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AN INTRODUCTION TO CHEMICALEQUILIBRIUM KNOCKHARDY PUBLISHING.

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1 AN INTRODUCTION TO CHEMICALEQUILIBRIUM KNOCKHARDY PUBLISHING

2 INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard KNOCKHARDY PUBLISHING CHEMICAL EQUILIBRIUM

3 CONTENTS Concentration change during a chemical reaction Dynamic equilibrium Equilibrium constants Le Chatelier’s Principle Haber process Acid - base theories Strong acids and bases Weak acids and bases Reactions of hydrochloric acid Check list CHEMICAL EQUILIBRIUM

4 In an ordinary reaction; all reactants end up as products; there is 100% conversion CONCENTRATION CHANGE IN A REACTION As the rate of reaction is dependant on the concentration of reactants... the forward reaction starts off fast but slows as the reactants get less concentrated FASTEST AT THE START SLOWS DOWN AS REACTANTS ARE USED UP TOTAL CONVERSION TO PRODUCTS THE STEEPER THE GRADIENT, THE FASTER THE REACTION

5 Initially, there is no backward reaction but, as products form, it speeds up and provided the temperature remains constant there will come a time when the backward and forward reactions are equal and opposite; the reaction has reached equilibrium. EQUILIBRIUM REACTIONS In an equilibrium reaction, not all the reactants end up as products; there is not a 100% conversion. BUT IT DOESN’T MEAN THE REACTION IS STUCK IN THE MIDDLE FASTEST AT THE START NO BACKWARD REACTION FORWARD REACTION SLOWS DOWN AS REACTANTS ARE USED UP BACKWARD REACTION STARTS TO INCREASE AT EQUILIBRIUM THE BACKWARD AND FORWARD REACTIONS ARE EQUAL AND OPPOSITE

6 IMPORTANT REMINDERS a reversible chemical reaction is a dynamic process everything may appear stationary but the reactions are moving both ways the position of equilibrium can be varied by changing certain conditions Trying to get up a “down” escalator gives an excellent idea of a non-chemical situation involving dynamic equilibrium. DYNAMIC EQUILIBRIUM

7 IMPORTANT REMINDERS a reversible chemical reaction is a dynamic process everything may appear stationary but the reactions are moving both ways the position of equilibrium can be varied by changing certain conditions Trying to get up a “down” escalator gives an excellent idea of a non-chemical situation involving dynamic equilibrium. Summary When a chemical equilibrium is established... both the reactants and the products are present at all times the equilibrium can be approached from either side the reaction is dynamic - it is moving forwards and backwards the concentrations of reactants and products remain constant DYNAMIC EQUILIBRIUM

8 Simply states “If the concentrations of all the substances present at equilibrium are raised to the power of the number of moles they appear in the equation, the product of the concentrations of the products divided by the product of the concentrations of the reactants is a constant, provided the temperature remains constant” There are several forms of the constant; all vary with temperature. K c the equilibrium values are expressed as concentrations of mol dm -3 K p the equilibrium values are expressed as partial pressures The partial pressure expression can be used for reactions involving gases THE EQUILIBRIUM LAW

9 for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C] c. [D] d = a constant, (K c ) [A] a. [B] b where [ ] denotes the equilibrium concentration in mol dm -3 K c is known as the Equilibrium Constant THE EQUILIBRIUM CONSTANT K c

10 for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C] c. [D] d = a constant, (K c ) [A] a. [B] b where [ ] denotes the equilibrium concentration in mol dm -3 K c is known as the Equilibrium Constant THE EQUILIBRIUM CONSTANT K c Example Fe 3+ (aq) + NCS¯(aq) FeNCS 2+ (aq) K c = [ FeNCS 2+ ]with units of dm 3 mol -1 [ Fe 3+ ] [ NCS¯ ]

11 for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C] c. [D] d = a constant, (K c ) [A] a. [B] b where [ ] denotes the equilibrium concentration in mol dm -3 K c is known as the Equilibrium Constant THE EQUILIBRIUM CONSTANT K c VALUE OF K c AFFECTED bya change of temperature NOT AFFECTED bya change in concentration of reactants or products a change of pressure adding a catalyst

12 ”When a change is applied to a system in dynamic equilibrium, the system reacts in such a way as to oppose the effect of the change.” Everyday example A rose bush grows with increased vigour after it has been pruned. Chemistry example If you do something to a reaction that is in a state of equilibrium, the equilibrium position will change to oppose what you have just done LE CHATELIER’S PRINCIPLE

13 CONCENTRATION FACTORS AFFECTING THE POSITION OF EQUILIBRIUM The equilibrium constant is not affected by a change in concentration at constant temperature. To maintain the constant, the composition of the equilibrium mixture changes. If you increase the concentration of a substance, the value of K c will theoretically be affected. As it must remain constant at a particular temperature, the concentrations of the other species change to keep the constant the same.

14 CONCENTRATION example CH 3 CH 2 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) the equilibrium constant K c = [CH 3 COOC 2 H 5 ] [H 2 O] = 4 (at 298K) [CH 3 CH 2 OH] [CH 3 COOH] FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

15 CONCENTRATION example CH 3 CH 2 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) the equilibrium constant K c = [CH 3 COOC 2 H 5 ] [H 2 O] = 4 (at 298K) [CH 3 CH 2 OH] [CH 3 COOH] increasing [CH 3 CH 2 OH]- will make the bottom line larger so K c will be smaller - to keep it constant, some CH 3 CH 2 OH reacts with CH 3 COOH - this reduces the value of the bottom line and increases the top - eventually the value of the constant will be restored FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

16 CONCENTRATION example CH 3 CH 2 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) the equilibrium constant K c = [CH 3 COOC 2 H 5 ] [H 2 O] = 4 (at 298K) [CH 3 CH 2 OH] [CH 3 COOH] increasing [CH 3 CH 2 OH]- will make the bottom line larger so K c will be smaller - to keep it constant, some CH 3 CH 2 OH reacts with CH 3 COOH - this reduces the value of the bottom line and increases the top - eventually the value of the constant will be restored decreasing [H 2 O]- will make the top line smaller - some CH 3 CH 2 OH reacts with CH 3 COOH to replace the H 2 O - more CH 3 COOC 2 H 5 is also produced - this reduces the value of the bottom line and increases the top FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

17 CONCENTRATION example CH 3 CH 2 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) the equilibrium constant K c = [CH 3 COOC 2 H 5 ] [H 2 O] = 4 (at 298K) [CH 3 CH 2 OH] [CH 3 COOH] increasing [CH 3 CH 2 OH]- will make the bottom line larger so K c will be smaller - to keep it constant, some CH 3 CH 2 OH reacts with CH 3 COOH - this reduces the value of the bottom line and increases the top - eventually the value of the constant will be restored decreasing [H 2 O]- will make the top line smaller - some CH 3 CH 2 OH reacts with CH 3 COOH to replace the H 2 O - more CH 3 COOC 2 H 5 is also produced - this reduces the value of the bottom line and increases the top FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

18 SUMMARY REACTANTS PRODUCTS INCREASE CONCENTRATION OF A REACTANTEQUILIBRIUM MOVES TO THE RIGHT THE EFFECT OF CHANGING THE CONCENTRATION ON THE POSITION OF EQUILIBRIUM DECREASE CONCENTRATION OF A REACTANTEQUILIBRIUM MOVES TO THE LEFT INCREASE CONCENTRATION OF A PRODUCTEQUILIBRIUM MOVES TO THE LEFT DECREASE CONCENTRATION OF A PRODUCTEQUILIBRIUM MOVES TO THE RIGHT FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

19 Predict the effect of increasing the concentration of O 2 on the equilibrium position 2SO 2 (g) + O 2 (g) 2SO 3 (g) SUMMARY REACTANTS PRODUCTS INCREASE CONCENTRATION OF A REACTANTEQUILIBRIUM MOVES TO THE RIGHT THE EFFECT OF CHANGING THE CONCENTRATION ON THE POSITION OF EQUILIBRIUM DECREASE CONCENTRATION OF A REACTANTEQUILIBRIUM MOVES TO THE LEFT INCREASE CONCENTRATION OF A PRODUCTEQUILIBRIUM MOVES TO THE LEFT DECREASE CONCENTRATION OF A PRODUCTEQUILIBRIUM MOVES TO THE RIGHT Predict the effect of decreasing the concentration of SO 3 on the equilibrium position

20 FACTORS AFFECTING THE POSITION OF EQUILIBRIUM Predict the effect of increasing the concentration of O 2 on the equilibrium position 2SO 2 (g) + O 2 (g) 2SO 3 (g) EQUILIBRIUM MOVES TO RHS SUMMARY REACTANTS PRODUCTS INCREASE CONCENTRATION OF A REACTANTEQUILIBRIUM MOVES TO THE RIGHT THE EFFECT OF CHANGING THE CONCENTRATION ON THE POSITION OF EQUILIBRIUM DECREASE CONCENTRATION OF A REACTANTEQUILIBRIUM MOVES TO THE LEFT INCREASE CONCENTRATION OF A PRODUCTEQUILIBRIUM MOVES TO THE LEFT DECREASE CONCENTRATION OF A PRODUCTEQUILIBRIUM MOVES TO THE RIGHT Predict the effect of decreasing the concentration of SO 3 on the equilibrium position EQUILIBRIUM MOVES TO RHS

21 PRESSURE When studying the effect of a change in pressure, we consider the number of gaseous molecules only. The more particles you have in a given volume, the greater the pressure they exert. If you apply a greater pressure they will become more crowded (i.e. they are under a greater stress). However, if the system can change it will move to the side with fewer gaseous molecules - it is less crowded. No change occurs when equal numbers of gaseous molecules appear on both sides. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

22 PRESSURE When studying the effect of a change in pressure, we consider the number of gaseous molecules only. The more particles you have in a given volume, the greater the pressure they exert. If you apply a greater pressure they will become more crowded (i.e. they are under a greater stress). However, if the system can change it will move to the side with fewer gaseous molecules - it is less crowded. No change occurs when equal numbers of gaseous molecules appear on both sides. INCREASE PRESSUREMOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES DECREASE PRESSUREMOVES TO THE SIDE WITH MORE GASEOUS MOLECULES THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

23 PRESSURE When studying the effect of a change in pressure, we consider the number of gaseous molecules only. The more particles you have in a given volume, the greater the pressure they exert. If you apply a greater pressure they will become more crowded (i.e. they are under a greater stress). However, if the system can change it will move to the side with fewer gaseous molecules - it is less crowded. No change occurs when equal numbers of gaseous molecules appear on both sides. INCREASE PRESSUREMOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES DECREASE PRESSUREMOVES TO THE SIDE WITH MORE GASEOUS MOLECULES THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM Predict the effect of an increase of pressure on the equilibrium position of.. 2SO 2 (g) + O 2 (g) 2SO 3 (g) H 2 (g) + CO 2 (g) CO(g) + H 2 O(g) FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

24 PRESSURE When studying the effect of a change in pressure, we consider the number of gaseous molecules only. The more particles you have in a given volume, the greater the pressure they exert. If you apply a greater pressure they will become more crowded (i.e. they are under a greater stress). However, if the system can change it will move to the side with fewer gaseous molecules - it is less crowded. No change occurs when equal numbers of gaseous molecules appear on both sides. INCREASE PRESSUREMOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES DECREASE PRESSUREMOVES TO THE SIDE WITH MORE GASEOUS MOLECULES THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM Predict the effect of an increase of pressure on the equilibrium position of.. 2SO 2 (g) + O 2 (g) 2SO 3 (g) MOVES TO RHS :- fewer gaseous molecules H 2 (g) + CO 2 (g) CO(g) + H 2 O(g) NO CHANGE:- equal numbers on both sides FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

25 TEMPERATURE temperature is the only thing that can change the value of the equilibrium constant. altering the temperature affects the rate of both backward and forward reactions it alters the rates to different extents the equilibrium thus moves producing a new equilibrium constant. the direction of movement depends on the sign of the enthalpy change. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

26 TEMPERATURE temperature is the only thing that can change the value of the equilibrium constant. altering the temperature affects the rate of both backward and forward reactions it alters the rates to different extents the equilibrium thus moves producing a new equilibrium constant. the direction of movement depends on the sign of the enthalpy change. REACTION TYPE  INCREASE TEMP DECREASE TEMP EXOTHERMIC - TO THE LEFTTO THE RIGHT ENDOTHERMIC+TO THE RIGHTTO THE LEFT FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

27 TEMPERATURE temperature is the only thing that can change the value of the equilibrium constant. altering the temperature affects the rate of both backward and forward reactions it alters the rates to different extents the equilibrium thus moves producing a new equilibrium constant. the direction of movement depends on the sign of the enthalpy change. REACTION TYPE  INCREASE TEMP DECREASE TEMP EXOTHERMIC - TO THE LEFTTO THE RIGHT ENDOTHERMIC+TO THE RIGHTTO THE LEFT Predict the effect of a temperature increase on the equilibrium position of... H 2 (g) + CO 2 (g) CO(g) + H 2 O(g)  = + 40 kJ mol -1 2SO 2 (g) + O 2 (g) 2SO 3 (g)  = - ive FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

28 TEMPERATURE temperature is the only thing that can change the value of the equilibrium constant. altering the temperature affects the rate of both backward and forward reactions it alters the rates to different extents the equilibrium thus moves producing a new equilibrium constant. the direction of movement depends on the sign of the enthalpy change. REACTION TYPE  INCREASE TEMP DECREASE TEMP EXOTHERMIC - TO THE LEFTTO THE RIGHT ENDOTHERMIC+TO THE RIGHTTO THE LEFT Predict the effect of a temperature increase on the equilibrium position of... H 2 (g) + CO 2 (g) CO(g) + H 2 O(g)  = + 40 kJ mol -1 moves to the RHS 2SO 2 (g) + O 2 (g) 2SO 3 (g)  = - ive moves to the LHS FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

29 CATALYSTS FACTORS AFFECTING THE POSITION OF EQUILIBRIUM EaEa MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULAR ENERGY Catalysts work by providing an alternative reaction pathway involving a lower activation energy.

30 CATALYSTS An increase in temperature is used to speed up chemical reactions but it can have an undesired effect when the reaction is reversible and exothermic. In this case you get to the equilibrium position quicker but with a reduced yield because the increased temperature moves the equilibrium to the left. In many industrial processes a compromise temperature is used (see Haber and Contact Processes). To reduce the problem one must look for a way of increasing the rate of a reaction without decreasing the yield i.e. with a catalyst. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

31 CATALYSTS An increase in temperature is used to speed up chemical reactions but it can have an undesired effect when the reaction is reversible and exothermic. In this case you get to the equilibrium position quicker but with a reduced yield because the increased temperature moves the equilibrium to the left. In many industrial processes a compromise temperature is used (see Haber and Contact Processes). To reduce the problem one must look for a way of increasing the rate of a reaction without decreasing the yield i.e. with a catalyst. Adding a catalyst DOES NOT AFFECT THE POSITION OF EQUILIBRIUM. However, it does increase the rate of attainment of equilibrium. This is especially important in reversible, exothermic industrial reactions such as the Haber or Contact Processes where economic factors are paramount. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

32 N 2 (g) + 3H 2 (g) 2NH 3 (g) :  = - 92 kJ mol -1 ConditionsPressure20000 kPa (200 atmospheres) Temperature °C Catalystiron HABER PROCESS

33 N 2 (g) + 3H 2 (g) 2NH 3 (g) :  = - 92 kJ mol -1 ConditionsPressure20000 kPa (200 atmospheres) Temperature °C Catalystiron Equilibrium theory favours low temperature exothermic reaction - higher yield at lower temperature high pressure decrease in number of gaseous molecules HABER PROCESS

34 N 2 (g) + 3H 2 (g) 2NH 3 (g) :  = - 92 kJ mol -1 ConditionsPressure20000 kPa (200 atmospheres) Temperature °C Catalystiron Equilibrium theory favours low temperature exothermic reaction - higher yield at lower temperature high pressure decrease in number of gaseous molecules Kinetic theory favours high temperature greater average energy + more frequent collisions high pressure more frequent collisions for gaseous molecules catalyst lower activation energy HABER PROCESS

35 N 2 (g) + 3H 2 (g) 2NH 3 (g) :  = - 92 kJ mol -1 ConditionsPressure20000 kPa (200 atmospheres) Temperature °C Catalystiron Equilibrium theory favours low temperature exothermic reaction - higher yield at lower temperature high pressure decrease in number of gaseous molecules Kinetic theory favours high temperature greater average energy + more frequent collisions high pressure more frequent collisions for gaseous molecules catalyst lower activation energy Compromise conditions Which is better? A low yield in a shorter time or a high yield over a longer period. The conditions used are a compromise with the catalyst enabling the rate to be kept up, even at a lower temperature. HABER PROCESS

36 IMPORTANT USES OF AMMONIA AND ITS COMPOUNDS MAKING FERTILISERS80% of the ammonia produced goes to make fertilisers such as ammonium nitrate (NITRAM) and ammonium sulphate NH 3 + HNO 3 ——> NH 4 NO 3 2NH 3 + H 2 SO 4 ——> (NH 4 ) 2 SO 4 MAKING NITRIC ACIDammonia can be oxidised to nitric acid nitric acid is used to manufacture... fertilisers (ammonium nitrate) explosives (TNT) polyamide polymers (NYLON) HABER PROCESS

37 LEWIS THEORY LEWIS ACIDelectron pair acceptorH +, AlCl 3 LEWIS BASEelectron pair donor NH 3, H 2 O, C 2 H 5 OH, OH¯ e.g.H 2 O: H + ——> H 3 O + base acid H 3 N: BF 3 ——> H 3 N + — BF 3 ¯see co-ordinate bonding base acid ACIDS AND BASES

38 BRØNSTED-LOWRY THEORY ACIDproton donorHCl ——> H + (aq) + Cl¯(aq) BASEproton acceptor NH 3 (aq) + H + (aq) ——> NH 4 + (aq) ACIDS AND BASES

39 BRØNSTED-LOWRY THEORY ACIDproton donorHCl ——> H + (aq) + Cl¯(aq) BASEproton acceptor NH 3 (aq) + H + (aq) ——> NH 4 + (aq) Conjugate systems Acids are related to bases ACID PROTON + CONJUGATE BASE Bases are related to acids BASE + PROTON CONJUGATE ACID ACIDS AND BASES

40 BRØNSTED-LOWRY THEORY ACIDproton donorHCl ——> H + (aq) + Cl¯(aq) BASEproton acceptor NH 3 (aq) + H + (aq) ——> NH 4 + (aq) Conjugate systems Acids are related to bases ACID PROTON + CONJUGATE BASE Bases are related to acids BASE + PROTON CONJUGATE ACID For an acid to behave as an acid, it must have a base present to accept a proton... HA + B BH + + A¯ acid base conjugate conjugate acid base example CH 3 COO¯ + H 2 O CH 3 COOH + OH¯ base acid acid base ACIDS AND BASES

41 STRONG ACIDS completely dissociate (split up) into ions in aqueous solution e.g. HCl ——> H + (aq) + Cl¯(aq) MONOPROTIC1 replaceable H HNO 3 ——> H + (aq) + NO 3 ¯(aq) H 2 SO 4 ——> 2H + (aq) + SO 4 2- (aq) DIPROTIC 2 replaceable H’s STRONG ACIDS AND BASES

42 STRONG ACIDS completely dissociate (split up) into ions in aqueous solution e.g. HCl ——> H + (aq) + Cl¯(aq) MONOPROTIC1 replaceable H HNO 3 ——> H + (aq) + NO 3 ¯(aq) H 2 SO 4 ——> 2H + (aq) + SO 4 2- (aq) DIPROTIC 2 replaceable H’s STRONG BASES completely dissociate into ions in aqueous solution e.g. NaOH ——> Na + (aq) + OH¯(aq) STRONG ACIDS AND BASES

43 Weak acids partially dissociate into ions in aqueous solution e.g. ethanoic acid CH 3 COOH(aq) CH 3 COO¯(aq) + H + (aq) When a weak acid dissolves in water an equilibrium is set upHA(aq) + H 2 O(l) A¯(aq) + H 3 O + (aq) The water stabilises the ions To make calculations easier the dissociation can be written... HA(aq) A¯(aq) + H + (aq) WEAK ACIDS

44 Weak acids partially dissociate into ions in aqueous solution e.g. ethanoic acid CH 3 COOH(aq) CH 3 COO¯(aq) + H + (aq) When a weak acid dissolves in water an equilibrium is set upHA(aq) + H 2 O(l) A¯(aq) + H 3 O + (aq) The water stabilises the ions To make calculations easier the dissociation can be written... HA(aq) A¯(aq) + H + (aq) The weaker the acid the less it dissociates the more the equilibrium lies to the left. The relative strengths of acids can be expressed as K a or pK a values The dissociation constant for the weak acid HA is K a = [H + (aq)] [A¯(aq)] mol dm -3 [HA(aq)] WEAK ACIDS

45 Partially react with water to give ions in aqueous solution e.g. ammonia When a weak base dissolves in water an equilibrium is set up NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH¯ (aq) as in the case of acids it is more simply written NH 3 (aq) + H + (aq) NH 4 + (aq) WEAK BASES

46 Partially react with water to give ions in aqueous solution e.g. ammonia When a weak base dissolves in water an equilibrium is set up NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH¯ (aq) as in the case of acids it is more simply written NH 3 (aq) + H + (aq) NH 4 + (aq) The weaker the basethe less it dissociates the more the equilibrium lies to the left The relative strengths of bases can be expressed as K b or pK b values. WEAK BASES

47 Is a typical acid in dilute aqueous solution HCl ——> H + (aq) + Cl¯(aq) REACTIONS OF HYDROCHLORIC ACID Hydrogen chloride is a colourless covalent gas; it is a poor conductor of electricity because there are no free electrons or ions present. It has no action on dry litmus paper because there are no aqueous hydrogen ions present.

48 Is a typical acid in dilute aqueous solution HCl ——> H + (aq) + Cl¯(aq) REACTIONS OF HYDROCHLORIC ACID Hydrogen chloride is a colourless covalent gas; it is a poor conductor of electricity because there are no free electrons or ions present. It has no action on dry litmus paper because there are no aqueous hydrogen ions present. If the gas is passed into water, the hydrogen chloride molecules dissociate into ions. The solution now conducts electricity showing ions are present. For each hydrogen chloride molecule that dissociates one hydrogen ion and one chloride ion are produced. The solution turns litmus paper red because of the H + (aq) ions.

49 Is a typical acid in dilute aqueous solution HCl ——> H + (aq) + Cl¯(aq) REACTIONS OF HYDROCHLORIC ACID Hydrogen chloride is a colourless covalent gas; it is a poor conductor of electricity because there are no free electrons or ions present. It has no action on dry litmus paper because there are no aqueous hydrogen ions present. If the gas is passed into water, the hydrogen chloride molecules dissociate into ions. The solution now conducts electricity showing ions are present. For each hydrogen chloride molecule that dissociates one hydrogen ion and one chloride ion are produced. The solution turns litmus paper red because of the H + (aq) ions.

50 Is a typical acid in dilute aqueous solution HCl ——> H + (aq) + Cl¯(aq) REACTIONS OF HYDROCHLORIC ACID HYDROGEN CHLORIDE HYDROCHLORIC ACID colourless gas Appearance colourless soln. covalent molecule Bonding aqueous ions HCl(g) Formula HCl(aq) poor Conductivity good no reaction Dry blue litmus goes red

51 Is a typical acid in dilute aqueous solution HCl ——> H + (aq) + Cl¯(aq) REACTIONS OF HYDROCHLORIC ACID Appearance Bonding and formula Conductivity Dry litmus hydrogen chloridecolourless gas covalent molecule HCl(g) poor no reaction hydrochloric acidcolourless soln. aqueous ions HCl(aq) good goes red Hydrogen chloride is a colourless covalent gas; it is a poor conductor of electricity because there are no free electrons or ions present. It has no action on dry litmus paper because there are no aqueous hydrogen ions present. If the gas is passed into water, the hydrogen chloride molecules dissociate into ions. The solution now conducts electricity showing ions are present. For each hydrogen chloride molecule that dissociates one hydrogen ion and one chloride ion are produced. The solution turns litmus paper red because of the H + (aq) ions.

52 Metalsmagnesium + dil. hydrochloric acid ——> magnesium chloride + hydrogen Mg(s) + 2HCl(aq)——> MgCl 2 (aq) + H 2 (g) REACTIONS OF HYDROCHLORIC ACID 1. 1.WRITE OUT THE BALANCED EQUATION FOR THE REACTION

53 Metalsmagnesium + dil. hydrochloric acid ——> magnesium chloride + hydrogen Mg(s) + 2HCl(aq)——> MgCl 2 (aq) + H 2 (g) Mg(s) + 2H + (aq) + 2Cl¯(aq) ——> Mg 2+ (aq) + 2Cl¯(aq) + H 2 (g) REACTIONS OF HYDROCHLORIC ACID 1. 1.WRITE OUT THE BALANCED EQUATION FOR THE REACTION 2. 2.DILUTE ACIDS AND SALTS CONTAIN IONS; WATER, HYDROGEN & CARBON DIOXIDE DON’T

54 Metalsmagnesium + dil. hydrochloric acid ——> magnesium chloride + hydrogen Mg(s) + 2HCl(aq)——> MgCl 2 (aq) + H 2 (g) Mg(s) + 2H + (aq) + 2Cl¯(aq) ——> Mg 2+ (aq) + 2Cl¯(aq) + H 2 (g) cancel ions Mg(s) + 2H + (aq) ——> Mg 2+ (aq) + H 2 (g) REACTIONS OF HYDROCHLORIC ACID 1. 1.WRITE OUT THE BALANCED EQUATION FOR THE REACTION 2. 2.DILUTE ACIDS AND SALTS CONTAIN IONS; WATER, HYDROGEN & CARBON DIOXIDE DON’T 3. 3.CANCEL OUT THE IONS WHICH APPEAR ON BOTH SIDES OF THE EQUATION

55 Metalsmagnesium + dil. hydrochloric acid ——> magnesium chloride + hydrogen Mg(s) + 2HCl(aq)——> MgCl 2 (aq) + H 2 (g) Mg(s) + 2H + (aq) + 2Cl¯(aq) ——> Mg 2+ (aq) + 2Cl¯(aq) + H 2 (g) cancel ions Mg(s) + 2H + (aq) ——> Mg 2+ (aq) + H 2 (g) Basic Oxidescopper(II) oxide + dil. hydrochloric acid ——> copper(II) chloride + water CuO(s) + 2HCl(aq) ——> CuCl 2 (aq) + H 2 O(l) Cu 2+ O 2- (s) + 2H + (aq) + 2Cl¯(aq) ——> Cu 2+ (aq) + 2Cl¯(aq) + H 2 O(l) cancel ions O H + (aq) ——> H 2 O(l) REACTIONS OF HYDROCHLORIC ACID

56 Alkalissodium hydroxide + dil. hydrochloric acid ——> sodium chloride + water NaOH(aq) + HCl(aq) ——> NaCl(aq) + H 2 O(l) Na + (aq) + OH¯(aq) + H + (aq) + Cl¯(aq) ——> Na + (aq) + Cl¯(aq) + H 2 O(l) cancel ions H + (aq) + OH¯(aq) ——> H 2 O(l) REACTIONS OF HYDROCHLORIC ACID

57 Alkalissodium hydroxide + dil. hydrochloric acid ——> sodium chloride + water NaOH(aq) + HCl(aq) ——> NaCl(aq) + H 2 O(l) Na + (aq) + OH¯(aq) + H + (aq) + Cl¯(aq) ——> Na + (aq) + Cl¯(aq) + H 2 O(l) cancel ions H + (aq) + OH¯(aq) ——> H 2 O(l) Carbonates calcium carbonate + hydrochloric acid ——> calcium chloride + carbon dioxide + water CaCO 3 (s) + 2HCl(aq) ——> CaCl 2 (aq) + CO 2 (g) + H 2 O(l) Ca 2+ CO 3 2- (s) + 2H + (aq) + 2Cl¯(aq) ——> Ca 2+ (aq) + 2Cl¯(aq) + CO 2 (g) + H 2 O(l) cancel ions CO H + (aq) ——> CO 2 (g) + H 2 O(l) REACTIONS OF HYDROCHLORIC ACID

58 Alkalissodium hydroxide + dil. hydrochloric acid ——> sodium chloride + water NaOH(aq) + HCl(aq) ——> NaCl(aq) + H 2 O(l) Na + (aq) + OH¯(aq) + H + (aq) + Cl¯(aq) ——> Na + (aq) + Cl¯(aq) + H 2 O(l) cancel ions H + (aq) + OH¯(aq) ——> H 2 O(l) Carbonates calcium carbonate + hydrochloric acid ——> calcium chloride + carbon dioxide + water CaCO 3 (s) + 2HCl(aq) ——> CaCl 2 (aq) + CO 2 (g) + H 2 O(l) Ca 2+ CO 3 2- (s) + 2H + (aq) + 2Cl¯(aq) ——> Ca 2+ (aq) + 2Cl¯(aq) + CO 2 (g) + H 2 O(l) cancel ions CO H + (aq) ——> CO 2 (g) + H 2 O(l) Hydrogen carbonates H + (aq) + HCO 3 ¯ ——> CO 2 (g) + H 2 O(l) REACTIONS OF HYDROCHLORIC ACID

59 SUMMARY METALS react to givea salt + hydrogen METAL OXIDES react to givea salt + water METAL HYDROXIDES react to givea salt + water CARBONATES react to give a salt + water + carbon dioxide HYDROGENCARBONATES react to give a salt + water + carbon dioxide AMMONIA reacts to givean ammonium salt REACTIONS OF HYDROCHLORIC ACID

60 REVISION CHECK What should you be able to do? Understand how concentration changes during chemical reactions Recall the the characteristics of dynamic equilibrium Construct an expression for the equilibrium constant K c Recall the factors which can affect the position of equilibrium Apply Le Chatelier’s Principle to predict changes in the position of equilibrium Recall and explain the conditions used in the Haber process Recall the importance of ammonia and its compounds Recall the definitions of and differences between weak acids and bases Recall the properties of hydrochloric acid and its role in salt formation CAN YOU DO ALL OF THESE? YES NO

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62 WELL DONE! Try some past paper questions

63 © 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING AN INTRODUCTION TO CHEMICALEQUILIBRIUM THE END


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