AN INTRODUCTION TO CHEMICAL EQUILIBRIUM KNOCKHARDY PUBLISHING KNOCKHARDY

AN INTRODUCTION TO CHEMICAL EQUILIBRIUM KNOCKHARDY PUBLISHING

KNOCKHARDY PUBLISHING CHEMICAL EQUILIBRIUM INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A 2 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 A 2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at. . . www. argonet. co. uk/users/hoptonj/sci. htm Navigation is achieved by. . . either clicking on the grey arrows at the foot of each page or using the left and right arrow keys on the keyboard

CHEMICAL EQUILIBRIUM 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

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 THE STEEPER THE GRADIENT, THE FASTER THE REACTION In an ordinary reaction; all reactants end up as products; there is 100% conversion SLOWS DOWN AS REACTANTS ARE USED UP TOTAL CONVERSION TO PRODUCTS

EQUILIBRIUM REACTIONS 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. FASTEST AT THE START NO BACKWARD REACTION FORWARD REACTION SLOWS DOWN AS REACTANTS ARE USED UP BACKWARD REACTION STARTS TO INCREASE 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 AT EQUILIBRIUM THE BACKWARD AND FORWARD REACTIONS ARE EQUAL AND OPPOSITE

DYNAMIC EQUILIBRIUM 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 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

THE EQUILIBRIUM LAW 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. Kc the equilibrium values are expressed as concentrations of mol dm -3 Kp the equilibrium values are expressed as partial pressures The partial pressure expression can be used for reactions involving gases

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

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

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

LE CHATELIER’S PRINCIPLE ”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

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM CONCENTRATION 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 Kc 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.

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

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM CONCENTRATION example CH 3 CH 2 OH(l) + CH 3 COOH(l) the equilibrium constant Kc = CH 3 COOC 2 H 5(l) + H 2 O(l) [CH 3 COOC 2 H 5] [H 2 O] = 4 (at 298 K) [CH 3 CH 2 OH] [CH 3 COOH] increasing [CH 3 CH 2 OH] - will make the bottom line larger so Kc 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 CONCENTRATION example CH 3 CH 2 OH(l) + CH 3 COOH(l) the equilibrium constant Kc = CH 3 COOC 2 H 5(l) + H 2 O(l) [CH 3 COOC 2 H 5] [H 2 O] = 4 (at 298 K) [CH 3 CH 2 OH] [CH 3 COOH] increasing [CH 3 CH 2 OH] decreasing [H 2 O] - will make the bottom line larger so Kc 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 - 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 SUMMARY REACTANTS PRODUCTS THE EFFECT OF CHANGING THE CONCENTRATION ON THE POSITION OF EQUILIBRIUM INCREASE CONCENTRATION OF A REACTANT EQUILIBRIUM MOVES TO THE RIGHT DECREASE CONCENTRATION OF A REACTANT EQUILIBRIUM MOVES TO THE LEFT INCREASE CONCENTRATION OF A PRODUCT EQUILIBRIUM MOVES TO THE LEFT DECREASE CONCENTRATION OF A PRODUCT EQUILIBRIUM MOVES TO THE RIGHT

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

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

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM 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 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. THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM INCREASE PRESSURE MOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES DECREASE PRESSURE MOVES TO THE SIDE WITH MORE GASEOUS MOLECULES

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM 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. THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM INCREASE PRESSURE MOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES DECREASE PRESSURE MOVES TO THE SIDE WITH MORE GASEOUS MOLECULES Predict the effect of an increase of pressure on the equilibrium position of. . 2 SO 2(g) + O 2(g) 2 SO 3(g) H 2(g) + CO 2(g) CO(g) + H 2 O(g)

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM 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. THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM INCREASE PRESSURE MOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES DECREASE PRESSURE MOVES TO THE SIDE WITH MORE GASEOUS MOLECULES Predict the effect of an increase of pressure on the equilibrium position of. . 2 SO 2(g) + O 2(g) 2 SO 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 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 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 DH INCREASE TEMP DECREASE TEMP EXOTHERMIC - TO THE LEFT TO THE RIGHT ENDOTHERMIC + TO THE RIGHT TO THE LEFT

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM 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 DH INCREASE TEMP DECREASE TEMP EXOTHERMIC - TO THE LEFT TO THE RIGHT ENDOTHERMIC + TO THE RIGHT TO THE LEFT Predict the effect of a temperature increase on the equilibrium position of. . . H 2(g) + CO 2(g) 2 SO 2(g) + O 2(g) CO(g) + H 2 O(g) 2 SO 3(g) DH = + 40 k. J mol-1 DH = - ive

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM 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 DH INCREASE TEMP DECREASE TEMP EXOTHERMIC - TO THE LEFT TO THE RIGHT ENDOTHERMIC + TO THE RIGHT TO THE LEFT Predict the effect of a temperature increase on the equilibrium position of. . . H 2(g) + CO 2(g) 2 SO 2(g) + O 2(g) CO(g) + H 2 O(g) 2 SO 3(g) DH = + 40 k. J mol-1 DH = - ive moves to the RHS moves to the LHS

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

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM 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 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.

HABER PROCESS N 2(g) + 3 H 2(g) Conditions Pressure Temperature Catalyst 2 NH 3(g) : DH = - 92 k. J mol-1 20000 k. Pa (200 atmospheres) 380 -450°C iron

HABER PROCESS N 2(g) + 3 H 2(g) Conditions Pressure Temperature Catalyst 2 NH 3(g) : DH = - 92 k. J mol-1 20000 k. Pa (200 atmospheres) 380 -450°C iron Equilibrium theory favours low temperature exothermic reaction - higher yield at lower temperature high pressure decrease in number of gaseous molecules

HABER PROCESS N 2(g) + 3 H 2(g) Conditions Pressure Temperature Catalyst 2 NH 3(g) : DH = - 92 k. J mol-1 20000 k. Pa (200 atmospheres) 380 -450°C iron 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 N 2(g) + 3 H 2(g) Conditions Pressure Temperature Catalyst 2 NH 3(g) : DH = - 92 k. J mol-1 20000 k. Pa (200 atmospheres) 380 -450°C iron 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 IMPORTANT USES OF AMMONIA AND ITS COMPOUNDS MAKING FERTILISERS 80% of the ammonia produced goes to make fertilisers such as ammonium nitrate (NITRAM) and ammonium sulphate NH 3 + HNO 3 ——> 2 NH 3 + H 2 SO 4 ——> MAKING NITRIC ACID NH 4 NO 3 (NH 4)2 SO 4 ammonia can be oxidised to nitric acid is used to manufacture. . . fertilisers (ammonium nitrate) explosives (TNT) polyamide polymers (NYLON)

ACIDS AND BASES LEWIS THEORY LEWIS ACID electron pair acceptor H+ , Al. Cl 3 LEWIS BASE electron pair donor NH 3 , H 2 O , C 2 H 5 OH , OH¯ e. g. H 2 O: base H+ acid H 3 N: base BF 3 ——> acid ——> H 3 O + H 3 N+— BF 3¯ see co-ordinate bonding

ACIDS AND BASES BRØNSTED-LOWRY THEORY ACID proton donor HCl ——> H+(aq) + Cl¯(aq) BASE proton acceptor NH 3 (aq) + H+(aq) ——> NH 4+(aq)

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

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

STRONG ACIDS AND BASES STRONG ACIDS e. g. completely dissociate (split up) into ions in aqueous solution HCl ——> H+(aq) + Cl¯(aq) MONOPROTIC 1 replaceable H DIPROTIC 2 replaceable H’s HNO 3 ——> H+(aq) + NO 3¯(aq) H 2 SO 4 ——> 2 H+(aq) + SO 42 -(aq)

STRONG ACIDS AND BASES STRONG ACIDS e. g. completely dissociate (split up) into ions in aqueous solution HCl ——> H+(aq) + Cl¯(aq) MONOPROTIC 1 replaceable H DIPROTIC 2 replaceable H’s HNO 3 ——> H+(aq) + NO 3¯(aq) H 2 SO 4 ——> 2 H+(aq) + SO 42 -(aq) STRONG BASES e. g. completely dissociate into ions in aqueous solution Na. OH ——> Na+(aq) + OH¯(aq)

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

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

WEAK BASES 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 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) + The weaker the base H+ (aq) NH 4+ (aq) the less it dissociates the more the equilibrium lies to the left The relative strengths of bases can be expressed as Kb or p. Kb values.

REACTIONS OF HYDROCHLORIC ACID Is a typical acid in dilute aqueous solution 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. HCl ——> H+ (aq) + Cl¯(aq)

REACTIONS OF HYDROCHLORIC ACID Is a typical acid in dilute aqueous solution 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. HCl ——> H+ (aq) + Cl¯(aq) 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.

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

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

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

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

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

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

REACTIONS OF HYDROCHLORIC ACID Alkalis sodium hydroxide + dil. hydrochloric acid Na. OH(aq) Na+(aq) + OH¯(aq) cancel ions Carbonates + HCl(aq) + H+(aq) + Cl¯(aq) ——> sodium chloride + water ——> Na. Cl(aq) + H 2 O(l) ——> Na+ (aq) + Cl¯(aq) H+(aq) + OH¯(aq) ——> + H 2 O(l) calcium carbonate + hydrochloric acid ——> calcium chloride + carbon dioxide + water Ca. CO 3(s) + 2 HCl(aq) Ca 2+CO 32 -(s) + 2 H+(aq) + 2 Cl¯(aq) cancel ions ——> Ca. Cl 2(aq) + CO 2(g) + H 2 O(l) ——> Ca 2+(aq) + 2 Cl¯(aq) + CO 2(g) + H 2 O(l) CO 32 - + 2 H+(aq) ——> CO 2(g) + H 2 O(l)

REACTIONS OF HYDROCHLORIC ACID Alkalis sodium hydroxide + dil. hydrochloric acid Na. OH(aq) Na+(aq) + OH¯(aq) cancel ions Carbonates + HCl(aq) + H+(aq) + Cl¯(aq) ——> sodium chloride + water ——> Na. Cl(aq) + H 2 O(l) ——> Na+ (aq) + Cl¯(aq) H+(aq) + OH¯(aq) ——> + H 2 O(l) calcium carbonate + hydrochloric acid ——> calcium chloride + carbon dioxide + water Ca. CO 3(s) + 2 HCl(aq) Ca 2+CO 32 -(s) + 2 H+(aq) + 2 Cl¯(aq) cancel ions Hydrogen carbonates ——> Ca. Cl 2(aq) + CO 2(g) + H 2 O(l) ——> Ca 2+(aq) + 2 Cl¯(aq) + CO 2(g) + H 2 O(l) CO 32 - + 2 H+(aq) ——> H+(aq) + HCO 3¯ ——> CO 2(g) + H 2 O(l)

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

REVISION CHECK What should you be able to do? Understand how concentration changes during chemical reactions Recall the characteristics of dynamic equilibrium Construct an expression for the equilibrium constant Kc 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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