JK-TK/Newdata huggingface dataset:
Biological catalysts are called enzymes. A catalyst does not alter the amount of products formed but itself may be altered physically e.g. from solid to powder to fine powder. Like biological enzymes, a catalyst only catalyse specific type of reactions Most industrial catalysts are transition metals or their compounds. Catalyst works by lowering the Enthalpy of activation(∆Ha)/activation energy (Ea) of the reactants .The catalyst lowers the Enthalpy of activation(∆Ha)/activation energy (Ea) by: (i) forming short lived intermediate compounds called activated complex that break up to form the final product/s (ii) being absorbed by the reactants thus providing the surface area on which reaction occurs. A catalyst has no effect on the enthalpy of reaction ∆Hr but only lowers the Enthalpy of activation(∆Ha)/activation energy (Ea)It thus do not affect/influence whether the reaction is exothermic or endothermic as shown in the energy level diagrams below. Energy level diagram showing the activation energy for exothermic processes /reactions. Activated complex Ea Catalysed A A B B Energy kJ Ea uncatalysed A-A B-B A-B A-B
Energy level diagram showing the activation energy for endothermic processes /reactions. Activated complex The following are some catalysed reaction processes. (a)The contact process Vanadium(V) Oxide(V2O5) or platinum(Pt) catalyses the oxidation of sulphur(IV)oxide during the manufacture of sulphuric(VI) acid from contact process. SO2(g) + O2(g) ----V2O5--> SO3(g) To reduce industrial cost of manufacture of sulphuric (VI) acid from contact process Vanadium(V) Oxide(V2O5) is used because it is cheaper though it is easily poisoned by impurities.
Activated complex The following are some catalysed reaction processes. (a)The contact process Vanadium(V) Oxide(V2O5) or platinum(Pt) catalyses the oxidation of sulphur(IV)oxide during the manufacture of sulphuric(VI) acid from contact process. SO2(g) + O2(g) ----V2O5--> SO3(g) To reduce industrial cost of manufacture of sulphuric (VI) acid from contact process Vanadium(V) Oxide(V2O5) is used because it is cheaper though it is easily poisoned by impurities. (b)Ostwalds process A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B ∆Hr
Platinum promoted with Rhodium catalyses the oxidation of ammonia to nitrogen(II)oxide and water during the manufacture of nitric(V)acid 4NH3(g) + 5O2(g) ----Pt/Rh--> 4NO (g) + 6H2O(l) (c)Haber process Platinum or iron catalyses the combination of nitrogen and hydrogen to form ammonia gas N2(g) + 3H2(g) ---Pt or Fe---> 2NH3(g) (d)Hydrogenation/Hardening of oil to fat Nickel (Ni) catalyses the hydrogenation of unsaturated compound containing - C=C- or –C=C- to saturated compounds without double or triple bond This process is used is used in hardening oil to fat. (e)Decomposition of hydrogen peroxide Manganese(IV)oxide speeds up the rate of decomposition of hydrogen peroxide to water and oxygen gas. This process/reaction is used in the school laboratory preparation of Oxygen. 2H2O2 (g) ----MnO2--> O2(g) + 2H2O(l) (f)Reaction of metals with dilute sulphuric(VI)acid Copper(II)sulphate(VI) speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuric(VI)acid. This process/reaction is used in the school laboratory preparation of Hydrogen.
This process/reaction is used in the school laboratory preparation of Oxygen. 2H2O2 (g) ----MnO2--> O2(g) + 2H2O(l) (f)Reaction of metals with dilute sulphuric(VI)acid Copper(II)sulphate(VI) speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuric(VI)acid. This process/reaction is used in the school laboratory preparation of Hydrogen. H2 SO4 (aq) + Zn(s) ----CuSO4--> ZnSO4 (aq) + H2(g) (g) Substitution reactions When placed in bright sunlight or U.V /ultraviolet light , a mixture of a halogen and an alkane undergo substitution reactions explosively to form halogenoalkanes. When paced in diffused sunlight the reaction is very slow. e.g. CH4(g) + Cl2(g) ---u.v. light--> CH3Cl(g) + HCl(g) (h)Photosynthesis Plants convert carbon(IV)oxide gas from the atmosphere and water from the soil to form glucose and oxygen as a byproduct using sunlight / ultraviolet light. 6CO2(g) + 6H2O(l) ---u.v. light--> C6H12O6(g) + O2(g) (i)Photography
Photographic film contains silver bromide emulsion which decomposes to silver and bromine on exposure to sunlight. 2AgBr(s) ---u.v/sun light--> 2Ag(s) + Br2(l) When developed, the silver deposits give the picture of the object whose photograph was taken depending on intensity of light. A picture photographed in diffused light is therefore blurred. Practical determination of effect of catalyst on decomposition of hydrogen peroxide Measure 5cm3 of 20 volume hydrogen peroxide and then dilute to make 40cm3 in a measuring cylinder by adding distilled water. Divide it into two equal portions. (i)Transfer one 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask. Cork and swirl for 2 minutes. Remove the cork. Test the gas produced using a glowing splint. Clean the conical/round bottomed/flat bottomed flask.
Remove the cork. Test the gas produced using a glowing splint. Clean the conical/round bottomed/flat bottomed flask. (ii)Put 2.0g of Manganese (IV) oxide into the clean conical/round bottomed/flat bottomed flask. Stopper the flask. Transfer the second portion of the 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask through the dropping/thistle funnel. Connect the delivery tube to a calibrated/graduated gas jar as in the set up below. Start off the stop watch and determine the volume of gas in the calibrated/graduated gas jar after every 30 seconds to complete Table 1. (iii)Weigh a filter paper .Use the filter paper to filter the contents of the conical conical/round bottomed/flat bottomed flask. Put the residue on a sand bath to dry. Weigh the dry filter paper again .Determine the new mass Manganese (IV) oxide. Mass of MnO2 before reaction(g) Mass of MnO2 after reaction(g) 2.0 2.0 Plot a graph of volume of gas produced against time(x-axes) b) On the same axes, plot a graph of the uncatalysed reaction. (c) Explain the changes in mass of manganese(IV)oxide before and after the reaction. The mass of MnO2 before and after the reaction is the same but a more fine powder after the experiment. A catalyst therefore remains unchanged chemically but may physically change. Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 270.0 Volume of gas (cm3) 0.0 20.0 40.0 60.0 80.0 90.0 95.0 96.0 96.0 96.0 Catalysed reaction Uncatalysed reaction
B.EQUILIBRIA (CHEMICAL CYBERNETICS) Equilibrium is a state of balance. Chemical equilibrium is state of balance between the reactants and products. As reactants form products, some products form back the reactants. Reactions in which the reactants form products to completion are said to be reversible i.e.
Chemical equilibrium is state of balance between the reactants and products. As reactants form products, some products form back the reactants. Reactions in which the reactants form products to completion are said to be reversible i.e. A + B -> C + D Reactions in which the reactants form products and the products can reform the reactants are said to be reversible. A + B C + D Reversible reactions may be: (a)Reversible physical changes (b)Reversible chemical changes (c)Dynamic equilibrium (a)Reversible physical changes Reversible physical change is one which involves: (i) change of state/phase from solid, liquid, gas or aqueous solutions. States of matter are interconvertible and a reaction involving a change from one state/phase can be reversed back to the original. (ii) colour changes. Some substances/compounds change their colours without change in chemical substance. Examples of reversible physical changes (i) colour change on heating and cooling: I. Zinc(II)Oxide changes from white when cool/cold to yellow when hot/heated and back. ZnO(s) ZnO(s) (white when cold) (yellow when hot) II. Lead(II)Oxide changes from yellow when cold/cool to brown when hot/heated and back. PbO(s) PbO(s)
(brown when hot) (yellow when cold) (ii)Sublimation I. Iodine sublimes from a grey crystalline solid on heating to purple vapour. Purple vapour undergoes deposition back to the grey crystalline solid. I2(s) I2(g) (grey crystalline solid (purple vapour undergo sublimation) undergo deposition) II. Carbon (IV)oxide gas undergoes deposition from a colourless gas to a white solid at very high pressures in a cylinder. It sublimes back to the colourless gas if pressure is reduced CO2(s) CO2(g) (white powdery solid (colourless/odourless gas undergo sublimation) undergo deposition) (iii)Melting/ freezing and boiling/condensation Ice on heating undergo melting to form a liquid/water. Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid.
Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid. Melting boiling Freezing condensing (iv)Dissolving/ crystallization/distillation Solid crystals of soluble substances (solutes) dissolve in water /solvents to form a uniform mixture of the solute and solvent/solution. On crystallization /distillation /evaporation the solvent evaporate leaving a solute back. e.g. NaCl(s) + aq NaCl(aq) (b)Reversible chemical changes These are reactions that involve a chemical change of the reactants which can be reversed back by recombining the new substance formed/products. Examples of Reversible chemical changes (i)Heating Hydrated salts/adding water to anhydrous salts. H2O(s) H2O(l) H2O(s)
When hydrated salts are heated they lose some/all their water of crystallization and become anhydrous.Heating an unknown substance /compound that forms a colourless liquid droplets on the cooler parts of a dry test/boiling tube is in fact a confirmation inference that the substance/compound being heated is hydrated. When anhydrous salts are added (back) some water they form hydrated compound/salts. Heating Copper(II)sulphate(VI)pentahydrate and cobalt(II)chloride hexahydrate (i)Heat about 5.0g of Copper(II)sulphate(VI) pentahydrate in a clean dry test tube until there is no further colour change on a small Bunsen flame. Observe any changes on the side of the test/boiling tube. Allow the boiling tube to cool.Add about 10 drops of distilled water. Observe any changes. (ii)Dip a filter paper in a solution of cobalt(II)chloride hexahydrate. Pass one end the filter paper to a small Bunsen flame repeatedly. Observe any changes on the filter paper. Dip the paper in a beaker containing distilled water. Observe any changes. Sample observations Hydrated compound Observation before heating Observation after heating Observation on adding water Copper(II)sulphate (VI) pentahydrate Blue crystalline solid (i)colour changes from blue to white.
Dip the paper in a beaker containing distilled water. Observe any changes. Sample observations Hydrated compound Observation before heating Observation after heating Observation on adding water Copper(II)sulphate (VI) pentahydrate Blue crystalline solid (i)colour changes from blue to white. (ii)colourless liquid forms on the cooler parts of boiling / test tube (i)colour changes from white to blue (ii)boiling tube becomes warm /hot. Cobalt(II)chloride hexahydrate Pink crystalline solid/solution (i)colour changes from pink to blue. (ii) colourless liquid forms on the cooler parts of boiling / test tube (if crystal are used) (i)colour changes from blue to pink (ii)boiling tube becomes warm/hot. When blue Copper(II)sulphate (VI) pentahydrate is heated, it loses the five molecules of water of crystallization to form white anhydrous Copper(II)sulphate (VI).Water of crystallization form and condenses as colourless droplets on the cooler parts of a dry boiling/test tube. This is a chemical change that produces a new substance. On adding drops of water to an anhydrous white copper(II)sulphate(VI) the hydrated compound is
formed back. The change from hydrated to anhydrous and back is therefore reversible chemical change.Both anhydrous white copper(II)sulphate(VI) and blue cobalt(II)chloride hexahydrate are therefore used to test for the presence of water when they turn to blue and pink respectively. CuSO4(s) + 5H2 O(l) CuSO4.5H2 O(s/aq) (white/anhydrous) (blue/hydrated) CoCl2(s) + 6H2 O(l) CoCl2.6H2 O(s/aq) (blue/anhydrous) (pink/hydrated) (ii)Chemical sublimation Some compounds sublime from solid to gas by dissociating into new different compounds. e.g. Heating ammonium chloride (i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made.
Heating ammonium chloride (i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made. When a glass rod containing hydrogen chloride gas is placed near ammonia gas, they react to form ammonium chloride solid that appear as white fumes. This experiment is used interchangeably to test for the presence of hydrogen chloride gas (and hence Cl- ions) and ammonia gas (and hence NH4+ ions) (ii)Put 2.0 g of ammonium chloride in a long dry boiling tube. Place wet / moist /damp blue and red litmus papers separately on the sides of the mouth of the boiling tube. Heat the boiling tube gently then strongly. Explain the observations made. When ammonium chloride is heated it dissociates into ammonia and hydrogen chloride gases. Since ammonia is less dense, it diffuses faster to turn both litmus papers blue before hydrogen chloride turn red because it is denser. The heating and cooling of ammonium chloride is therefore a reversible chemical change. NH3(g) + HCl(g) NH4Cl(s) (Turns moist (Turns moist (forms white fumes) litmus paper blue) litmus paper red)
(c)Dynamic equilibria For reversible reactions in a closed system: (i) at the beginning; -the reactants are decreasing in concentration with time -the products are increasing in concentration with time (ii) after some time a point is reached when as the reactants are forming products the products are forming reactants. This is called equilibrium. Sketch showing the changes in concentration of reactants and products in a closed system For a system in equilibrium: (i) a reaction from left to right (reactants to products) is called forward reaction. (ii) a reaction from right to left (products to reactants) is called backward reaction. (iii)a reaction in which the rate of forward reaction is equal to the rate of backward reaction is called a dynamic equilibrium. A dynamic equilibrium is therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered. Reactants concentration decreases to form products Products concentration increases from time=0.0 Equilibrium established /rate of formation of products equal to rate of formation of reactants.
A dynamic equilibrium is therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered. Reactants concentration decreases to form products Products concentration increases from time=0.0 Equilibrium established /rate of formation of products equal to rate of formation of reactants. Concentration Mole dm-3 Reaction progress/path/coordinate
The influence of different factors on a dynamic equilibrium was first investigated from 1850-1936 by the French Chemist Louis Henry Le Chatellier. His findings were called Le Chatelliers Principle which states that: “if a stress/change is applied to a system in dynamic equilibrium, the system readjust/shift/move/behave so as to remove/ reduce/ counteract/ oppose the stress/change” Le Chatelliers Principle is applied in determining the effect/influence of several factors on systems in dynamic equilibrium. The following are the main factors that influence /alter/ affect systems in dynamic equilibrium: (a)Concentration (b)Pressure (c)Temperature (d)Catalyst (a)Influence of concentration on dynamic equilibrium An increase/decrease in concentration of reactants/products at equilibrium is a stress. From Le Chatelliers principle the system redjust so as to remove/add the excessreduced concentration. Examples of influence of concentration on dynamic equilibrium (i)Chromate(VI)/CrO42- ions in solution are yellow. Dichromate(VI)/Cr2O72- ions in solution are orange. The two solutions exist in equilibrium as in the equation: 2H+ (aq) + 2CrO42- (aq) Cr2O72- (aq) + H2O(l) (Yellow) (Orange) I. If an acid is/H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already H+ ions. The equilibrium shift forward to the right to remove/reduce the excess H+ ions added. Solution mixture becomes More Cr2O72- ions formed in the solution mixture make it to be more orange in colour. II. If a base/OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ ions. H+ ions react with OH- (aq) to form water.
II. If a base/OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ ions. H+ ions react with OH- (aq) to form water. H+ (aq) +OH- (aq) -> H2O(l) The equilibrium shift backward to the left to add/replace the H+ ions that have reacted with the OH- (aq) ions . More of the CrO42- ions formed in the solution mixture makes it to be more yellow in colour. 2OH- (aq) + 2Cr2O72- (aq) CrO42- (aq) + H2O(l) (Orange) (Yellow) I. If an acid/ H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side on the OH- (aq). H+ ions react with OH- (aq) to form water. H+ (aq) +OH- (aq) -> H2O(l) The equilibrium shift backward to the left to add/replace the 2OH- (aq) that have reacted with the H+ (aq) ions . More Cr2O72- (aq)ions formed in the solution mixture makes it to be more Orange in colour. II. If a base /OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already OH- (aq) ions. The equilibrium shift forward to the right to remove/reduce the excess OH- (aq) ions added. More of the Cr2O72- ions are formed in the solution mixture making it to be more orange in colour. (i)Practical determination of the influence of alkali/acid on Cr2O72- / CrO42- equilibrium mixture Measure about 2 cm3 of Potassium dichromate (VI) solution into a test tube. Note that the solution mixture is orange. Add three drops of 2M sulphuric(VI) acid. Shake the mixture carefully. Note that the solution mixture is remains orange. Add about six drops of 2M sodium hydroxide solution. Shake carefully. Note that the solution mixture is turns yellow. Explanation The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium.
Shake carefully. Note that the solution mixture is turns yellow. Explanation The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium. Dichromate(VI) ions are stable in acidic solutions while chromate(VI)ions are stable in basic solutions. An equilibrium exist thus:
OH- H+ When an acid is added, the equilibrium shift forward to the right and the mixture become more orange as more Cr2O72- ions exist. When a base is added, the equilibrium shift backward to the left and the mixture become more yellow as more CrO42- ions exist. (ii)Practical determination of the influence of alkali/acid on bromine water in an equilibrium mixture Measure 2cm3 of bromine water into a boiling tube. Note its colour. Bromine water is yellow Add three drops of 2M sulphuric(VI)acid. Note any colour change Colour becomes more yellow Add seven drops of 2M sodium hydroxide solution. Note any colour change. Solution mixture becomes colourless/Bromine water is decolourized. Explanation When added distilled water,an equilibrium exist between bromine liquid (Br2(aq)) and the bromide ion(Br-), hydrobromite ion(OBr-) and hydrogen ion(H+) as in the equation: H2O(l) + Br2(aq) OBr- (aq) + H+ (aq) + Br- (aq) If an acid (H+)ions is added to the equilibrium mixture, it increases the concentration of the ions on the product side which shift backwards to the left to remove the excess H+ ions on the product side making the colour of the solution mixture more yellow. If a base/alkali OH- is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water. H+ (aq)+ OH-(aq) -> H2O(l) This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/less yellow (Bromine water is decolorized) (iii)Practical determination of the influence of alkali/acid on common acidbase indicators. Cr2O72- CrO42-
Place 2cm3 of phenolphthalein ,methyl orange and litmus solutions each in three separate test tubes.
If a base/alkali OH- is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water. H+ (aq)+ OH-(aq) -> H2O(l) This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/less yellow (Bromine water is decolorized) (iii)Practical determination of the influence of alkali/acid on common acidbase indicators. Cr2O72- CrO42-
Place 2cm3 of phenolphthalein ,methyl orange and litmus solutions each in three separate test tubes. To each test tube add two drops of water. Record your observations in Table 1 below. To the same test tubes, add three drops of 2M sulphuric(VI)acid. Record your observations in Table 1 below. To the same test tubes, add seven drops of 2M sodium hydroxide solution. Record your observations in Table 1 below. To the same test tubes, repeat adding four drops of 2M sulphuric(VI)acid. Table 1 Indicator Colour of indicator in Water Acid(2M sulphuric (VI) acid) Base(2M sodium hydroxide) Phenolphthalein Colourless Colourless Pink Methyl orange Yellow Red Orange Litmus solution Colourless Red Blue Explanation An indicator is a substance which shows whether another substance is an acid , base or neutral. Most indicators can be regarded as very weak acids that are partially dissociated into ions.An equilibrium exist between the undissociated molecules and the dissociated anions. Both the molecules and anions are coloured. 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.