry/AS 2. Changing Temperature Equilibrium reactions are exothermic in one direction and endothermic in the opposite direction. Increasing the temperature will favour the endothermic reaction. Decreasing the temperature will favour the exothermic side. What are we doing to Equilibrium Position Moves in the exothermic/ the system? will shift to try to endothermic direction increasing temperature decrease the temperature endothermic direction of system of the system decreasing temperature increase temperature of exothermic direction of system the system  Example 4: The Haber Process H = -92kJ N2 + 3H2 ⇌ 2NH3 The forward direction has an enthalpy change that is negative, so it is exothermic. Thus the backward reaction will have an enthalpy change that is positive and be endothermic. Increasing the temperature will cause the equilibrium position to shift to the left towards the endothermic side of the equilibrium reaction. Decreasing the temperature will cause the equilibrium position to shift to the right towards the exothermic side of the equilibrium reaction.  Example 5: Reacting a carboxylic acid with an alcohol (esterification) CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O H = 0 kJ Changing the temperature will have no effect on this reaction as there is no change in enthalpy. 122

NES/Chemistry/AS 3. Changing Pressure If there are gases in the reaction then changing the pressure will affect the equilibrium position. Increasing the pressure will favour the side with less moles of gas. Decreasing the pressure will favour the side with more moles of gas. What are we doing to Equilibrium Position Shifts to the side with more mole of gas or the system? shifts to try to: less mole of gas increasing the pressure decrease the pressure less mole of gas decreasing the pressure increase the pressure more mole of gas  Example 6: Thermal decomposition of calcium carbonate CaCO3(s) ⇌ CaO(s) + CO2(g) The left hand side has 0 moles of gas The right hand side has 1 mole of gas Increasing the pressure will shift the equilibrium position to the left. Decreasing the pressure will shift the equilibrium position to the right.  Example 7: Formation of hydrogen iodide H2(g) + I2(g) ⇌ 2HI(g) The left hand side has 2 moles of gas The right hand side has 2 moles of gas Changing the pressure will have no effect on the equilibrium position as the number of moles left and right are equal. 123

NES/Chemistry/AS 4. Adding / Removing some Reagent / Product  Adding reagent - this will cause the equilibrium position to move to the right hand side.  Removing reagent - this will cause the equilibrium position to move to the left hand side.  Adding product - this will cause the equilibrium position to move to the left hand side.  Removing product - this will cause the equilibrium position to move to the right hand side. In each case the equilibrium position will shift to try to negate the changes we do to the system. 124

NES/Chemistry/AS 5. Adding Acid / Alkali to a Reaction with H+ / OH- Ions Equilibrium reactions with H+ ions: Adding acid will increase the concentration of H+ ions and the equilibrium position will shift to the side without H+ ions. Adding alkali will decrease the concentration of H+ ions and the equilibrium position will shift to the side with H+ ions. Equilibrium reactions with OH- ions: Adding acid will decrease the concentration of OH- ions and the equilibrium position will shift to the side with the OH- ions. Adding alkali will increase the concentration of OH- ions and the equilibrium position will shift to the side without the OH- ions. 125

NES/Chemistry/AS  Example 8: Bromine reacting with water Br2(aq) + H2O(l) ⇌ Br–(aq) + 2H+(aq) + OBr–(aq) orange colourless Adding acid will shift the equilibrium position to the left. Adding alkali will shift the equilibrium position to the right. You would see the following colour changes: intense orange colourless orange H+ OH– What are we Equilibrium From left to [Reagents] [Products] doing? position shifts to right or right to increase decrease decrease increase Increasing [H+] try to: left? Decreasing [H+] decrease [H+] right to left by adding OH– increase [H+] left to right 126

NES/Chemistry/AS Adding a Catalyst A catalyst does not change the equilibrium position, but the equilibrium reaction will reach dynamic equilibrium faster. A catalyst increases the rate of both the forward and backward reaction by equal amounts. Different metals, or metal compounds, affect different reactions. So you need to get the right catalyst for a specific reaction. Some reactions are affected by more than one catalyst, and each catalyst will have a different affect on the rate of reaction. Enzymes are biological catalysts and affect specific reactions only. For example protease affects the rate of decomposition of protein. 127

NES/Chemistry/AS The Haber Process for Making Ammonia N2 + 3H2 ⇌ 2NH3 H = -92kJ Temperature Lowering the temperature decreases the rate of reaction but increases the yield of ammonia as equilibrium position moves in the exothermic direction. Increasing the temperature increases the rate of reaction, but decreases the yield of ammonia equilibrium position moves in the endothermic direction. The 450oC temperature used is high enough to give a good rate of reaction, but low enough to give a good yield. Pressure Decreasing the pressure decreases the rate of reaction and decreases the yield of ammonia as equilibrium position moves to the side with more gas mole. Increasing the pressure increases the rate of reaction and increases the yield of ammonia as equilibrium position moves to the side with less gas mole, but the cost of the process increases. The 200 atm pressure used is a compromise between obtaining a reasonable yield with a reasonable rate of reaction at a reasonable cost. Catalyst An iron catalyst is used to increase the rates of the forward and backward reactions, which means the ammonia will be made faster. Recycling the Unused Hydrogen and Nitrogen The hot gases are cooled to liquefy the ammonia. Ammonia has a higher boiling point than nitrogen or hydrogen and so condenses to form a liquid. The unreacted nitrogen and hydrogen gas are recycled by passing over the catalyst again. The yield of ammonia is about 15% of the total gas provided. The liquid ammonia is run off from the reaction vessel. 128

NES/Chemistry/AS The Contact Process for Making Sulphuric Acid One of the steps in the Contact process involves the oxidation of sulphur dioxide. 2SO2 + O2 ⇌ 2SO3 (H negative) Again, optimum conditions are used to balance rate of reaction, %yield and cost. The conditions used are: Temperature: 450oC Pressure: 3atm Catalyst: vanadium(V) oxide, V2O5 Note - although higher pressures could be used, there is considerable extra risk of accidents/harm by having an acidic, corrosive gas at higher pressures. 129

NES/Chemistry/AS Equilibrium Constants Now, the new AS content starts. There are two equilibrium constants: 1. Kc for equations where the concentrations of reagents and products are given. 2. Kp for equations with gases 1. Kc Equilibrium Constant - concentration This is a ratio of the concentration of the products over the reagents. Each concentration is raised to the power of the number of moles in the balanced equation. For an equation: wA + xB ⇌ yC + zD We get the equilibrium expression: The units for Kc are calculated by cancelling down the concentration units. 130

NES/Chemistry/AS  Example 9: Write the equilibrium expression for the following reaction: CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O Note - as the mole ratio is 1:1:1:1, then the values for w,x,y and z are all 1 and not written. The units for concentration moldm-3, so the units for Kc are calculated by cancelling out: The units cancel out completely, this is called dimensionless.  Example 10: Write the equilibrium expression, with units, for the following reaction: H2 + I2 ⇌ 2HI Equilibrium expression: Units: dimensionless 131

NES/Chemistry/AS  Example 11: Write the equilibrium expression, with units, for the Haber process: Equation: N2 + 3H2 ⇌ 2NH3 Equilibrium expression: Units: mol-2dm6 Calculations Using Kc If you are given the concentrations of the reagents and products at dynamic equilibrium, then you can calculate the value for Kc.  Example 12: Calculate Kc, with units, for the hydrolysis of propyl methanoate where the concentration of the reagents is 0.1moldm-3 and the products is 0.6moldm-3. Equation: HCOOC3H7 + H2O ⇌ HCOOH + C3H7OH Equilibrium expression: Units: dimensionless 132

NES/Chemistry/AS If you are given starting concentration, or number of moles and container volume, then Kc can be calculated using the equilibrium expression.  Example 13: Calculate Kc, with units, for the hydrolysis of propyl methanoate where the starting concentration of the reagents is 0.9moldm-3 and the products is 0.0moldm-3. Equation: HCOOC3H7 + H2O ⇌ HCOOH + C3H7OH Start concentrations/moldm-3: 0.9 0.0 0.0 0.9 0.0+X Dynamic equilibrium concentrations/moldm-3: 0.9-X 0.9-X 0.0+X Note - each reagent will decrease in concentration, by an unknown amount, X. As the reagents are in a 1:1 mole ration, then both will decrease by X. The products will increase by an unknown amount and as they are in a 1:1 ratio with the reagents then the unknown value will again be X. Equilibrium expression: Note - for this question, this is as far as we can solve the equation. Units: dimensionless 133

NES/Chemistry/AS  Example 14: Calculate Kc, with units, for the Haber process in a 2dm3 sealed container with the following starting moles: Nitrogen = 0.50mol Hydrogen = 0.80mol Ammonia = 0.10mol Calculate the concentrations of each chemical using the equation Nitrogen = 0.25mol Hydrogen = 0.40mol Ammonia = 0.05mol Write the equation for the reaction: N2 + 3H2 ⇌ 2NH3 Work out expressions for the concentrations at dynamic equilibrium: N2 H2 NH3 0.25-X 0.40-3X 0.05+2X Note - the reagents go down by the mole ratio from the balanced equation. Note - the products go up by the mole ratio from the balanced equation. Write the equilibrium expression: Calculate Kc Note - again, this is as far as we can solve the equation. Units = mol-2dm6 134

NES/Chemistry/AS  Example 15: Use the following information to calculate the concentration of the chemicals at dynamic equilibrium: HCOOC3H7 + H2O ⇌ HCOOH + C3H7OH Kc = 4.00 dimensionless Temperature = 300K Start concentrations : 1 1 0 0 moldm-3 From this information we can use the following steps to calculate the concentrations at dynamic equilibrium: Dynamic equilibrium concentrations 0+X 0+X 1-X 1-X Equilibrium expression: At this point we can take the square root of the whole expression continued on the next page... 135

NES/Chemistry/AS We still have not finished - we must now put this back into the dynamic equilibrium concentrations. HCOOC3H7 + H2O ⇌ HCOOH + C3H7OH 1-0.667 1-0.667 0.667 0.667 0.333 0.333 0.667 0.667 moldm-3 Note - the temperature is not used at all in the calculation. It is given because that is the value of Kc at that specific temperature. If the temperature is changed then the value of Kc will also change. Kc is not affected by any other factor. 136

NES/Chemistry/AS 2. Kp Equilibrium Constant - pressure This is very similar to Kc, however, this equilibrium constant only applies to gases in the chemical equation.  Example 16: Write the equilibrium expression for the following reaction: ⇌H2 (g) + Cl2 (g) 2HCl (g) Equilibrium expression: Units: dimensionless  Example 17: Write the equilibrium expression for the Haber process: N2 + 3H2 ⇌ 2NH3 Equilibrium expression: Units: atm-2 137

NES/Chemistry/AS Calculations Using Kp The sequence of calculations is slightly more complicated and follows the steps: 1. Write the chemical equation for the reaction 2. Write the moles of gas for each chemical at the start of the reaction 3. Write the moles of gas for each chemical at dynamic equilibrium 4. Calculate total moles of gas 5. Calculate the mole fraction for each chemical 6. Calculate the partial pressures for each chemical 7. Use the equilibrium expression to calculate Kp with units. 138

NES/Chemistry/AS  Example 18: When 1.00mol of hydrogen gas and 1.00mol of iodine are allowed to reach equilibrium in a 1dm3 sealed container at 4500C at 3atm, the amount of hydrogen iodide at equilibrium is 1.56mol. Calculate Kp. 1. equation H2 (g) + I2 (g) ⇌ 2HI (g) 2. mol gas start 1.00 1.00 0.00 3. mol gas equil 1-X 1-X 1.56 As the product has increased by 2X (using the mole ratio), then we know that 2X = 1.56, therefore X = 0.78, so mol gas equil 0.22 0.22 1.56 4. total mol gas 2mol 5. mol fraction 0.11 0.11 0.78 using: mol gas / total mol gas 6. partial pressure 0.33 0.33 2.34 using: mol fraction x total pressure 7. dimensionless Again, just like Kc the values for Kp are only affected by changes in temperature. The temperature is only given in the question as a reference to the value of Kp, or Kc. 139

NES/Chemistry/AS 7.2 Ionic Equilibria Concentration of Acids / Alkalis Concentration of an acid is the number of moles of acid molecules per unit volume. The higher the number of moles of acid molecules per unit volume, the higher the concentration. A 1 mol/dm3 solution of HCl(aq) contains 1 mole of HCl in 1dm3 (see Topic 4.2). Concentration of an alkali is the number of moles of alkali molecules per unit volume. The higher the number of moles of alkali molecules per unit volume, the higher the concentration. A 1 mol/dm3 solution of NaOH(aq) contains 1 mole of NaOH in 1dm3. Strength of Acids Acidity is caused by the presence of H+ ions (hydrogen ions) in a solution. Strength is a measure of the degree of ionisation of the acid molecules. The stronger the acid, the greater the degree of ionisation, producing a higher concentration of H+ ions and the lower the pH number. Strong Acids Strong acids completely ionise in water forming H+ ions. Hydrochloric acid is formed by dissolving hydrogen chloride gas in water. All the HCl molecules fully ionise forming the ions H+(aq) and Cl–(aq). HCl(g) + H2O(l)  H+(aq) + Cl–(aq) Weak Acids Weak acids only partially ionise in water forming H+ ions. A weak acid forms fewer H+(aq) ions in comparison to a strong acid of the same concentration. Ethanoic acid is an example of a weak acid. This is an example of a reversible process. CH3COOH ⇌ H+ + CH3COO– 140

NES/Chemistry/AS Concentration and pH A strong acid and a weak acid of the same concentration can have different concentrations of H+ ions and therefore a different pH. This is because a strong acid fully ionised in water producing more H+ ions per unit volume compared to a weak acid, which only partially ionises in water producing fewer H+ ions per unit volume. This means a strong acid is a better electrical conductor than a weak acid due to a higher concentration of ions. Strong and Weak Alkalis Alkalinity is caused by the presence of OH– ions (hydroxide ions) in a solution. When sodium hydroxide is dissolved in water all the NaOH molecules ionise forming the ions OH–(aq) and Na+(aq), which will have a high pH number (12-14). NaOH(s) + H2O(l)  Na+(aq) + OH–(aq) When ammonia dissolves, it only partly ionises, forming less OH- ions. The pH will be lower than that of a strong alkali, from 8-11. NH3(g) + H2O(l) ⇌ NH4+(aq) + OH–(aq) Monoprotic and Diprotic Acids Monoprotic acids (e.g. HCl) produce 1 hydrogen ion per acid molecule, whereas diprotic acids (e.g. H2SO4) produce two hydrogen ions per acid molecule. HCl(g) + H2O(l)  H+(aq) + Cl–(aq) H2SO4(aq) + H2O(l)  2H+(aq) + SO42–(aq) 141

NES/Chemistry/AS Conjugate Acid-Base Pairs As an acid loses a proton, then the species formed must be a base, because under different conditions that base can accept a proton to reform the original acid. This means that every acid will have a conjugate base and every base will have a conjugate acid.  Example 19: Hydrogen chloride HCl (g) + H2O (l) ⇌ H3O+ (aq) + Cl– (aq) The hydronium ion, H3O+ is formed when the H+ ion form a dative covalent bond with a water molecule. It is usually written as H+ (aq). The HCl is behaving as an acid (giving away a proton) so its conjugate base is the Cl- ion. The H2O is behaving as a base (taking in a proton) so its conjugate acid is H3O+.  Example 20: Ammonia NH3 (g) + H2O (l) ⇌ NH4+ (aq) + OH– (aq) The NH3 is behaving as a base (taking in a proton) so its conjugate acid is NH4+. The H2O is behaving as an acid (giving away a proton) so its conjugate base is the OH- ion. You will notice that water is behaving as a base (example 19) and an acid (example 20). Water is in fact amphoteric. 142

NES/Chemistry/AS Topic 8 – Reaction Kinetics The investigation of the factors that affect the rate of a chemical reaction is important in the study of physical chemistry. The temperature and the addition of a catalyst can both affect the progression of a chemical reaction. 8.1 Simple rate equations a) explain and use the term rate of reaction b) explain qualitatively, in terms of collisions, the effect of concentration changes on the rate of a reaction c) -h) --A2 Only-- 8.2 Effect of temperature on reaction rates and the concept of activation energy a) explain and use the term activation energy, including reference to the Boltzmann distribution b) explain qualitatively, in terms both of the Boltzmann distribution and of collision frequency, the effect of temperature change on the rate of a reaction c) --A2 Only-- 8.3 Homogeneous and heterogeneous catalysts including enzymes a) explain and use the term catalysis b) explain that catalysts can be homogenous or heterogeneous c) (i) explain that, in the presence of a catalyst, a reaction has a different mechanism, i.e. one of lower activation energy (ii) interpret this catalytic effect in terms of the Boltzmann distribution d) describe enzymes as biological catalysts (proteins) which may have specificity e) --A2 Only-- 143

NES/Chemistry/AS Rate of Reaction is a measure of how fast a reaction is. It is the rate of change in the concentration of reagents, or products. The units are moldm-3s-1. Activation Energy is the minimum amount of energy required for a successful collision. It is used to break bonds in reagents. Boltzmann Distribution is a graphical way of showing the distribution of energies of particles at a given temperature. It is similar to a distribution curve. Catalysis is the acceleration of a chemical reaction by a catalyst. Enzyme is a biological catalyst (a protein). 144

NES/Chemistry/AS 8.1 Simple Rate Equations and 8.2 Effect of Temperature This topic is very similar to IGCSE level. To explain rates in a sequence that makes sense I have combined these two syllabus statements together, rather than cover each individually. Collision Theory For a reaction to occur, the reagent particles must have a successful collision. A collision is successful if the particles have enough energy to break the bonds of the reagents when they collide, otherwise the collision is unsuccessful. Boltzmann Distribution and Activation Energy The temperature of a substance is proportional to the average kinetic energy of a substance. For any given chemical, there will be a wide range of kinetic energies that the particles have even though the temperature is the same. Think of the temperature as just being an average value.  The x-axis is the kinetic energy of the particles  The y-axis is how many particles have this much energy  The graph starts at 0,0 as no particles have 0J kinetic energy  The peak of the graph is the mode average kinetic energy of the particles  The graph does not come back to the x-axis as there is no maximum value for kinetic energy The particles are constantly colliding, changing direction and kinetic energy. However, each time the kinetic energies of the particles is measured you will get the same shaped graph at that temperature. 145

NES/Chemistry/AS The activation energy, EA is the minimum amount of energy a particle must have to have a successful collision. So, on the graph:  particles to the left of the EA line do not have enough energy to break the bonds in the reagents, so they will have unsuccessful collisions – they will not react  particles to the right of the EA line do have enough energy to break the bonds in the reagents, so they will have successful collisions – they will react The value for activation energy is different for each reaction and is constant. It will not change with changing concentration, or temperature. Rate of Reaction The rate of reaction can be measured experimentally and this data is used to help make deductions about the mechanism for the reaction. The rate of reaction equation is: Or: The units for rate of reaction are moldm-3s-1. 146

NES/Chemistry/AS Effect of Concentration (and Pressure) The rate of reaction is proportional to the concentration of aqueous reagents. The greater the concentration of a reagent, the faster the rate of reaction is. This is because there are more particles per unit volume, so there will be more successful collisions per second. The effect of pressure on a gas is the same as the concentration of a solution. In all reactions, the number of reagent particles reduces during a reaction as they get converted into product. This means that the concentration of reagents decreases with time, therefore the rate of reaction will also decrease with time. Reactions are usually the fastest at the beginning. In some reactions, the product acts as a catalyst for the reaction, so the reaction will actually speed up as the concentration of the catalyst (product) increases. This is called auto-catalysis. 147

Number of Particles NES/Chemistry/AS Effect of Temperature The rate of reaction is also proportional to temperature. For every 100C increase in temperature the rate of reaction will double. This is because there are double the number of particles with kinetic energy greater than activation energy.  T2 is a higher temperature than T1  As the temperature increases, the peak shifts to the right (increase in mode average kinetic energy)  As the temperature increases, the peak gets lower, this is because the area under the graph (total number of particles) must remain constant 148

NES/Chemistry/AS 8.3 Catalysts A catalyst is a substance that speeds up a chemical, but remains unchanged at the end. Most catalysts are transition metals. As they have variable valencies, they can provide an alternative route for the reaction with a lower activation energy. You can see from the shaded area on the graph that with a catalyst there are more particles with energy greater than activation energy, so there are more successful collisions per second. A catalysed reaction usually takes place in two steps: Step 1 The catalyst will react with the reagents, with a lower activation energy than the uncatalysed reaction, forming a transition state. Step 2 The catalyst will change back to its original form and the products are made. 149