dHydrogenationHardening of oil to fat


Nickel Ni catalyses the hydrogenation of unsaturated compound containing  CC or CC to saturated compounds without double or triple bond


This process is used is used in hardening oil to fat


eDecomposition of hydrogen peroxide


ManganeseIVoxide speeds up the rate of decomposition of hydrogen peroxide to water and oxygen gas


This processreaction is used in the school laboratory preparation of Oxygen


2H2O2 g MnO2 O2g  2H2Ol


fReaction of metals with dilute sulphuricVIacid


CopperIIsulphateVI speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuricVIacid


This processreaction is used in the school laboratory preparation of Hydrogen


H2 SO4 aq  Zns CuSO4 ZnSO4 aq  H2g


g Substitution reactions


When placed in bright sunlight or UV 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


eg CH4g  Cl2g uv light CH3Clg  HClg


hPhotosynthesis


Plants convert carbonIVoxide gas from the atmosphere and water from the soil to form glucose and oxygen as a byproduct using sunlight  ultraviolet light


6CO2g  6H2Ol uv light C6H12O6g  O2g


iPhotography


Photographic film contains silver bromide emulsion which decomposes to silver and bromine on exposure to sunlight


2AgBrs uvsun light 2Ags  Br2l


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


iTransfer one 20cm3volume hydrogen peroxide into a conicalround bottomedflat bottomed flask Cork and swirl for 2 minutes Remove the cork Test the gas produced using a glowing splint Clean the conicalround bottomedflat bottomed flask


iiPut 20g of Manganese IV oxide into the clean conicalround bottomedflat bottomed flask Stopper the flask


Transfer the second portion of the 20cm3volume hydrogen peroxide into a conicalround bottomedflat bottomed flask through the droppingthistle funnel Connect the delivery tube to a calibratedgraduated gas jar as in the set up below


Start off the stop watch and determine the volume of gas in the calibratedgraduated gas jar after every 30 seconds to complete Table 1


iiiWeigh a filter paper Use the filter paper to filter the contents of the conical conicalround bottomedflat bottomed flask Put the residue on a sand bath to dry Weigh the dry filter paper again Determine the new mass Manganese IV oxide


Plot a graph of volume of gas produced against timexaxes


b On the same axes plot a graph of the uncatalysed reaction


c Explain the changes in mass of manganeseIVoxide 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


BEQUILIBRIA 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 ie


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


aReversible physical changes


bReversible chemical changes


cDynamic equilibrium


Reversible physical change is one which involves


i change of statephase from solid liquid gas or aqueous solutions States of matter are interconvertible and a reaction involving a change from one statephase can be reversed back to the original


ii colour changes Some substancescompounds change their colours without change in chemical substance


Examples of reversible physical changes


i colour change on heating and cooling


I ZincIIOxide changes from white when coolcold to yellow when hotheated and back


ZnOs ZnOs


white when cold yellow when hot


II LeadIIOxide changes from yellow when coldcool to brown when hotheated and back


PbOs PbOs


brown when hot yellow when cold


iiSublimation


I Iodine sublimes from a grey crystalline solid on heating to purple vapour Purple vapour undergoes deposition back to the grey crystalline solid


I2s I2g


grey crystalline solid purple vapour


undergo sublimation undergo deposition


II Carbon IVoxide 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


CO2s CO2g


white powdery solid colourlessodourless gas


iiiMelting freezing and boilingcondensation


Ice on heating undergo melting to form a liquidwater Liquidwater on further heating boilvaporizes to form gaswater vapour Gaswater vapour on cooling condensesliquidifies to waterliquid On further cooling liquid water freezes to icesolid


Melting boiling


Freezing condensing


ivDissolving crystallizationdistillation


Solid crystals of soluble substances solutes dissolve in water solvents to form a uniform mixture of the solute and solventsolution On crystallization distillation evaporation the solvent evaporate leaving a solute back eg


NaCls  aq NaClaq


bReversible chemical changes


These are reactions that involve a chemical change of the reactants which can be reversed back by recombining the new substance formedproducts


Examples of Reversible chemical changes


iHeating Hydrated saltsadding water to anhydrous salts


When hydrated salts are heated they lose someall their water of crystallization and become anhydrousHeating an unknown substance compound that forms a colourless liquid droplets on the cooler parts of a dry testboiling tube is in fact a confirmation inference that the substancecompound being heated is hydrated


When anhydrous salts are added back some water they form hydrated compoundsalts


Heating CopperIIsulphateVIpentahydrate and cobaltIIchloride hexahydrate


iHeat about 50g of CopperIIsulphateVI 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 testboiling tube Allow the boiling tube to coolAdd about 10 drops of distilled water Observe any changes


iiDip a filter paper in a solution of cobaltIIchloride 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


When blue CopperIIsulphate VI pentahydrate is heated it loses the five molecules of water of crystallization to form white anhydrous CopperIIsulphate VIWater of crystallization form and condenses as colourless droplets on the cooler parts of a dry boilingtest tube


This is a chemical change that produces a new substance On adding drops of water to an anhydrous white copperIIsulphateVI the hydrated compound is formed back The change from hydrated to anhydrous and back is therefore reversible chemical changeBoth anhydrous white copperIIsulphateVI and blue cobaltIIchloride hexahydrate are therefore used to test for the presence of water when they turn to blue and pink respectively


CuSO4s  5H2 Ol CuSO45H2 Osaq


whiteanhydrous bluehydrated


CoCl2s  6H2 Ol CoCl26H2 Osaq


blueanhydrous pinkhydrated


iiChemical sublimation


Some compounds sublime from solid to gas by dissociating into new different compounds eg


Heating ammonium chloride


iDip 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


iiPut 20 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


NH3g  HClg NH4Cls


Turns moist Turns moist forms white fumes


litmus paper blue litmus paper red


cDynamic 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


iiia 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 disturbedchanged altered


The influence of different factors on a dynamic equilibrium was first investigated from 18501936 by the French Chemist Louis Henry Le Chatellier His findings were called Le Chatelliers Principle which states that


if a stresschange is applied to a system in dynamic equilibrium the system readjustshiftmovebehave so as to remove reduce counteract oppose the stresschange


Le Chatelliers Principle is applied in determining the effectinfluence of several factors on systems in dynamic equilibrium The following are the main factors that influence alter affect systems in dynamic equilibrium


aConcentration


bPressure


cTemperature


dCatalyst


aInfluence of concentration on dynamic equilibrium


An increasedecrease in concentration of reactantsproducts at equilibrium is a stress From Le Chatelliers principle the system redjust so as to removeadd the excessreduced concentration


Examples of influence of concentration on dynamic equilibrium


iChromateVICrO42 ions in solution are yellow DichromateVICr2O72 ions in solution are orange The two solutions exist in equilibrium as in the equation


2H aq  2CrO42 aq Cr2O72 aq  H2Ol


Yellow Orange


I If an acid isH 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 removereduce 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 baseOH 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  H2Ol


The equilibrium shift backward to the left to addreplace 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  H2Ol


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


The equilibrium shift backward to the left to addreplace the 2OH aq that have reacted with the H aq ions  More Cr2O72 aqions 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 removereduce 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


iPractical determination of the influence of alkaliacid 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 sulphuricVI 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 dichromateVIand chromateVI exist in equilibrium DichromateVI ions are stable in acidic solutions while chromateVIions 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


iiPractical determination of the influence of alkaliacid 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 sulphuricVIacid Note any colour change


Colour becomes more yellow


Add seven drops of 2M sodium hydroxide solution Note any colour change


Solution mixture becomes colourlessBromine water is decolourized


When added distilled wateran equilibrium exist between bromine liquid Br2aq and the bromide ionBr hydrobromite ionOBr and hydrogen ionH as in the equation


H2Ol  Br2aq OBr aq  H aq  Br aq


If an acid Hions 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 basealkali OH is added to the equilibrium mixture it reacts with H ions on the product side to form water


H aq OHaq  H2Ol


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 colourlessless yellow Bromine water is decolorized


iiiPractical determination of the influence of alkaliacid on common acidbase indicators


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 sulphuricVIacid 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


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 ionsAn equilibrium exist between the undissociated molecules and the dissociated anions Both the molecules and anions are coloured ie


HInaq H aq  In aq


undissociated indicator dissociated indicator


moleculecoloured moleculecoloured


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 showsappears


When a basealkali OH is added to the indicator the OH reacts with H ions from the dissociated indicator to form water


H aq  OHaq  H2Ol


from indicator from alkalibase


The equilibrium shift forward to the right to replace the H ion and therefore the colour of dissociated In molecule showsappears


The following examples illustrate the above


iPhenolphthalein indicator exist as


HPh H aq  Phaq


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


The equilibrium shift forward to the right to replace the removedreduced H ions The pink colour of dissociated Ph molecule showsappears


iiMethyl Orange indicator exists as


HMe H aq  Meaq


Red molecule YellowOrange anion


On adding an acid equilibrium shift backward to the left to remove excess H ions and the solution mixture is therefore red


The equilibrium shift forward to the right to replace the removedreduced H ions The Orange colour of dissociated Me molecule showsappears


bInfluence of Pressure on dynamic equilibrium


Pressure affects gaseous reactantsproducts Increase in pressure shiftfavours the equilibrium towards the side with less volumemolecules Decrease in pressure shift the equilibrium towards the side with more volumemolecules More yield of products is obtained if high pressures produce less molecules  volume of products are formed


If the products and reactants have equal volumemolecules then pressure has no effect on the position of equilibrium


The following examples show the influence of pressure on dynamic equilibrium


iNitrogenIVoxide Dinitrogen tetroxide mixture


NitrogenIVoxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube NitrogenIVoxide is a brown gas Dinitrogen tetraoxide is a yellow gas


Chemical equation  2NO2g  N2 O4 g


Gay Lussacs law 2Volume 1Volume


Avogadros law 2molecule 1molecule


2 volumesmolecules of NitrogenIVoxide form 1 volumesmolecules of dinitrogen tetraoxide


Increase in pressure shift the equilibrium forward to the left where there is less volumemoleculesThe equilibrium mixture become more yellow


Decrease in pressure shift the equilibrium backward to the right where there is more volumemolecules The equilibrium mixture become more brown


iiIodine vapourHydrogen gasHydrogen Iodide mixture


Pure hydrogen gas reacts with Iodine vapour to form Hydrogen Iodide gas


Chemical equation  I2g  H2g  2HI g


Gay Lussacs law 1Volume 1Volume 2Volume


Avogadros law 1molecule 1molecule 2molecule


11 2 volumesmolecules of Iodine and Hydrogen gasform 2 volumesmolecules of Hydrogen Iodide gas


Change in pressure thus has no effect on position of equilibrium


iiiHaber process


Increase in pressure of the NitrogenHydrogen 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  N2g  3H2 g  2NH3 g


Gay Lussacs law 1Volume 3Volume 2Volume


Avogadros law 1molecule 3molecule 2molecule


1  3 4 volumesmolecules of Nitrogen and Hydrogen react to form 2 volumesmolecules of ammonia


Increase in pressure shift the equilibrium forward to the left where there is less volumemolecules


The yield of ammonia increase


Decrease in pressure shift the equilibrium backward to the right where there is more volumemolecules


The yield of ammonia decrease


ivContact process


Increase in pressure of the SulphurIVoxideOxygen mixture favours the formation of more molecules of SulphurVIoxide gas in Contact process The yield of SulphurVIoxide gas is thus favoured by high pressures


Chemical equation  2SO2g  O2 g  2SO3 g


Gay Lussacs law 2Volume 1Volume 2Volume


Avogadros law 2molecule 1molecule 2molecule


2  1 3 volumesmolecules of SulphurIVoxideOxygen mixture react to form 2 volumesmolecules of SulphurVIoxide gas


Increase in pressure shift the equilibrium forward to the left where there is less volumemolecules The yield of SulphurVIoxide gas increase


Decrease in pressure shift the equilibrium backward to the right where there is more volumemolecules The yield of SulphurVIoxide gas decrease


vOstwalds process


Increase in pressure of the AmmoniaOxygen mixture favours the formation of more molecules of NitrogenIIoxide gas and water vapour in Ostwalds process The yield of NitrogenIIoxide gas and water vapour is thus favoured by low pressures


Chemical equation  4NH3g  5O2 g  4NOg  6H2O g


Gay Lussacs law 4Volume 5Volume 4Volume 6Volume


Avogadros law 4molecule 5molecule 4molecule 6Molecule


4  5 9 volumesmolecules of AmmoniaOxygen mixture react to form 10 volumesmolecules of NitrogenIIoxide gas and water vapour


Increase in pressure shift the equilibrium backward to the left where there is less volumemolecules The yield of NitrogenIIoxide gas and water vapour decrease


Decrease in pressure shift the equilibrium forward to the right where there is more volumemolecules The yield of NitrogenIIoxide gas and water vapour increase


Note


If the water vapour is condensed on cooling then


Chemical equation  4NH3g  5O2 g  4NOg  6H2O l


Gay Lussacs law 4Volume 5Volume 4Volume 0Volume


Avogadros law 4molecule 5molecule 4molecule 0Molecule


4  5 9 volumesmolecules of AmmoniaOxygen mixture react to form 4 volumesmolecules of NitrogenIIoxide gas and no vapour


Increase in pressure shift the equilibrium forward to the right where there is less volumemolecules The yield of NitrogenIIoxide gas increase


Decrease in pressure shift the equilibrium backward to the left where there is more volumemolecules The yield of NitrogenIIoxide gas decrease


cInfluence of Temperature on dynamic equilibrium


A decrease in temperature favours the reaction that liberategenerate more heat thus exothermic reactionH


An increase in temperature favours the reaction that do not liberate generate more heat thus endothermic reactionH


Endothermic reaction are thus favoured by high temperatureheating


Exothermic reaction are favoured by low temperaturecooling


If a reactionequilibrium mixture is neither exothermic or endothermic then a change in temperaturecoolingheating has no effect on the equilibrium position


iNitrogenIVoxide Dinitrogen tetroxide mixture


NitrogenIVoxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube NitrogenIVoxide is a brown gas Dinitrogen tetraoxide is a yellow gas


Chemical equation  2NO2g  N2 O4 g


On heating increasing temperature the mixture becomes more brown On cooling the mixture become more yellow


This show that


ithe forward reaction to the right is exothermicH


On heating an exothermic process the equilibrium shifts to the side that generate liberate less heat


iithe backward reaction to the right is endothermicH


On cooling an endothermic process the equilibrium shifts to the side that do not generate liberate heat


cInfluence of Catalyst on dynamic equilibrium


A catalyst has no effect on the position of equilibrium It only speeds up the rate of attainment eg


Esterification of alkanols and alkanoic acids naturally take place in fruitsIn the laboratory concentrated sulphuricVIacid catalyse the reactionThe equilibrium mixture forms the ester faster but the yield does not increase


CH3CH2OHlCH3COOHl ConcH2SO4 CH3COOCH2CH3aq  H2Ol


dInfluence 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


irate of reactionChemical kinetics


iidynamic equilibriumChemical cybernetics


1Optimum condition in Haber process


Chemical equation


N2 g  3H2 g FePt 2NH3 g H  92kJ


EquilibriumReaction rate considerations


iRemoving ammonia gas once formed shift the equilibrium forward to the right to replace the ammonia Morehigher yield of ammonia is attained


iiIncrease in pressure shift the equilibrium forward to the right where there is less volumemolecules  Morehigher 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


iiiIncrease in temperature shift the equilibrium backward to the left because the reaction is exothermicH  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


ivIron 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 AluminiumOxideAl2O3 to increase its surface areaarea of contact with reactants and thus efficiencyThe catalyst does not increase the yield of ammonia but it speed up its rate of formation


2Optimum condition in Contact process


2SO2 g  O2 g V2O5Pt 2SO3 g H  197kJ


iRemoving sulphurVIoxide gas once formed shift the equilibrium forward to the right to replace the sulphurVIoxide Morehigher yield of sulphurVI oxide is attained


iiIncrease in pressure shift the equilibrium forward to the right where there is less volumemolecules  Morehigher yield of sulphurVIoxide is attained Very high pressures raises the cost of production because they are expensive to produce and maintain An optimum pressure of about 12 atmospheres is normally used to attain about 96 yield of SO3


iiiIncrease in temperature shift the equilibrium backward to the left because the reaction is exothermicH  197kJ  SulphurVIoxide formed decomposes back to SulphurIVoxide and Oxygen to remove excess heat therefore a less yield of SulphurVIoxide is attained Very low temperature decrease the collision frequency of SulphurIVoxide and Oxygen and thus the rate of reaction too slow and uneconomical


ivVanadiumVOxide and platinum can be used as catalyst Platinum is a better catalyst and less easily poisoned by impurities but more expensive VanadiumVOxide is very cheap even if it is easily poisoned by impurities The catalyst does not increase the yield of Sulphur VIOxide but it speed up its rate of formation


3Optimum condition in Ostwalds process


4NH3 g  5O2 g PtRh 4NO g  6H2O g H  950kJ


iRemoving NitrogenIIoxide gas once formed shift the equilibrium forward to the right to replace the NitrogenIIoxide Morehigher yield of NitrogenII oxide is attained


iiIncrease in pressure shift the equilibrium backward to the left where there is less volumemolecules  Lesslower yield of NitrogenIIoxide is attained Very low pressures increases the distance between reacting NH3and O2 molecules


An optimum pressure of about 9 atmospheres is normally used


iiiIncrease in temperature shift the equilibrium backward to the left because the reaction is exothermicH  950kJ  NitrogenIIoxide and water vapour formed decomposes back to Ammonia and Oxygen to remove excess heat therefore a less yield of NitrogenIIoxide 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


ivPlatinum can be used as catalyst Platinum is very expensiveIt is


promoted with Rhodium to increase the surface areaarea of contact


addedcoated on the surface of asbestos to form platinized asbestos to reduce the amountquantity used


The catalyst does not increase the yield of Nitrogen IIOxide but it speed up its rate of formation


CSAMPLE REVISION QUESTIONS


1State two distinctive features of a dynamic equilibrium


ithe rate of forward reaction is equal to the rate of forward reaction


iiat equilibrium the concentrations of reactants and products do not change


2 Explain the effect of increase in pressure on the following


i N2g  O2g  2NOg


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 2H2g  COg  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


2SO2g  O2 g  2SO3 g H  189kJ Forward reaction is exothermic Increase in temperature shift the equilibrium backward to reduce the excess heat


5120g 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


aComplete the table by calculating the loss in mass


bWhat does the 600 gram reading on the balance represent


The initial mass of brass and the acid before any reaction take place


cPlot a graph of Time xaxes against loss in mass


dExplain 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 flaskAfter 120 secondsthe react is complete No more hydrogen is evolvedThe mass of flask remain constant


dAt what time was the loss in mass equal to


Reading from a correctly plotted graph


Readng from a correctly plotted graph


eWhat was the loss in mass at


ii 70oC


iii 90oC g


THERMOCHEMISTRY


1Introduction to Energy changes


Energy is the capacity to do work There are manyvarious forms of energy like heat electric mechanical and or chemical energyThere are two types of energy


iKinetic EnergyKE the energy in motion


iiPotential EnergyPE the storedinternal energy


Energy like matter  is neither created nor destroyed but can be transformed changed from one form to the other is interconvertible This is the principle of conservation of energy eg Electrical energy into heat through a filament in bulb


Chemical and physical processes take place with absorption or evolutionproduction of energy mainly in form of heat


The study of energy changes that accompany physicalchemical reactionchanges is called Thermochemistry Physicalchemical reactionchanges that involve energy changes are called thermochemical reactions The SI unit of energy is the JouleJKilo JouleskJand megaJoulesMJ are also used The JouleJ is defined as the


i quantity of energy transferred when a force of one newton acts through a distance of one metre


ii quantity of energy transferred when one coulomb of electric charge is passed through a potential difference of one volt


All thermochemical reactions should be carried out at standard conditions of


i 298K 25oC temperature


ii101300Pa101300Nm2 760mmHg1 atmosphere pressure


2Exothermic and endothermic processesreactions


Some reactions  processes take place with evolutionproduction of energy They are said to be exothermic while others take place with absorption of energy They are said to be endothermic


Practically exothermic reactions  processes cause a rise in temperature by a rise in thermometer readingmercury or alcohol level rise


Practically endothermic reactions  processes cause a fall in temperature by a fall in thermometer readingmercury or alcohol level decrease


To demonstrateillustrate exothermic and endothermic processesreactions


Dissolving Potassium nitrateVammonium chloride crystals


Procedure


Measure 20cm3 of water in a beaker Determine and record its temperature T1Put about 10g of Potassium nitrateV crystals into the beaker Stir the mixture carefully and note the highest temperature rise fall T2Repeat the whole procedure by using ammonium chloride in place of Potassium nitrate V crystals


Sample results


Note


iInitialT1 temperature of dissolution of both potassium nitrateV crystals and ammonium chloride crystals is higher than the final temperatureT2


ii Change in temperatureT2 T1 is not a mathematical 40 or 30


iiiDissolution of both potassium nitrateV and ammonium chloride crystals is an endothermic process because initialT1 temperature is higher than the final temperatureT2 thus causes a falldrop in temperature


Dissolving concentrated sulphuricVI acidsodium hydroxide crystals


Measure 20cm3 of water in a beaker Determine and record its temperature T1Carefully put about 10gfour pellets of sodium hydroxide crystals into the beaker Stir the mixture carefully and note the highest temperature rise fall T2Repeat the whole procedure by using 2cm3 of concentrated sulphuricVI acid in place of sodium hydroxide crystals


CAUTION


iSodium hydroxide crystals are caustic and cause painful blisters on contact with skin


ii Concentrated sulphuric VI acid is corrosive and cause painful wounds on contact with skin


iInitial T1 temperature of dissolution of both concentrated sulphuric VI acid and sodium hydroxide pellets is lower than the final temperature T2


iiDissolution of both Sodium hydroxide pellets and concentrated sulphuric VI acid is an exothermic process because final T2 temperature is higher than the initial temperature T1 thus causes a rise in temperature


The above reactions show heat loss to and heat gain from the surrounding as illustrated by a rise and fall in temperaturethermometer readings


Dissolving both potassium nitrateV and ammonium chloride crystals causes heat gain from the surrounding that causes fall in thermometer reading


Dissolving both Sodium hydroxide pellets and concentrated sulphuric VI acid causes heat loss to the surrounding that causes rise in thermometer reading


At the same temperature and pressure heat absorbed and released is called enthalpy heat content denoted H


Energy change is measured from the heat contententhalpy of the final and initial products It is denoted Hdelta Hie


Enthalpyenergy change in heat content H  Hfinal  Hinitial


For chemical reactions


H  Hproducts  Hreactants


For exothermic reactions the heat contents of the reactants is more thanhigher than the heat contents of products therefore the H is negative H


For endothermic reactions the heat contents of the reactants is less thanlower than the heat contents of products therefore the H is negative H


Graphically in a sketch energy level diagram


iFor endothermic reactions the heat content of the reactants should be relativelyslightly lower than the heat content of the products


iiFor exothermic reactions the heat content of the reactants should be relativelyslightly higher than the heat content of the products


Sketch energy level diagrams for endothermic dissolution


Energy


kJ H2 KNO3aq


H  H2  H1


H1 KNO3s


Reaction pathcoordinateprogress


kJ H2 NH4Cl aq


H1 NH4Cl s


Sketch energy level diagrams for exothermic dissolution


H2 NaOH s


EnergykJ


H1 NaOH aq


H2 H2SO4 l


kJ


H1 H2SO4 aq


3Energy changes in physical processes


Meltingfreezingfusionsolidification and boilingvaporizationevaporation are the two physical processes Melting freezing point of pure substances is fixed constant The boiling point of pure substance depend on external atmospheric pressure


Meltingfusion is the physical change of a solid to liquid Freezing is the physical change of a liquid to solid


Meltingfreezingfusionsolidification are therefore two opposite but same reversible physical processes ie


A s Al


Boilingvaporizationevaporation is the physical change of a liquid to gasvapour Condensationliquidification is the physical change of gasvapour to liquid Boilingvaporizationevaporation and condensationliquidification are therefore two opposite but same reversible physical processes ie


B l Bg


Practically


i Meltingliquidificationfusion involves heating a solid to weaken the strong bonds holding the solid particles together Solids are made up of very strong bonds holding the particles very close to each other Kinetic Theory of matterOn heating these particles gain energyheat from the surrounding heat source to form a liquid with weaker bonds holding the particles close together but with some degree of freedom Meltingfreezingfusion is an endothermic Hprocess that requireabsorb energy from the surrounding


iiFreezingfusionsolidification involves cooling a a liquid to reform rejoin the very strong bonds to hold the particles very close to each other as solid and thus lose their degree of freedom Kinetic Theory of matter Freezing fusion  solidification is an exothermic Hprocess that require particles holding the liquid together to lose energy to the surrounding


iiiBoilingvaporizationevaporation involves heating a liquid to completely breakfree the bonds holding the liquid particles together Gaseous particles have high degree of freedom Kinetic Theory of matter Boiling vaporization  evaporation is an endothermic H process that requireabsorb energy from the surrounding


ivCondensationliquidification is reverse process of boiling vaporization  evaporationIt involves gaseous particles losing energy to the surrounding to form a liquidIt is an exothermicH process


The quantity of energy required to change one mole of a solid to liquid or to form one mole of a solid from liquid at constant temperature is called molar enthalpylatent heat of fusion eg


H2Os  H2Ol H  60kJ mole1 endothermic process


H2Ol  H2Os H  60kJ mole1 exothermic process


The quantity of energy required to change one mole of a liquid to gasvapour or to form one mole of a liquid from gasvapour at constant temperature is called molar enthalpylatent heat of vapourization eg


H2Ol  H2Og H  440kJ mole1 endothermic process


H2Og  H2Ol H  440kJ mole1 exothermic process


The following experiments illustratedemonstrate practical determination of melting and boiling


To determine the boiling point of water


Procedure


Measure 20cm3 of tap water into a 50cm3 glass beaker Determine and record its temperatureHeat the water on a strong Bunsen burner flame and record its temperature after every thirty seconds for four minutes


Sample results


Questions


1Plot a graph of temperature against timeyaxis


Sketch graph of temperature against time


boiling point


96 oC


timeseconds


2From the graph show and determine the boiling point of water


Note


Water boils at 100oC at sea levelone atmosphere pressure101300Pa but boils at below 100oC at higher altitudes The sample results above are from Kiriari Girls High SchoolEmbu County on the slopes of Mt Kenya in Kenya Water here boils at 96oC


3Calculate the molar heat of vaporization of waterH 10O 16O


Working


Mass of water  density x volume  20 x 1 1000  002kg


Quantity of heat produced


mass of water x specific heat capacity of water x temperature change


002kg x 42 x  96  25   5964kJ


Heat of vaporization of one mole H2O  Quantity of heat


Molar mass of H2O


5964kJ  03313 kJ mole 1


To determine the melting point of candle wax


Procedure


Weigh exactly 50 g of candle wax into a boiling tube Heat it on a strongly Bunsen burner flame until it completely melts Insert a thermometer and remove the boiling tube from the flame Stir continuously Determine and record the temperature after every 30seconds for four minutes


Sample results


Questions


93 oC


Temperature0C melting point


timeseconds


2From the graph show and determine the melting point of the candle wax


4Energy changes in chemical processes


Thermochemical reactions measured at standard conditions of 298K25oC and 101300Pa101300Nm2 1 atmospheres760mmHg76cmHg produce standard enthalpies denoted H


Thermochemical reactions are named from the type of reaction producing the energy change Below are some thermochemical reactions


Standard enthalpyheat of reaction Hr


Standard enthalpyheat of combustion Hc


Standard enthalpyheat of displacement Hd


Standard enthalpyheat of neutralization Hn


Standard enthalpyheat of solutiondissolution Hs


Standard enthalpyheat of formation Hf


aStandard enthalpyheat of reaction Hr


The molar standard enthalpyheat of reaction may be defined as the energyheat change when one mole of products is formed at standard conditions


A chemical reaction involves the reactants forming products For the reaction to take place the bonds holding the reactants must be broken so that new bonds of the products are formed ie


AB  CD  AC  BD


Old Bonds broken AB and CD on reactants


New Bonds formed AC and BD on products


The energy required to break one mole of a covalent bond is called bond dissociation energy The SI unit of bond dissociation energy is kJmole1


The higher the bond dissociation energy the stronger the covalentbond


Bond dissociation energies of some covalentbonds


The molar enthalpy of reaction can be calculated from the bond dissociation energy by


iadding the total bond dissociation energy of the reactantsendothermic processH and total bond dissociation energy of the productsexothermic processH


iisubtracting total bond dissociation energy of the reactants from the total bond dissociation energy of the productsexothermic processH lessminus endothermic processH


Practice examplesCalculating Hr


1Calculate Hr from the following reaction


H2g  Cl2g  2HClg


Working


Old bonds broken endothermic processH


HH  ClCl  431   239   670kJ


New bonds broken exothermic processH


2HCl    428 x 2  856kJ


Hr   670kJ  856kJ  186 kJ  93kJ mole1


The above reaction has negative H enthalpy change and is therefore practically exothermic


The thermochemical reaction is thus


H2g   Cl2g  HClg Hr  93kJ


CH4g  Cl2g  CH3Cl  HClg


4CH  ClCl


4 x 413   239   1891kJ


3CH  HCl  CCl


3 x  413  428  346  2013 kJ


Hr   1891kJ  2013 kJ  122 kJ mole1


CH4g  Cl2g  CH3Clg  HClg H  122 kJ


CH2CH2g  Cl2g  CH3Cl CH3Cl g


Working


Old bonds broken endothermic processH


4CH  ClCl  CC


4 x 413   239 612   2503kJ


New bonds broken exothermic processH


4CH  CC  2CCl


3 x  413  436 2 x 346  2367 kJ


Hr   2503kJ  2367 kJ  136 kJ mole1


The above reaction has negative H enthalpy change and is therefore practically endothermic


The thermochemical reaction is thus


CH2CH2g  Cl2g  CH3Cl CH3Cl g H  136 kJ


Note that


ia reaction is exothermic if the bond dissociation energy of reactants is more than bond dissociation energy of products


iia reaction is endothermic if the bond dissociation energy of reactants is less than bond dissociation energy of products


bStandard enthalpyheat of combustion Hc


The molar standard enthalpyheat of combustionHc is defined as the energyheat change when one mole of a substance is burnt in oxygenexcess air at standard conditions


Burning is the reaction of a substance with oxygenair It is an exothermic process producing a lot of energy in form of heat


A substance that undergoes burning is called a fuel A fuel is defined as the combustible substance which burns in air to give heat energy for domestic or industrial use A fuel may be solid eg coal wood charcoal liquid eg petrol paraffin ethanol kerosene or gas eg liquefied petroleum gasLPG Water gasCO2H2 biogasmethane Natural gasmixture of hydrocarbons


To determine the molar standard enthalpyheat of combustionHc of ethanol


Procedure


Put 20cm3 of distilled water into a 50cm3 beaker Clamp the beaker Determine the temperature of the water T1Weigh an empty burnerempty tin with wick


Record its mass M1Put some ethanol into the burner Weigh again the burner with the ethanol and record its mass M2 Ignite the burner and place it below the clamped 50cm3 beaker Heat the water in the beaker for about one minute Put off the burner Record the highest temperature rise of the water T2 Weigh the burner again and record its mass M3


Sample results


Sample calculations


a T the change in temperature


T  T2  T1  350oC  250oC  100oC


b the mass of ethanol used in burning


mass of ethanol used  M2  M1  291g  287g  04g


c the number of moles of ethanol used in burning


moles of ethanol  mass used  04  00087 87 x 103 moles


molar mass of ethanol 46


2 Given that the specific heat capacity of water is 42 kJ1kg1K1determine the heat produced during the burning


Heat produced H  mass of waterm x specific heat capacity cx T


20 x 42 x 10  840 Joules  084 kJ


3Calculate the molar heat of combustion of ethanol


Molar heat of combustion Hc  Heat produced H


Number of moles of fuel


084 kJ  965517 kJmole1


00087 87 x 103 moles


4List two sources of error in the above experiment


iHeat loss to the surrounding lowers the practical value of the molar heat of combustion of ethanol


A draught shield tries to minimize the loss by protecting wind from wobbling the flame


ii Heat gain by reaction vesselsbeaker lowers T and hence Hc


5Calculate the heating value of the fuel


Heating value  molar heat of combustion  965517 kJmole1  20989 kJg1


Molar mass of fuel 46 g


Heating value is the enrgy produced when a unit massgram of a fuel is completely burnt


6Explain other factors used to determine the choice of fuel for domestic and industrial use


i availability and affordabilitysome fuels are more available cheaply in rural than in urban areas at a lower cost


iicost of storage and transmissiona fuel should be easy to transport and store safely eg LPG is very convenient to store and use Charcoal and wood are bulky


iiienvironmental effects Most fuels after burning produce carbonIV oxide gas as a byproduct CarbonIV oxide gas is green house gas that causes global warming Some other fuel produce acidic gases like sulphurIV oxide and nitrogenIV oxide These gases cause acid rain Internal combustion engines exhaust produce lead vapour from leaded petrol and diesel Lead is carcinogenic


ivignition pointThe temperature at which a fuel must be heated before it burns in air is the ignition point Fuels like petrol have very low ignition point making it highly flammable Charcoal and wood have very high ignition point


7Explain the methods used to reduce pollution from common fuels


iPlanting treesPlants absorb excess carbonIVoxide for photosynthesis and release oxygen gas to the atmosphere


iiusing catalytic converters in internal combustion engines that convert harmfultoxicpoisonous gases like carbonIIoxide and nitrogenIVoxide to harmless nonpoisonous carbonIVoxide water and nitrogen gas by using platinumrhodium catalyst along the engine exhaust pipes


Further practice calculations


1Calculate the heating value of methanol CH3OH given that 087g of the fuel burn in air to raise the temperature of 500g of water from 20oC to 27oCC120H10 O160


Moles of methanol used  Mass of methanol used  087 g  002718 moles


Molar mass of methanol 32


Heat produced H  mass of waterm x specific heat capacity cx T


500 x 42 x 7  14700 Joules  147 kJ


Molar heat of combustion Hc  Heat produced H


Number of moles of fuel


147 kJ  5408389 kJmole1


002718 moles


Heating value  molar heat of combustion  5408389 kJmole1  169012 kJg1


Molar mass of fuel 32 g


2 10 g of carbon burn in excess air to raise the temperature of 400g of water by 18oCDetermine the molar heat of combustion and hence the heating value of carbonC120


Moles of carbon used  Mass of carbon used  10 g  00833 moles


Molar mass of carbon 12


400 x 42 x 18  30240 Joules  3024 kJ


3024 kJ  3630252 kJmole1


00833 moles


Heating value  molar heat of combustion  3630252 kJmole1 302521 kJg1


Molar mass of fuel 12 g


cStandard enthalpyheat of displacement Hd


The molar standard enthalpyheat of displacement Hd is defined as the energyheat change when one mole of a substance is displaced from its solution


A displacement reaction takes place when a more reactive elementwith less electrode potential E  negative E higher in the reactivityelectrochemical series removedisplace another with less reactive elementwith higher electrode potential E  positive E lower in the reactivityelectrochemical series from its solutioneg


iZns  CuSO4aq  Cus  ZnSO4aq


Ionically Zns  Cu2aq  Cus  Zn2 aq


iiFes  CuSO4aq  Cus  FeSO4aq


Ionically Fes  Cu2aq  Cus  Fe2 aq


iiiPbs  CuSO4aq  Cus  PbSO4s


This reaction stops after some time as insoluble PbSO4s coatcover unreacted lead


ivCl2g  2NaBraq  Br2aq  2NaClaq


Ionically Cl2g 2Br aq  Br2aq  2Cl aq


Practically a displacement reaction takes place when a known amount volume of a solution is added excess of a more reactive metal


To determine the molar standard enthalpyheat of displacementHd of copper


Procedure


Place 20cm3 of 02M copperIIsulphateVIsolution into a 50cm3 plastic beakercalorimeter Determine and record the temperature of the solution T1Put all the Zinc powder provided into the plastic beaker Stir the mixture using the thermometer Determine and record the highest temperature change to the nearest 05oC T2  Repeat the experiment to complete table 1 below


Table 1


Questions


1a Calculate


iaverage T


AverageT  change in temperature in experiment I and II


50  60  55oC


iithe number of moles of solution used


Moles used  molarity x volume of solution  02 x 20  0004 moles


1000 1000


iiithe enthalpy change H for the reaction


Heat produced H  mass of solutionm x specific heat capacity cx T


20 x 42 x 55  462 Joules  0462 kJ


ivState two assumptions made in the above calculations


Density of solution  density of water  1gcm3


Specific heat capacity of solutionSpecific heat capacity of solution42 kJ1kg1K


This is because the solution is assumed to be infinite dilute


2 Calculate the enthalpy change for one mole of displacement of Cu2 aq ions


Molar heat of displacement Hd  Heat produced H


Number of moles of fuel


0462 kJ  1155 kJmole1


3Write an ionic equation for the reaction taking place


Zns  Cu2aq  Cus  Zn2aq


4State the observation made during the reaction


Blue colour of copperIIsulphateVI fadesbecomes less bluecolourless


Brown solid deposits are formed at the bottom of reaction vessel beaker


5Illustrate the above reaction using an energy level diagram


Zns  Cu2aq


Energy H  1155 kJmole1


kJ


Cus  Zn2aq


Reaction progresspathcoordinates


6 Iron is less reactive than Zinc Explain the effect of using iron instead of Zinc on the standard molar heat of displacement Hd of copperIIsulphate VI solution


No effect Cu2 aq are displaced from their solutionThe element used to displace it does not matterThe reaction however faster if a more reactive metal is used


7aIf the standard molar heat of displacement Hd of copperIIsulphate VI solution is 209kJmole1 calculate the temperature change if 50cm3 of 02M solution was displaced by excess magnesium


Moles used  molarity x volume of solution  02 x 50  001 moles


1000 1000


Heat produced H  Molar heat of displacement Hd x Number of moles


209kJmole1x 001 moles  209 kJ


T change in temperature  Heat produced H Molar heat of displacement Hd x Number of moles


209 kJ  99524Kelvin


001 moles


bDraw an energy level diagram to show the above energy changes


Mgs  Cu2aq


Energy H  209 kJmole1


Cus  Mg2aq


8 The enthalpy of displacement Hd of copperIIsulphate VI solution is 12k6kJmole1Calculate the molarity of the solution given that 40cm3 of this solution produces 2204kJ of energy during a displacement reaction with excess iron filings


Number of moles  Heat produced H Molar heat of displacement Hd


2204 kJ  00206moles


126 moles


Molarity of the solution  moles x 1000 Volume of solution used


00206moles x 1000  05167 M


9 If the molar heat of displacement of ZincIInitrateVby magnesium powder is 2505kJmole1 calculate the volume of solution which must be added 05 moles solution if there was a 30K rise in temperature


2508kJmole1x 05 moles  1254 kJ x 1000 1254J


Mass of solution m  Heat produced H


specific heat capacity cx T


1254J  995238 g


42 x 3


Volume  mass x density  995238 g x 1  995238cm3


Note The solution assumes to be too dilute infinite dilute such that the density and specific heat capacity is assumed to be that of water


Graphical determination of the molar enthalpy of displacement of copper


Procedure


Place 20cm3 of 02M copperIIsulphate VI solution into a calorimeter50cm3 of plastic beaker wrapped in cotton wooltissue paper


Record its temperature at time T 0


Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds


Place all the 15g Zinc powder provided


Stir the solution with the thermometer carefully and continue recording the temperature after every 30 seconds for five minutes


Determine the highest temperature change to the nearest 05oC


Sample results


Sketch graph of temperature against time


Extrapolation


Temperature point T


oC


130 Timeseconds


Questions


Show and determine the change in temperature T


From a well constructed graph T T2 T1 at 150 second by extrapolation


T  365  250  115oC


2Calculate the number of moles of copperII sulphateVIused given the molar heat of displacement of Cu2 aqions is 125kJmole1


Heat produced H  mass of solutionm x specific heat capacity cx T


20 x 42 x 115  966 Joules  0966 kJ


966 kJ  0007728moles


125 moles 7728 x 103moles


What was the concentration of copperIIsulphateVI in moles per litre


Molarity  moles x 1000  7728 x 103moles x 1000  03864M


Volume used 20


4The actual concentration of copperIIsulphateVI solution was 04MExplain the differences between the two


Practical value is lower than theoretical Heatenergy loss to the surrounding and that absorbed by the reaction vessel decreases T hence lowering the practical number of moles and molarity against the theoretical value


5a In an experiment to determine the molar heat of reaction when magnesium displaces copper 015g of magnesium powder were added to 250cm3 of 20M copper II chloride solution The temperature of copper II chloride solution was 25oCWhile that of the mixture was 43oC


iOther than increase in temperature state and explain the observations which were made during the reaction3mks


iiCalculate the heat change during the reaction specific heat capacity of the solution  42jg1k1and the density of the solution  1gcm32mks


iiiDetermine the molar heat of displacement of copper by magnesiumMg240


ivWrite the ionic equation for the reaction1mk


vSketch an energy level diagram for the reaction2mks


cStandard enthalpyheat of neutralization Hn


The molar standard enthalpyheat of neutralization Hn is defined as the energyheat change when one mole of a H H3Oions react completely with one mole of OH ions to form one mole of H2Owater


Neutralization is thus a reaction of an acid H H3Oions with a basealkali OH ions to form salt and water only


Strong acidsbasesalkalis are completely dissociated to many free ionsH H3O and OH ions


Weak acidsbasesalkalis are partially dissociated to few free ionsH H3O and OH ions and exist more as molecules


Neutralization is an exothermicH processThe enrgy produced during neutralization depend on the amount of free ions H H3O and OHions existing in the acidbasealkali reactant


ifor weak acidbasealkali neutralizationsome of the energy is used to dissociate ionize the molecule into free H H3O and OH ions therefore the overall energy evolved is comparatively lowerlessersmaller than strong acid  base alkali neutralizations


ii ifor strong acidbasealkali neutralization no energy is used to dissociate ionize since molecule is whollyfully dissociatedionized into free H H3O and OH ionsThe overall energy evolved is comparatively highermore than weak acidbase alkali neutralizations For strong acidbasealkali neutralization the enthalpy of neutralization is constant at about 573kJmole1 irrespective of the acidbase used This is because ionically


OHaq Haq  H2Ol for any wholly dissociated acidbasealkali


Practically Hn can be determined as in the examples below


To determine the molar enthalpy of neutralization Hn of Hydrochloric acid


Procedure


Place 50cm3 of 2M hydrochloric acid into a calorimeter200cm3 plastic beaker wrapped in cotton wooltissue paper Record its temperature T1Using a clean measuring cylinder measure another 50cm3 of 2M sodium hydroxide Rinse the bulb of the thermometer in distilled water Determine the temperature of the sodium hydroxide T2Average T2 andT1 to get the initial temperature of the mixture T3


Carefully add all the alkali into the calorimeter200cm3 plastic beaker wrapped in cotton wooltissue paper containing the acid Stir vigorously the mixture with the thermometer


Determine the highest temperature change to the nearest 05oC T4 as the final temperature of the mixture Repeat the experiment to complete table 1


Table I  Sample results


aCalculate T6 the average temperature change T6  1325 1375  135 oC 2


bWhy should the apparatus be very clean


Impurities present in the apparatus reacts with acid base lowering the overall temperature change and hence Hn


cCalculate the


inumber of moles of the acid used


number of moles  molarity x volume  2 x 50  01moles 1000 1000


iienthalpy change H of neutralization


H  mmass of solutionacidbase x cspecific heat capacity of solution x TT6  50 50 x 42 x 135  5670Joules  567kJ


iii the molar heat of neutralization the acid


Hn  Enthalpy change H  567kJ  567kJ mole1


Number of moles 01moles


cWrite the ionic equation for the reaction that takes place


OHaq Haq  H2Ol


dThe theoretical enthalpy change is 574kJ Explain the difference with the results above


The theoretical value is higher


Heatenergy loss to the surroundingenvironment lowers TT6 and thus Hn


Heatenergy is absorbed by the reaction vesselcalorimeterplastic cup lowers T and hence Hn


eCompare the Hn of the experiment above with similar experiment repeated with neutralization of a solution of


i potassium hydroxide with nitricV acid


The results would be the samesimilar


Both are neutralization reactions of strong acids and basesalkalis that are fully wholly dissociated into many free H  H3O and OH ions


ii ammonia with ethanoic acid


The results would be lowerHn would be less


Both are neutralization reactions of weak acids and basesalkalis that are partially partly dissociated into few free H  H3O and OH ions Some energy is used to ionize the molecule


fDraw an energy level diagram to illustrate the energy changes


H2 H aqOH aq


Energy


kJ


H  567kJ


H1 H2O l


Reaction pathcoordinateprogress


Theoretical examples


1The molar enthalpy of neutralization was experimentary shown to be 515kJ per mole of 05M hydrochloric acid and 05M sodium hydroxide If the volume of sodium hydroxide was 20cm3 what was the volume of hydrochloric acid used if the reaction produced a 50oC rise in temperature


Working


Moles of sodium hydroxide  molarity x volume  05 M x 20cm3  001 moles


1000 1000


Enthalpy change H  Hn  515  0515kJ


Moles sodium hydroxide 001 moles


Mass of base  acid  Enthalpy change H in Joules


Specific heat capacity x T


0515kJ x 1000  245238g


42 x 5


Massvolume of HCl  Total volume  volume of NaOH


245238  200  45238 cm3


Hn of potassium hydroxide was practically determined to be 567kJmole1Calculate the molarity of 500 cm3 potassium hydroxide used to neutralize 250cm3 of dilute sulphuricVI acid raising the temperature of the solution from 100oC to 165oC


H  mmass of solutionacidbase x cspecific heat capacity of solution x T


50 25 x 42 x 65  20475Joules


Moles potassium hydroxide Enthalpy change H


Hn


20475Joules  00361 moles


Molarity of KOH  moles x 1000  00361 moles x 1000  0722M


Volume used 50cm3


3Determine the specific heat capacity of a solution of a solution mixture of 500cm3 of 2M potassium hydroxide neutralizing 500cm3 of 2M nitricV acid if a 1325oC rise in temperature is recorded1mole of potassium hydroxide produce 554kJ of energy


Moles of potassium hydroxide  molarity KOH x volume


2 M x 50cm3  01 moles


Enthalpy change H  Hn x Moles potassium hydroxide  554kJ x 01 moles  554kJ x 10005540Joules


Specific heat capacity  Enthalpy change H in Joules


Mass of base  acid x T


5540  41811J1g1K1


5050 x 1325


Graphically Hn can be determined as in the example below


Procedure


Place 8 test tubes in a test tube rack Put 5cm3 of 2M sodium hydroxide solution into each test tube


Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker


Record its initial temperature at volume of base 0 Put one portion of the base into the beaker containing the acid


Stir carefully with the thermometer and record the highest temperature change to the nearest 05oC


Repeat the procedure above with other portions of the base to complete table 1 below


Table 1Sample results


aComplete the table to determine the change in temperature


bPlot a graph of volume of sodium hydroxide against temperature change


T oC


From the graph show and determine


ithe highest temperature change T


T T2T1  highest temperatureT2 from extrapolating a correctly plotted graph less lowest temperature at volume of base0 T1


T  67  00  670C


iithe volume of sodium hydroxide used for complete neutralization


From a correctly plotted graph  1675cm3


cCalculate the number of moles of the alkali used


Moles NaOH  molarity x volume 2M x 1675cm3  00335 moles


1000 1000


dCalculate H for the reaction


H  mass of solutionacidbase x c x T


250  1675 x 42 x 67  1174845 J  1174845kJ


eCalculate the molar enthalpy of neutralization of the alkali


Hn  Hn  1174845kJ  350701kJ


Number of moles 00335


dStandard enthalpyheat of solution Hs


The standard enthalpy of solution Hsis defined as the energy change when one mole of a substance is dissolve in excess distilled water to form an infinite dilute solution An infinite dilute solution is one which is too dilute to be diluted further


Dissolving a solid involves two processes


i breaking the crystal of the solid into free ionscations and anionThis process is the opposite of the formation of the crystal itself The energy required to form one mole of a crystal structure from its gaseous ions is called Lattice energyheatenthalpy of lattice Hl Lattice energy heatenthalpy of lattice Hl is an endothermic process Hl


The table below shows some Hl in kJ for the process MXs  M g  X g


iisurrounding the free ions by polar water molecules This process is called hydration The energy produced when one mole of ions are completely hydrated is called hydration energy heatenthalpy of hydrationHhHydration energy enthalpy of hydrationHh is an exothermic processHh


The table below shows some Hh in kJ for some ions


The sum of the lattice energy Hl endothermic and hydration energy Hh exothermic gives the heat of solutionHs


Hs  Hl Hh


Note


Since Hl is an endothermic process and Hh is an exothermic process then Hs is


iexothermic if Hl is less than Hh and hence a solid dissolve easily in water


iiendothermic if Hl is more than Hh and hence a solid does not dissolve easily in water


aDissolving sodium chloride crystals


i NaCl breaking the crystal into free ions Na g Clg Hl 771 kJ


ii Hydrating the ions


Na g  aq  Naaq Hh   406 kJ


Clg  aq  Claq Hh   364 kJ


Hs Hh Hs   406 kJ   364 kJ  771 kJ   10 kJmole1


NaCl does not dissolve easily in water because overall Hs is endothermic


Solubility of NaCl therefore increases with increase in temperature


Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na g Clg


bDissolving magnesium chloride crystals MgCl2 s MgCl2 aq


i MgCl2 breaking the crystal into free ionsMg 2g 2Clg Hl 2493 kJ


Mg 2g  aq  Mg 2g aq Hh   1920 kJ


2Clg  aq  2Claq Hh   364 x 2 kJ


Hs Hh Hs   1920 kJ   364 x 2 kJ  2493 kJ  1550 kJmole1


MgCl2 s dissolve easily in water because overall Hs is exothermic


Solubility of MgCl2 s therefore decreases with increase in temperature


cDissolving Calcium floride crystals CaF2 s  CaF2 aq


i CaF2 Ca 2g 2Fg Hl 760 kJ


Ca 2g  aq  Ca 2g aq Hh   1650 kJ


2Fg  aq  2Faq Hh   506 x 2 kJ


Hs Hh Hs   1650 kJ   506 x 2 kJ  760 kJ  19020 kJmole1


CaF2 s dissolve easily in water because overall Hs is exothermic


Solubility of CaF2 s therefore decreases with increase in temperature


dDissolving magnesium bromide crystals MgBr2 s MgBr2 aq


i MgCl2 breaking the crystal into free ionsMg 2g 2Brg Hl 2226 kJ


ii Hydrating the ions


2Brg  aq  2Braq Hh   335x 2 kJ


Hs Hh Hs   1920 kJ   335 x 2 kJ  2226 kJ  3640 kJmole1


MgBr2 s dissolve easily in water because overall Hs is exothermic


Solubility of MgBr2s therefore decreases with increase in temperature


Practically the heat of solution can be determined from dissolving known amount massvolume of solute in known mass volume of watersolvent


From the temperature of solvent before and after dissolving the change in temperatureT during dissolution is determined


To determine the Hs ammonium nitrate


Place 100cm3 of distilled water into a plastic beakercalorimeter Determine its temperature and record it at time 0 in table I below


Put all the 50g of ammonium nitrate potassium nitrateammonium chloride can also be usedprovided into the plastic beakercalorimeter stir using a thermometer and record the highest temperature change to the nearest 05oCafter every  minute to complete table I


Continue stirring the mixture throughout the experiment


Sample results Table I


aPlot a graph of temperature against timexaxis


temperatureoC T


bFrom the graph show and determine the highest temperature change T


T T2T1  lowest temperatureT2 from extrapolating a correctly plotted graph less highest temperature at volume of base0 T1


T 187  220  330C


cCalculate the number of moles of ammonium nitrateV used


Moles NH4NO3  mass used  50  00625 moles


Molar mass 80


dCalculate H for the reaction


H  mass of water x c x T


100 x 42 x 33  1386 J  1386kJ


eCalculate the molar enthalpy of dissolution of ammonium nitrateV


Hs  H  1386kJ   22176kJ mole1


Number of moles 00625 moles


fWhat would happen if the distilled water was heated before the experiment was performed


The ammonium nitrateVwould take less time to dissolves Increase in temperature reduces lattice energy causing endothermic dissolution to be faster


gIllustrate the process above in an energy level diagram


NH4 g  NO3g


H NH4 aqNO3aq EnergykJ H H  22176kJ NH4NO3s Reaction path progresscoordinate


h 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuricVIacid of density 184gcm3The temperature of the mixture rose from 250C to 38oCCalculate the molar heat of solution of sulphuricVIacid S320H100160


Working


Molar mass of H2SO4  98g


Mass of H2SO4 Density x volume  184gcm3 x 3cm3  552 g


Mass of H2O  Density x volume  100gcm3 x 100cm3  100 g


Moles of H2SO4 mass  552 g  00563 moles


Molar mass of H2SO4 98g


Enthalpy change H mass of acid  water x specific heat capacity of water x T  100 552 g x 42 x 13oC  5761392 J  5761392 kJ


Hs of H2SO4 H  5761392 kJ  10233378kJmoles1


Moles of H2SO4 00563 moles


eStandard enthalpyheat of formation Hf


The molar enthalpy of formation Hf is defined as the energy change when one mole of a compound is formed from its elements at 298K25oC and 101325Paone atmospherepressure Hf is practically difficult to determine in a school laboratory


It is determined normally determined by applying Hess law of constant heat summation


Hess law of constant heat summation states that the total enthalpyheatenergy change of a reaction is the same regardless of the route taken from reactants to products at the same temperature and pressure


Hess law of constant heat summation is as a result of a series of experiments done by the German Scientist Henri Hess18021850


He found that the total energy change from the reactants to products was the same irrespective of the intermediate products between ie


As H1Cs  As H2BsH3Cs


Applying Hess law of constant heat summation then


As H2 Bs


H1 H3


Cs


The above is called an energy cycle diagram It can be used to calculate any of the missing energy changes since


i H1 H2  H3


ii H2 H1  H3


iii H3   H1  H2


Examples of applying Hess law of constant heat summation


1Calculate the molar enthalpy of formation of methane CH4 given that Hc of carbongraphite is 3935kJmole1Hydrogen is 2857 kJmole1 and that of methane is 890 kJmole1


Working


Carbongraphite hydrogen and oxygen can react to first form methane


Methane will then burn in the oxygen present to form carbonIVoxide and water Carbongraphite can burn in the oxygen to form carbonIVoxide


Hydrogen can burn in the oxygen to form water


Cs 2H2 g2O2 g H1 CH4g 2O2g H2 CO2g2H2Ol


Cs 2H2 g2O2 g H3 CO2g2H2Ol


Energy cycle diagram


Cs  2H2 g  2O2g H1Hc 8904kJ CH4g2O2g


H3Hc 3935kJ H3Hc 2857kJ x 2 H2 Hf x


CO2g  2H2Ol


Substituting


H3  H1  H2


3935  2857 x 2  8904kJ  x


x  745 kJ


Heat of formation Hf CH4  745 kJmole1


2 Calculate the molar enthalpy of formation of ethyne C2H2 given  Hc of carbongraphite  394kJmole1Hydrogen  286 kJmole1  C2H2  1300 kJmole1


Carbongraphite hydrogen and oxygen can react to first form ethyne


Ethyne will then burn in the oxygen present to form carbonIVoxide and water Carbongraphite can burn in the oxygen to form carbonIVoxide


2Cs H2 g2  O2 g H1 C2 H2 g 2  O2g H2 CO2gH2Ol


2Cs H2 g 2  O2 g H3 2CO2gH2Ol


2Cs  H2 g 2O2g H1Hf x C2 H22O2g


H3Hc 394kJx 2 H3Hc 286kJ H2 Hc 1300kJ


2CO2g  H2Ol


394 x 2  286  1300kJ  x


x  244 kJ


Heat of formation Hf CH4  244 kJmole1


3 Calculate the molar enthalpy of formation of carbonIIoxide CO given  Hc of carbongraphite  3935kJmole1 Hc of carbonIIoxide CO 283 kJmole1


Working


Carbongraphite reacts with oxygen first to form carbon IIoxide CO


CarbonIIoxide CO then burn in the excess oxygen to form carbonIVoxide Carbongraphite can burn in excess oxygen to form carbon IV oxide


Cs O2 g H1 CO g   O2g H2 CO2g


Cs O2 g H3 CO2g


Energy cycle diagram


Cs  O2g H1Hf x COO2g


H3Hc 3935kJ H2 Hc 283kJ


Substituting


H3  H1  H2


3935kJ  283kJ  x


x  110 kJ


Heat of formation Hf CO  110 kJmole1


4Study the information below


H2g   O2g  H2Ol H1286 kJmole1


Cs  O2g  CO2g H2393 kJmole1


2Cs  H2g   O2g C2H5OHl H3277 kJmole1


Use the information to calculate the molar enthalpy of combustion H4 of ethanol


2Cs  3H2 g 3O2g H3Hf 227kJ C2 H5OH 3O2g


H2Hc 394kJx 2 H1Hc 286kJx 3 H4 Hc x


2CO2g  3H2Ol


H1  H2  H3  H4


394 x 2  286 x 3  277  x


H4  1369 kJ


Heat of combustion Hc C2H5OH  1369 kJmole1


5Given the following information below


CuSO4s  aq  CuSO4aq H661 kJmole1


CuSO4s  aq  5H2Ol CuSO4 5H2O aq H774 kJmole1


Calculate H for the reaction


CuSO4aq  5H2O CuSO4 5H2O aq H774 kJmole1


Working


CuSO4s  aq  5H2Ol CuSO4aq 5H2Ol CuSO4 5H2O aq


CuSO4s  aq  5H2Ol CuSO4 5H2O aq


Energy cycle diagram


CuSO4s  aq  5H2Ol H1661kJ CuSO4aq 5H2Ol


H3 774kJ H2 x


CuSO4 5H2O aq


Substituting


H3  H2  H1


774kJ  x  661kJ


H4  109 kJ


Heat of dissolution of CuSO4  109kJmole1


Practically Hess law can be applied practically as in the following examples


aPractical example 1


Determination of the enthalpy of formation of CuSO45H2O


Experiment I


Weigh accurately 125 g of copperIIsulphateVIpentahydrate Measure 100cm3 of distilled water into a beaker Determine its temperature T1 Put all the crystals of the copperIIsulphateVIpentahydrate carefully into the beaker Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table 1


Table 1Sample results


Experiment II


Weigh accurately 80g of anhydrous copperIIsulphateVI Measure 100cm3 of distilled water into a beaker Determine its temperature T1 Put all the crystals of the copperIIsulphateVIpentahydrate carefully into the beaker Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II


Table II Sample results


Questions


aCalculate the average T in


iTable I


T T2 T1  30 40  35 oC


iiTable II


T T2 T1  10 20  15 oC


bCalculate the number of moles of solid used in


iExperiment I


Moles of CuSO45H2O  Mass  125  005 moles


Molar mass 250


iiExperiment II


Moles of CuSO4  Mass  80  005 moles


Molar mass 160


cCalculate the enthalpy change for the reaction in


Enthalpy change of CuSO45H2O mass of Waterm x c x T


100cm3 x 42 x 35 oC  147kJ


Enthalpy change of CuSO4  mass of waterm x c x T


100cm3 x 42 x 15 oC  063kJ


cCalculate the molar enthalpy of solution CuSO4 5H2O s form the results in iexperiment I


Hs  CuSO45H2O H  147kJ  294kJ


Number of Moles 005 moles


iiexperiment II


Hs  CuSO4 H  063kJ  126kJ


d Using an energy level diagram calculate the molar enthalpy change for the reaction


CuSO4 5H2O s  CuSO4s  5H2Ol


Energy cycle diagram


CuSO4s  aq  5H2Ol H1x CuSO4 5H2O s aq


H3 294kJ H2 126kJ


CuSO4 5H2O aq


H3  H1 H2


294kJ  126kJ  x


x  168kJ


bPractical example II


Determination of enthalpy of solution of ammonium chloride


Theoretical information


Ammonium chloride dissolves in water to form ammonium chloride solution Aqueous ammonia can react with excess dilute hydrochloric acid to form ammonium chloride solution The heat change taking place can be calculated from the heat of reactions


i NH3aq  HClaq  NH4Cls


ii NH4Cls  aq  NH4Claq


iii NH3aq  HClaq  NH4Claq


Experiment procedure I


Measure 50cm3 of water into a 100cm3 beaker Record its temperature T1 as initial temperature to the nearest 05oC in table I Add exactly 50g of ammonium chloride crystals weighed carefully into the water Stir and record the highest temperature change T2 as the final temperature change Repeat the above procedure to complete table I


Sample results TableI


Experiment procedure II


Measure 25cm3 of 2M aqueous ammonia into a 100cm3 beaker Record its temperature T1 as initial temperature to the nearest 05oC in table II Measure 25cm3 of 2M hydrochloric acid solution Add the acid into the beaker containing aqueous ammonia Stir and record the highest temperature change T2 as the final temperature change Repeat the above procedure to complete table II


Sample resultsTable II


Sample Calculations


aCalculate the average T in


iTable I


T T2 T1  30 20  25 oC


iiTable II


T T2 T1  70 70  70 oC


bCalculate the enthalpy change for the reaction in


iExperiment I


Enthalpy change H  mass of Waterm x c x T


50cm3 x 42 x 25 oC  0525kJ


iiExperiment II


2525cm3 x 42 x 7 oC  147kJ


cWrite the equation for the reaction taking place in


NH4Cls  aq  NH4Claq


iiExperiment I


NH3aq  HClaq  NH4Claq


dCalculate the enthalpy change H for the reaction


NH3g  HClg  NH4Cls given that


i NH3g  aq  NH3aq H 403kJ


ii aq  HClg  HClaq H 1645kJ


eApplying Hess Law of constant heat summation


Energy level diagram


N2g  1 H2g   Cl2 Hf NH4Cls  aq


aq aq


403kJH1 1643kJH2


NH3 aq  HClaq 147kJH3 NH4Cls


H1  H2  H3  H4  Hf


403kJ  1643kJ  147kJ  0525kJ  Hf


Hf  58865kJ


Practice theoretical examples


1 Using an energy level diagram calculate the Hs of ammonium chloride crystals given that


Hf of NH3 aq  8054kJ mole1


Hf of HCl aq  16446kJ mole1


Hf of NH4Cl aq  2617483kJ mole1


Hs of NH4Cl aq  168517kJ mole1


N2g  1 H2g   Cl2 Hf2617483kJ NH4Cls  aq


xHs


8054kJH1 16446kJH2


NH3 aq  HClaq 168517kJH3 NH4Cls


8054kJ  16446kJ  168517kJ  2617483kJ  Hf


Hf  336kJmole1


Study the energy cycle diagram below and use it to


aIdentify the energy changes H1 H2 H3 H4 H5 H6


H1  enthalpyheat of formation of sodium chloride Hf


H2 enthalpyheat of atomization of sodium Hat


H3 enthalpyheat of ionizationionization energy of sodium H i


H4 enthalpyheat of atomization of chlorine Hat


H5 enthalpyheat of electron affinity of chlorine He


H6 enthalpyheat of latticeLattice energy of sodium chlorideH l


b Calculate H1 given that H2 108kJ  H3500kJ H4 121kJ H5 364kJ and H6 766kJ


Working


H1 H2 H3 H4 H5 H6


Substituting


H1 108kJ  500kJ  121kJ 364kJ  766kJ


H1 401kJmole1


c Given the that


i Ionization energy of sodium   500kJmole1


iiHat of sodium   110kJmole1


iii Electron affinity of chlorine   363kJmole1


ivHat of chlorine   120kJmole1


v Hf of sodium chloride  411kJ  calculate the lattice energy of sodium chloride using an energy cycle diagram


Applying Hess law then


Hf Ha Hi Ha He Hl


411 108kJ  500kJ  121kJ 364kJ  x


411  108kJ  500kJ  121kJ  364kJ  x


x 776kJmole1