JK-TK/Newdata huggingface dataset:
i.e. HIn(aq) H+ (aq) + In- (aq) (undissociated indicator (dissociated indicator molecule(coloured)) molecule(coloured)) When an acid H+ is added to an indicator, the H+ ions increase and equilibrium shift backward to remove excess H+ ions and therefore the colour of the undissociated (HIn) molecule shows/appears. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the H+ ion and therefore the colour of dissociated (In-) molecule shows/appears. The following examples illustrate the above. (i)Phenolphthalein indicator exist as: HPh H+ (aq) + Ph-(aq) (colourless molecule) (Pink anion) On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore colourless. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The pink colour of dissociated (Ph-) molecule shows/appears. (ii)Methyl Orange indicator exists as: HMe H+ (aq) + Me-(aq) (Red molecule) (Yellow/Orange anion) On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore red. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The Orange colour of dissociated (Me-) molecule shows/appears. (b)Influence of Pressure on dynamic equilibrium Pressure affects gaseous reactants/products. Increase in pressure shift/favours the equilibrium towards the side with less volume/molecules.
The Orange colour of dissociated (Me-) molecule shows/appears. (b)Influence of Pressure on dynamic equilibrium Pressure affects gaseous reactants/products. Increase in pressure shift/favours the equilibrium towards the side with less volume/molecules. Decrease in pressure shift the equilibrium towards the side with more volume/molecules. More yield of products is obtained if high pressures produce less molecules / volume of products are formed. If the products and reactants have equal volume/molecules then pressure has no effect on the position of equilibrium The following examples show the influence of pressure on dynamic equilibrium: (i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture
Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas. Chemical equation : 2NO2(g) ===== N2 O4 (g) Gay Lussacs law 2Volume 1Volume Avogadros law 2molecule 1molecule 2 volumes/molecules of Nitrogen(IV)oxide form 1 volumes/molecules of dinitrogen tetraoxide Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules.The equilibrium mixture become more yellow. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The equilibrium mixture become more brown. (ii)Iodine vapour-Hydrogen gas/Hydrogen Iodide mixture. Pure hydrogen gas reacts with Iodine vapour to form Hydrogen Iodide gas. Chemical equation : I2(g) + H2(g) ===== 2HI (g) Gay Lussacs law 1Volume 1Volume 2Volume Avogadros law 1molecule 1molecule 2molecule (1+1) 2 volumes/molecules of Iodine and Hydrogen gasform 2 volumes/molecules of Hydrogen Iodide gas. Change in pressure thus has no effect on position of equilibrium. (iii)Haber process. Increase in pressure of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process.
Change in pressure thus has no effect on position of equilibrium. (iii)Haber process. Increase in pressure of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process. The yield of ammonia is thus favoured by high pressures Chemical equation : N2(g) + 3H2 (g) -> 2NH3 (g) Gay Lussacs law 1Volume 3Volume 2Volume Avogadros law 1molecule 3molecule 2molecule (1 + 3) 4 volumes/molecules of Nitrogen and Hydrogen react to form 2 volumes/molecules of ammonia. Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules. The yield of ammonia increase. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The yield of ammonia decrease. (iv)Contact process. Increase in pressure of the Sulphur(IV)oxide/Oxygen mixture favours the formation of more molecules of Sulphur(VI)oxide gas in Contact process. The yield of Sulphur(VI)oxide gas is thus favoured by high pressures. Chemical equation : 2SO2(g) + O2 (g) -> 2SO3 (g) Gay Lussacs law 2Volume 1Volume 2Volume Avogadros law 2molecule 1molecule 2molecule (2 + 1) 3 volumes/molecules of Sulphur(IV)oxide/Oxygen mixture react to form 2 volumes/molecules of Sulphur(VI)oxide gas. Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules. The yield of Sulphur(VI)oxide gas increase. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The yield of Sulphur(VI)oxide gas decrease. (v)Ostwalds process. Increase in pressure of the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures.
(v)Ostwalds process. Increase in pressure of the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures. Chemical equation : 4NH3(g) + 5O2 (g) -> 4NO(g) + 6H2O (g) Gay Lussacs law 4Volume 5Volume 4Volume 6Volume Avogadros law 4molecule 5molecule 4molecule 6Molecule (4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 10 volumes/molecules of Nitrogen(II)oxide gas and water vapour. Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour decrease. Decrease in pressure shift the equilibrium forward to the right where there is more volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour increase. Note If the water vapour is condensed on cooling, then: Chemical equation : 4NH3(g) + 5O2 (g) -> 4NO(g) + 6H2O (l) Gay Lussacs law 4Volume 5Volume 4Volume 0Volume Avogadros law 4molecule 5molecule 4molecule 0Molecule (4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 4 volumes/molecules of Nitrogen(II)oxide gas and no vapour. Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules. The yield of Nitrogen(II)oxide gas increase. Decrease in pressure shift the equilibrium backward to the left where there is more volume/molecules. The yield of Nitrogen(II)oxide gas decrease. (c)Influence of Temperature on dynamic equilibrium A decrease in temperature favours the reaction that liberate/generate more heat thus exothermic reaction(-ΔH). An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reaction(+ΔH).
The yield of Nitrogen(II)oxide gas decrease. (c)Influence of Temperature on dynamic equilibrium A decrease in temperature favours the reaction that liberate/generate more heat thus exothermic reaction(-ΔH). An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reaction(+ΔH). Endothermic reaction are thus favoured by high temperature/heating Exothermic reaction are favoured by low temperature/cooling. If a reaction/equilibrium mixture is neither exothermic or endothermic, then a change in temperature/cooling/heating has no effect on the equilibrium position. (i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas. Chemical equation : 2NO2(g) ===== N2 O4 (g) On heating /increasing temperature, the mixture becomes more brown. On cooling the mixture become more yellow. This show that (i)the forward reaction to the right is exothermic(-ΔH). On heating an exothermic process the equilibrium shifts to the side that generate /liberate less heat. (ii)the backward reaction to the right is endothermic(+ΔH). On cooling an endothermic process the equilibrium shifts to the side that do not generate /liberate heat. (c)Influence of Catalyst on dynamic equilibrium A catalyst has no effect on the position of equilibrium. It only speeds up the rate of attainment. e.g. Esterification of alkanols and alkanoic acids naturally take place in fruits.In the laboratory concentrated sulphuric(VI)acid catalyse the reaction.The equilibrium mixture forms the ester faster but the yield does not increase. CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l) (d)Influence of rate of reaction and dynamic equilibrium (Optimum conditions) on industrial processes Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium. For manufacturers, obtaining the highest yield at minimum cost and shortest time is paramount.
CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l) (d)Influence of rate of reaction and dynamic equilibrium (Optimum conditions) on industrial processes Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium. For manufacturers, obtaining the highest yield at minimum cost and shortest time is paramount. The conditions required to obtain the highest yield of products within the shortest time at minimum cost are called optimum conditions Optimum condition thus require understanding the effect of various factors on: (i)rate of reaction(Chemical kinetics) (ii)dynamic equilibrium(Chemical cybernetics)
1.Optimum condition in Haber process Chemical equation N2 (g) + 3H2 (g) ===Fe/Pt=== 2NH3 (g) ΔH = -92kJ Equilibrium/Reaction rate considerations (i)Removing ammonia gas once formed shift the equilibrium forward to the right to replace the ammonia. More/higher yield of ammonia is attained. (ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of ammonia is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 500atmospheres is normally used. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -92kJ) . Ammonia formed decomposes back to Nitrogen and Hydrogen to remove excess heat therefore a less yield of ammonia is attained. Very low temperature decrease the collision frequency of Nitrogen and Hydrogen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 450oC is normally used. (iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation.
(iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation. 2.Optimum condition in Contact process Chemical equation 2SO2 (g) + O2 (g) ===V2O5/Pt=== 2SO3 (g) ΔH = -197kJ Equilibrium/Reaction rate considerations
(i)Removing sulphur(VI)oxide gas once formed shift the equilibrium forward to the right to replace the sulphur(VI)oxide. More/higher yield of sulphur(VI) oxide is attained. (ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of sulphur(VI)oxide is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 1-2 atmospheres is normally used to attain about 96% yield of SO3. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -197kJ) . Sulphur(VI)oxide formed decomposes back to Sulphur(IV)oxide and Oxygen to remove excess heat therefore a less yield of Sulphur(VI)oxide is attained. Very low temperature decrease the collision frequency of Sulphur(IV)oxide and Oxygen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 450oC is normally used. (iv)Vanadium(V)Oxide and platinum can be used as catalyst. Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation.
Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation. 3.Optimum condition in Ostwalds process Chemical equation 4NH3 (g) + 5O2 (g) ===Pt/Rh=== 4NO (g) + 6H2O (g) ΔH = -950kJ Equilibrium/Reaction rate considerations (i)Removing Nitrogen(II)oxide gas once formed shift the equilibrium forward to the right to replace the Nitrogen(II)oxide. More/higher yield of Nitrogen(II) oxide is attained. (ii)Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules . Less/lower yield of Nitrogen(II)oxide is attained. Very low pressures increases the distance between reacting NH3and O2 molecules. An optimum pressure of about 9 atmospheres is normally used. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -950kJ) . Nitrogen(II)oxide and water vapour formed decomposes back to Ammonia and Oxygen to remove excess heat therefore a less yield of Nitrogen(II)oxide is attained. Very low temperature decrease the collision frequency of Ammonia and Oxygen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 900oC is normally used. (iv)Platinum can be used as catalyst. Platinum is very expensive.It is: -promoted with Rhodium to increase the surface area/area of contact. -added/coated on the surface of asbestos to form platinized –asbestos to reduce the amount/quantity used. The catalyst does not increase the yield of Nitrogen (II)Oxide but it speed up its rate of formation. C.SAMPLE REVISION QUESTIONS 1.State two distinctive features of a dynamic equilibrium. (i)the rate of forward reaction is equal to the rate of forward reaction (ii)at equilibrium the concentrations of reactants and products do not change. 2.
C.SAMPLE REVISION QUESTIONS 1.State two distinctive features of a dynamic equilibrium. (i)the rate of forward reaction is equal to the rate of forward reaction (ii)at equilibrium the concentrations of reactants and products do not change. 2. Explain the effect of increase in pressure on the following: (i) N2(g) + O2(g) ===== 2NO(g) Gay Lussacs law 1Volume 1Volume 2 Volume Avogadros law 1 molecule 1 molecule 2 molecule 2 volume on reactant side produce 2 volume on product side. Increase in pressure thus have no effect on position of equilibrium. (ii) 2H2(g) + CO(g) ===== CH3OH (g)
Gay Lussacs law 2Volume 1Volume 1 Volume Avogadros law 2 molecule 1 molecule 1 molecule 3 volume on reactant side produce 1 volume on product side. Increase in pressure shift the equilibrium forward to the left. More yield of CH3OH is formed. 4. Explain the effect of increasing temperature on the following: 2SO2(g) + O2 (g) ===== 2SO3 (g) ΔH = -189kJ Forward reaction is exothermic. Increase in temperature shift the equilibrium backward to reduce the excess heat. 5.120g of brass an alloy of copper and Zinc was put it a flask containing dilute hydrochloric acid. The flask was placed on an electric balance. The readings on the balance were recorded as in the table below Time(Seconds) Mass of flask(grams) Loss in mass(grams) 0 600 20 599.50 40 599.12 60 598.84 80 598.66 100 598.54 120 598.50 140 598.50 160 598.50 (a)Complete the table by calculating the loss in mass (b)What does the “600” gram reading on the balance represent The initial mass of brass and the acid before any reaction take place. (c)Plot a graph of Time (x-axes) against loss in mass. (d)Explain the shape of your graph The reaction produce hydrogen gas as one of the products that escape to the atmosphere. This decreases the mass of flask.After 120 seconds,the
react is complete. No more hydrogen is evolved.The mass of flask remain constant.
(d)Explain the shape of your graph The reaction produce hydrogen gas as one of the products that escape to the atmosphere. This decreases the mass of flask.After 120 seconds,the
react is complete. No more hydrogen is evolved.The mass of flask remain constant. (d)At what time was the loss in mass equal to: (i)1.20g Reading from a correctly plotted graph = (ii)1.30g Reading from a correctly plotted graph = (iii)1.40g Reading from a correctly plotted graph = (e)What was the loss in mass at: (i)50oC Reading from a correctly plotted graph = (ii) 70oC Reading from a correctly plotted graph = (iii) 90oC g Reading from a correctly plotted graph =
1 21.0.0 ELECTROCHEMISTRY (25 LESSONS) Electrochemistry can be defined as the study of the effects of electricity on a substance/ compound and how chemical reactions produce electricity. Electrochemistry therefore deals mainly with: i) Reduction and oxidation ii) Electrochemical (voltaic) cell iii) Electrolysis (electrolytic) cell (i)REDUCTION AND OXIDATION (REDOX) 1. In teams of oxygen transfer: i) Reduction is removal of oxygen. ii) Oxidation is addition of oxygen. iii) Redox is simultaneous addition and removal of oxygen. iv) Reducing agent is the species that undergoes oxidation, therefore gains oxygen. v) Oxidizing agent is the species that undergoes reduction, therefore looses/donates oxygen. e.g. When hydrogen is passed through heated copper (II) oxide, it is oxidised to copper metal as in the equation below: CuO (s) + H2 (g) -> Cu (s) + H2O (l)
2 (Oxidising agent) (Reducing agent) 2. In terms of hydrogen transfer: i) Oxidation is the removal of hydrogen. ii) Reduction is the addition of hydrogen. iii) Redox is simultaneous addition and removal of hydrogen. iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates hydrogen. v) Oxidizing agent is the species that undergoes reduction, therefore gains hydrogen. e.g. When hydrogen sulphide gas is bubbled into a gas jar containing chlorine gas it is oxidized (loose the hydrogen) to sulphur (yellow solid).
v) Oxidizing agent is the species that undergoes reduction, therefore gains hydrogen. e.g. When hydrogen sulphide gas is bubbled into a gas jar containing chlorine gas it is oxidized (loose the hydrogen) to sulphur (yellow solid). The chlorine is reduced (gain hydrogen) to hydrogen chlorine gas. Cl2 (g) + H2S (g) -> S(S) + 2HCl (g) (Oxidizing agent) (Reducing agent) 3. In terms of electron transfer: i) Oxidation is donation/ loss/ removal of electrons. ii) Reduction is gain/ accept/ addition of electrons. iii) Redox is simultaneous gain/ accept/ addition and donation/ loss/ removal of electrons. iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates electrons. v) Oxidizing agent is the species that undergoes reduction, therefore gains/ accepts electrons. Example a) Displacement of metals from their solutions: Place 5cm3 each of Iron (II) sulphate (VI) solution into three different test tubes. Add about 1g of copper tunings / powder into one test tube then zinc and magnesium powders separately into the other test tubes. Shake thoroughly for 2 minutes each. Record any colour changes in the table below. Metal added to Iron (II) sulphate (VI) solution Colour changes Copper Solution remains green Zinc Green colour fades Magnesium Green colour fades Explanation -When a more reactive metal is added to a solution of less reactive metal, it displaces it from its solution. 3 -When a less reactive metal is added to a solution of a more reactive metal, it does not displace it from its solution. -Copper is less reactive than iron therefore cannot displace iron its solution. -Zinc is more reactive than iron therefore can displace iron from its solution. -Magnesium is more reactive than iron therefore can displace iron from its solution. In terms of electron transfer: - the more reactive metal undergoes oxidation (reducing agent) by donating/loosing electrons to form ions -the less reactive metal undergoes reduction (oxidizing agent) by its ions in solution gaining /accepting/acquiring the electrons to form the metal. -displacement of metals involves therefore electron transfer from a more reactive metal to ions of another less reactive metal. Examples 1.
In terms of electron transfer: - the more reactive metal undergoes oxidation (reducing agent) by donating/loosing electrons to form ions -the less reactive metal undergoes reduction (oxidizing agent) by its ions in solution gaining /accepting/acquiring the electrons to form the metal. -displacement of metals involves therefore electron transfer from a more reactive metal to ions of another less reactive metal. Examples 1. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Fe2+(aq) + 2e -> Fe(s) (reduction/gain of electrons) Fe2+(aq) + Zn(s) -> Zn2+(aq) + Fe(s) (redox/both donation and gain of electrons) 2. Mg(s) -> Mg2+(aq) + 2e (oxidation/donation of electrons) Fe2+(aq) + 2e -> Fe(s) (reduction/gain of electrons) Fe2+(aq) + Mg(s) -> Mg2+(aq) + Fe(s) (redox/both donation and gain of electrons) 3. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cu2+(aq) + 2e -> Cu(s) (reduction/gain of electrons) Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) (redox/both donation and gain of electrons) 4. Fe(s) -> Fe2+(aq) + 2e (oxidation/donation of electrons) 2Ag+(aq) + 2e -> 2Ag(s) (reduction/gain of electrons) 2Ag+(aq) + Fe(s) -> Fe2+(aq) + 2Ag(s) (redox/both donation and gain of electrons) 5. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cl2(g) + 2e -> 2Cl-(aq) (reduction/gain of electrons) Cl2(g) + Zn(s) -> Zn2+(aq) + 2Cl-(aq) (redox/both donation and gain of electrons) 6.